The Divergence Theorem

Section 4.2 The Divergence Theorem

The rest of this chapter concerns three theorems: the divergence theorem, Green’s theorem and Stokes’ theorem. Superficially, they look quite different from each other. But, in fact, they are all very closely related and all three are generalizations of the fundamental theorem of calculus

\begin{equation*} \int_a^b \diff{f}{t}(t)\ \dee{t} = f(b) -f(a) \end{equation*}

The left hand side of the fundamental theorem of calculus is the integral of the derivative of a function. The right hand side involves only values of the function on the boundary of the domain of integration. The divergence theorem, Green’s theorem and Stokes’ theorem also have this form, but the integrals are in more than one dimension. So the derivatives are multidimensional, like the curl and divergence, and the integrands can involve vector fields.

For the divergence theorem, the integral on the left hand side is over a (three dimensional) volume and the right hand side is an integral over the boundary of the volume, which is a surface.
For Green’s and Stokes’ theorems, the integral on the left hand side is over a (two dimensional) surface and the right hand side is an integral over the boundary of the surface, which is a curve.

The divergence theorem is going to relate a volume integral over a solid $V$ to a flux integral over the surface of $V\text{.}$ First we need a couple of definitions concerning the allowed surfaces. In many applications solids, for example cubes, have corners and edges where the normal vector is not defined. On the other hand, to be able to compute a flux integral over a surface, we certainly need that the set of points where the normal vector is not well-defined is small enough that the existence of the flux integral is not jeopardized. This is the case for “piecewise smooth” surfaces, which we now define.

Definition 4.2.1.

A surface is smooth if it has a parametrization $\vr(u,v)$ with continuous partial derivatives $\frac{\partial\vr}{\partial u}$ and $\frac{\partial\vr}{\partial v}$ and with $\frac{\partial\vr}{\partial u}\times\frac{\partial\vr}{\partial v}$ nonzero.
A surface is piecewise smooth if it consists of a finite number of smooth pieces that meet along sharp curves and at sharp corners.

Here are sketches of a smooth surface (a sausage) and a piecewise smooth surface (an ice-cream cone), followed by the divergence theorem

It is also known as Gauss’s theorem. Johann Carl Friedrich Gauss (1777–1855) was a German mathematician. Throughout the 1990’s Gauss’s portrait appeared on the German ten-mark banknote. In addition to Gauss’s theorem, the Gaussian distribution (the bell curve), degaussing and the CGS unit for the magnetic field, and the crater Gauss on the Moon are named in his honour.

Theorem 4.2.2. Divergence Theorem.

Let

$V$ be a bounded solid with a piecewise smooth surface
²
We are going to consistently use the notation $\partial$(thing) to denote the boundary of (thing).
$\partial V$
$\vF$ be a vector field that has continuous first partial derivatives at every point of $V\text{.}$

Then

\begin{align*} \dblInt_{\partial V} \vF\cdot\hn\,\dee{S} &=\tripInt_V\vnabla\cdot\vF\ \dee{V} \end{align*}

where $\hn$ is the outward unit normal of $\partial V\text{.}$

Like the fundamental theorem of calculus, the divergence theorem expresses the integral of a derivative of a function (in this case a vector-valued function) over a region in terms of the values of the function on the boundary of the region.

Warning 4.2.3.

Note that in Theorem 4.2.2 we are assuming that the vector field $\vF$ has continuous first partial derivatives at every point of $V\text{.}$ If that is not the case, for example because $\vF$ is not defined on all of $V\text{,}$ then the conclusion of the divergence theorem can fail. An example is $\vF = \frac{\vr}{|\vr|^3}\text{,}$ $V=\Set{(x,y,z)}{x^2+y^2+z^2\le 1}\text{.}$ See Example 4.2.7.

Proof.

We have to show that

\begin{align*} \dblInt_{\partial V} \Big( \vF_1\,\hi + \vF_2\,\hj + \vF_3\,\hk\Big) \cdot\hn\,\dee{S} &=\tripInt_V\Big( \frac{\,\partial \vF_1}{\partial x} +\frac{\partial \vF_2}{\partial y} +\frac{\partial \vF_3}{\partial z}\Big) \ \dee{V} \end{align*}

Note that the left hand side is a sum of three terms — one involving $\vF_1\text{,}$ one involving $\vF_2$ and one involving $\vF_3$ — and the right hand side is a sum of three terms — one involving $\vF_1\text{,}$ one involving $\vF_2$ and one involving $\vF_3\text{.}$ We’ll just show that the $\vF_3$ terms on the left hand side and right hand side are equal, i.e. that

\begin{align*} \dblInt_{\partial V} \vF_3\,\hk \cdot\hn\,\dee{S} &=\tripInt_V \frac{\partial \vF_3}{\partial z} \ \dee{V} \end{align*}

Showing that the $\vF_1$ terms match and the $\vF_2$ terms match is done in the same way

Mutatis mutandis.

Special Geometry

We’ll first assume that the solid has the special form

\begin{equation*} V = \Set{(x,y,z)}{ B(x,y)\le z\le T(x,y),\ \ (x,y)\in R_{xy}} \end{equation*}

where $R_{x,y}$ is some subset of the $xy$-plane. We can further assume that, for each $(x,y) \in R_{xy}\text{,}$ we have $B(x,y)\le T(x,y)\text{.}$ After we’re finished with this special case, we’ll handle the general case.

Let’s work on $\dblInt_{\partial V} \vF_3\,\hk \cdot\hn\,\dee{S}$ first. As in the figure below,

the surface $\partial V$ consists of three pieces — the top, the bottom and the side. We’ll consider each in turn.

The top is $\cT=\Set{(x,y,z)}{ z= T(x,y),\ \ (x,y)\in R_{xy}}\text{.}$ By 3.3.2, on $\cT$
\begin{equation*} \hn\, \dee{S} = +\big[-T_x(x,y)\,\hi - T_y(x,y)\,\hj + \hk\big]\ \dee{x}\dee{y} \end{equation*}
As $\hn$ is to be the outward normal, it must point upwards on $\cT\text{.}$ That’s why we have chosen, and emphasised, the “$+$” sign. So $\hk\cdot\hn\,\dee{S} = \dee{x}\dee{y}$ and
\begin{equation*} \dblInt_{\cT} \vF_3\,\hk \cdot\hn\,\dee{S} =\dblInt_{R_{xy}} \vF_3(x,y,T(x,y))\,\dee{x}\dee{y} \end{equation*}
The bottom is $\cB=\Set{(x,y,z)}{ z= B(x,y),\ \ (x,y)\in R_{xy}}\text{.}$ By 3.3.2, on $\cB$
\begin{equation*} \hn\, \dee{S} = -\big[-B_x(x,y)\,\hi - B_y(x,y)\,\hj + \hk\big]\ \dee{x}\dee{y} \end{equation*}
As $\hn$ is to be the outward normal, it must point downwards on $\cB\text{.}$ That’s why we have chosen the “$-$” sign. So $\hk\cdot\hn\,\dee{S} = -\dee{x}\dee{y}$ and
\begin{equation*} \dblInt_{\cB} \vF_3\,\hk \cdot\hn\,\dee{S} =-\dblInt_{R_{xy}} \vF_3(x,y,B(x,y))\,\dee{x}\dee{y} \end{equation*}
The side is $\cS=\Set{(x,y,z)}{(x,y)\in\partial R_{xy},\ B(x,y)\le z\le T(x,y)}\text{.}$ It runs vertically. Hence on $\cS$ the normal vector to $\partial V$ is parallel to the $xy$-plane so that $\hk\cdot\hn=0$ and
\begin{equation*} \dblInt_{\cS} \vF_3\,\hk \cdot\hn\,\dee{S} = 0 \end{equation*}

So all together

\begin{align*} &\dblInt_{\partial V}\!\! \vF_3\,\hk \cdot\hn\,\dee{S} =\!\dblInt_{\cT}\! \vF_3\,\hk \cdot\hn\,\dee{S} \!+\!\dblInt_{\cB}\! \vF_3\,\hk \cdot\hn\,\dee{S} \!+\!\dblInt_{\cS}\! \vF_3\,\hk \cdot\hn\,\dee{S} \notag\\ &\hskip0.5in=\dblInt_{R_{xy}} \big[\vF_3(x,y,T(x,y)) -\vF_3(x,y,B(x,y))\big]\,\dee{x}\dee{y} +0 \tag{$\partial V$} \end{align*}

Now let us examine

\begin{align*} \tripInt_V\frac{\partial \vF_3}{\partial z}\ \dee{V} &= \dblInt_{R_{xy}}\dee{x}\dee{y}\int_{B(x,y)}^{T(x,y)}\dee{z}\ \frac{\partial \vF_3}{\partial z}(x,y,z) \notag\\ &=\dblInt_{R_{xy}} \big[\vF_3(x,y,T(x,y)) -\vF_3(x,y,B(x,y))\big]\,\dee{x}\dee{y} \tag{$V$} \end{align*}

by the fundamental theorem of calculus. That’s exactly what we had to show. The integrals ($\partial V$) and ($V$) are equal.

General Geometry

Now we’ll drop the assumption on $V$ that we imposed in the “Special Geometry” section above. The key idea that makes the proof work is that we can cut up any

⁴

We are assuming that $V$ is “reasonable”.

$V$ into pieces, each of which does obey the special assumption that we just considered. Consider, for example, the sausage shaped solid in the figure on the left below.

Call the sausage $V\text{.}$ Cut it into two halves by running a cleaver horizontally through its centre. This splits the solid $V$ into two halves, $V_1$ and $V_2$ as in the figure on the right above. It also splits the boundary $\partial V$ of $V$ into two halves $S_1$ and $S_2\text{,}$ also as in the figure on the right above. Note that

the boundary, $\partial V_1\text{,}$ of $V_1$ is the union of $S_1$ and the shaded disk $S_c$ (the cut introduced by the cleaver). On the cut $S_c\text{,}$ the outward pointing normal to $V_1$ is $-\hk\text{.}$
The boundary, $\partial V_2\text{,}$ of $V_2$ is the union of $S_2$ and the shaded disk $S_c\text{.}$ On the cut $S_c\text{,}$ the outward pointing normal to $V_2$ is $+\hk\text{.}$

Now both $V_1$ and $V_2$ do satisfy the assumption of the “Special Geometry” section above. So

\begin{align*} &\tripInt_V\frac{\partial \vF_3}{\partial z}\ \dee{V} = \tripInt_{V_1}\frac{\partial \vF_3}{\partial z}\ \dee{V} +\tripInt_{V_2}\frac{\partial \vF_3}{\partial z}\ \dee{V}\\ &\hskip0.5in=\dblInt_{\partial V_1} \vF_3\,\hk \cdot\hn\,\dee{S} +\dblInt_{\partial V_2} \vF_3\,\hk \cdot\hn\,\dee{S}\\ &\hskip0.5in=\dblInt_{S_1} \vF_3\,\hk \cdot\hn\,\dee{S} +\dblInt_{S_c\downarrow} \vF_3\,\hk \cdot\hn\,\dee{S} +\dblInt_{S_2} \vF_3\,\hk \cdot\hn\,\dee{S}\\ &\hskip1.5in +\dblInt_{S_c\uparrow} \vF_3\,\hk \cdot\hn\,\dee{S} \end{align*}

Here $S_c\downarrow$ is the surface $S_c$ with normal vector $-\hk$ and $S_c\uparrow$ is the surface $S_c$ with normal vector $+\hk\text{.}$ So the second and fourth integrals are identical except that $\hn=-\hk$ in the second integral and $\hn=+\hk$ in the fourth integral. So they cancel exactly and

\begin{align*} \tripInt_V\frac{\partial \vF_3}{\partial z}\ \dee{V} &=\dblInt_{S_1} \vF_3\,\hk \cdot\hn\,\dee{S} +\dblInt_{S_2} \vF_3\,\hk \cdot\hn\,\dee{S} =\dblInt_{\partial V} \vF_3\,\hk \cdot\hn\,\dee{S} \end{align*}

as desired.

Example 4.2.4.

Evaluate the flux integral $\dblInt_S \vF\cdot\hn\,\dee{S}$ where $\hn $ is the outward normal to $S\text{,}$ which is the surface of the hemispherical region

\begin{equation*} V=\Set{(x,y,z)}{x^2+y^2+z^2\le a^2,\ z\ge 0} \end{equation*}

and

\begin{equation*} \vF = xz^2\,\hi + (x^2y-z^3)\,\hj + \big(2xy + y^2 z +e^{\cos y}\big)\hk \end{equation*}

Solution.

The $e^{\cos y}$ in $\vF$ suggests that a direct evaluation of the integral is difficult. So we’ll use a little trickery to to evaluate it. Not surprisingly, considering that we have just proven the divergence theorem, the trick is to apply the divergence theorem

⁵

It’s almost as though someone rigged the example with this in mind.

. Since

\begin{align*} \vnabla\cdot\vF &= \frac{\partial \vF_1}{\partial x} +\frac{\partial \vF_2}{\partial y} +\frac{\partial \vF_3}{\partial z}\\ &=\frac{\partial }{\partial x}\big(xz^2\big) +\frac{\partial }{\partial y}\big(x^2y-z^3\big) +\frac{\partial }{\partial z}\big(2xy + y^2 z +e^{\cos y}\big)\\ &= z^2 + x^2 +y^2 \end{align*}

The divergence theorem tell us that

\begin{align*} \dblInt_S \vF\cdot\hn\,\dee{S} &=\tripInt_V\big(x^2+y^2+z^2\big)\ \dee{V} \end{align*}

Spherical coordinates are perfect for this integral. (See Appendix A.6.3, if you need to refresh your memory.)

\begin{align*} \tripInt_V\big(x^2+y^2+z^2\big)\ \dee{V} &=\int_0^{2\pi}\dee{\theta}\int_0^{\frac{\pi}{2}}\dee{\varphi} \int_0^a \dee{\rho}\,\rho^2\sin\varphi\ \rho^2\\ &=\bigg[\int_0^{2\pi}\dee{\theta}\bigg] \bigg[\int_0^{\frac{\pi}{2}}\sin\varphi\,\dee{\varphi}\bigg] \bigg[\int_0^{a}\rho^4\,\dee{\rho}\bigg]\\ &=\big[2\pi\big]\Big[-\cos\varphi\Big]_0^{\frac{\pi}{2}} \left[\frac{\rho^5}{5}\right]_0^a\\ &=\frac{2\pi a^5}{5} \end{align*}

Example 4.2.5.

Evaluate the flux integral $\dblInt_S \vF\cdot\hn\,\dee{S}$ where $\hn $ is the outward normal to $S\text{,}$ which is the part of the surface $z^2=x^2+y^2$ with $1\le z\le 2\text{,}$ and where

\begin{equation*} \vF = 3x\,\hi + (5y+e^{\cos x})\,\hj + z\,\hk \end{equation*}

Solution.

Again the $e^{\cos x}$ in $\vF$ suggests that a direct evaluation is difficult

⁶

In fact, it is possible to evaluate this integral directly, if one recognizes that the ugly part of the integrand is odd under $y\rightarrow-y$ and integrates to exactly zero.

and again we’ll apply the divergence theorem. But this time $S$ is not the boundary of a solid $V\text{.}$ It is the portion of the cone outlined in red in the figure on the left below and does not have a top or bottom “cap”.

Fortunately, there is a solid $V$ whose boundary, while not being equal to $S\text{,}$ at least contains $S\text{.}$ It is (unsurprisingly)

\begin{equation*} V = \Set{(x,y,z)}{x^2+y^2\le z^2,\ \ 1\le z\le 2} \end{equation*}

and is sketched in the figure on the right above. The boundary, $\partial V\text{,}$ is the union of $S$ and the two disks

\begin{align*} D_1 &= \Set{(x,y,z)}{x^2+y^2\le z^2,\ \ z=1}\\ D_2 &= \Set{(x,y,z)}{x^2+y^2\le z^2,\ \ z=2} \end{align*}

So the divergence theorem gives

\begin{align*} \tripInt_V\vnabla\cdot\vF\ \dee{V} &= \dblInt_{\partial V} \vF\cdot\hn\,\dee{S}\\ &= \dblInt_S \vF\cdot\hn\,\dee{S} +\dblInt_{D_1} \vF\cdot\hn\,\dee{S} +\dblInt_{D_2} \vF\cdot\hn\,\dee{S} \end{align*}

which implies

\begin{align*} \dblInt_S \vF\cdot\hn\,\dee{S} &= \tripInt_V\vnabla\cdot\vF\ \dee{V} -\dblInt_{D_1} \vF\cdot\hn\,\dee{S} -\dblInt_{D_2} \vF\cdot\hn\,\dee{S} \end{align*}

The point of this exercise is that the left hand side, which is not easy to evaluate directly, is the integral we want, while the three integrals on the right hand side are all easy to evaluate. We do so now. The outward normal to (the horizontal disk) $D_2$ is $+\hk\text{.}$ So

\begin{align*} \dblInt_{D_2} \vF\cdot\hn\,\dee{S} &=\dblInt_{D_2} \vF\cdot\hk\,\dee{S} =\dblInt_{D_2} z\,\dee{S} \end{align*}

As $z=2$ on $D_2\text{,}$ and $D_2$ is a disk of radius $2\text{,}$

\begin{equation*} \dblInt_{D_1} \vF\cdot\hn\,\dee{S} =2\text{Area}(D_2) =2\pi 2^2 = 8\pi \end{equation*}

Similarly, the outward normal to (the horizontal disk) $D_1$ is $-\hk\text{.}$ So

\begin{align*} \dblInt_{D_1} \vF\cdot\hn\,\dee{S} &=-\dblInt_{D_1} \vF\cdot\hk\,\dee{S} =-\dblInt_{D_1} z\,\dee{S} \end{align*}

As $z=1$ on $D_1\text{,}$ and $D_1$ is a disk of radius $1\text{,}$

\begin{equation*} \dblInt_{D_1} \vF\cdot\hn\,\dee{S} =\text{Area}(D_1) =-\pi 1^2 = -\pi \end{equation*}

Finally, as $\vnabla\cdot\vF = 3+5+1 = 9$

\begin{gather*} \tripInt_V\vnabla\cdot\vF\ \dee{V} =9\,\text{Vol}(V) \end{gather*}

The volume of $V$ can be easily computed using the first year technique

⁷

You can review in §1.6 of the CLP-2 text.

of slicing $V$ into thin horizontal pancakes like that sketched in the figure below.

The pancake at height $z$ has

thickness $\dee{z}\text{,}$
a circular cross-section of radius $z$ (remember that the outer boundary of $V$ has equation $x^2+y^2=z^2$), and hence has
cross-sectional area $\pi z^2$ and
volume $\pi z^2\,\dee{z}\text{.}$

\begin{gather*} \tripInt_V\vnabla\cdot\vF\ \dee{V} =9\,\text{Vol}(V) =9\int_1^2 \pi z^2\,\dee{z} =9\left[\frac{\pi z^3}{3}\right]_1^2 =9\times \pi\frac{7}{3} =21\pi \end{gather*}

and, all together

\begin{align*} \dblInt_S \vF\cdot\hn\,\dee{S} &= \tripInt_V\vnabla\cdot\vF\ \dee{V} -\dblInt_{D_1} \vF\cdot\hn\,\dee{S} -\dblInt_{D_2} \vF\cdot\hn\,\dee{S}\\ &= 21\pi - (-\pi) -8\pi =14\pi \end{align*}

Example 4.2.6.

Evaluate the flux integral $\dblInt_S \vF\cdot\hn\,\dee{S}$ where $\hn $ is the upward normal to $S\text{,}$ which is the part of $z={\big(x^2+y^2\big)}^2$ with $0\le z\le 1\text{,}$ and

\begin{equation*} \vF = \big(x+e^{y^2}\big)\,\hi + (y+\cos z)\,\hj + \hk \end{equation*}

Solution.

This integral can be evaluated in much the same way as we evaluated the integral of Example 4.2.5. We first define a solid $V$ whose boundary $\partial V$ contains $S\text{.}$ A good, and hopefully obvious, choice is

\begin{gather*} V = \Set{(x,y,z)}{{\big(x^2+y^2\big)}^2\le z,\ \ 0\le z\le 1 } \end{gather*}

The boundary of $V$ is the union of $S\text{,}$ with outward pointing normal $-\vn$ (recall that the problem specifies that the symbol $\hn$ refers to the upward pointing normal) and the disk

\begin{equation*} D = \Set{(x,y,z)}{z= 1,\ \ {\big(x^2+y^2\big)}^2\le 1} \end{equation*}

with outward pointing normal $\hk\text{.}$

So the divergence theorem gives

\begin{gather*} \tripInt_V\vnabla\cdot\vF\ \dee{V} = -\dblInt_S \vF\cdot\hn\,\dee{S} +\dblInt_{D} \vF\cdot\hk\,\dee{S} \end{gather*}

which implies

\begin{align*} \dblInt_S \vF\cdot\hn\,\dee{S} &= -\tripInt_V\vnabla\cdot\vF\ \dee{V} +\dblInt_{D} \vF\cdot\hk\,\dee{S}\\ &= -\tripInt_V 2\ \dee{V} +\dblInt_{D} \dee{S} \end{align*}

$D$ is a circular disk of radius $1\text{,}$ and so has area $\pi\text{.}$ To evaluate the volume integral we slice $V$ into horizontal pancakes with the pancake at height $z$ having a circular cross-section of radius $z^{\frac{1}{4}}\text{.}$ (Recall that the boundary of $V$ has ${\big(x^2+y^2\big)}^2 = z\text{.}$) So

\begin{align*} \dblInt_S \vF\cdot\hn\,\dee{S} &= -2\int_0^1\pi \sqrt{z}\ \dee{z} +\pi =-2\pi\times\frac{2}{3} +\pi =-\frac{\pi}{3} \end{align*}

Again, you can see that the actual integration is quite easy. All of the work (or at least all of the thinking) happens in the setup.

Example 4.2.7.

In Warning 4.2.3 we emphasised that the conclusion of the divergence Theorem 4.2.2 can fail if the vector field $\vF$ is not defined at even a single point of $V\text{.}$ Here is an example. Set

\begin{equation*} \vF = \frac{\vr}{|\vr|^3} \qquad\text{where } \vr=x\,\hi+y\,\hj +z\,\hk \end{equation*}

and $V=\Set{(x,y,z)}{x^2+y^2+z^2\le 1}\text{.}$ Then, if $(x,y,z)\ne\vZero\text{,}$

\begin{align*} \vnabla\cdot\vF(x,y,z) &=\frac{\partial }{\partial x}\frac{x}{{\big[x^2+y^2+z^2\big]}^{3/2}} +\frac{\partial }{\partial y}\frac{y}{{\big[x^2+y^2+z^2\big]}^{3/2}}\\ &\hskip1in+\frac{\partial }{\partial z}\frac{z}{{\big[x^2+y^2+z^2\big]}^{3/2}}\\ &=\frac{\big[x^2+y^2+z^2\big]-x\frac{3}{2}(2x)}{{\big[x^2+y^2+z^2\big]}^{5/2}} +\frac{\big[x^2+y^2+z^2\big]-y\frac{3}{2}(2y)}{{\big[x^2+y^2+z^2\big]}^{5/2}}\\ &+\frac{\big[x^2+y^2+z^2\big]-z\frac{3}{2}(2z)}{{\big[x^2+y^2+z^2\big]}^{5/2}}\\ &=0 \end{align*}

On the other hand, the boundary of $V$ is the unit sphere $\partial V = \Set{(x,y,z)}{x^2+y^2+z^2 = 1}\text{.}$ The outward unit normal to $\partial V$ is $\hn = \frac{\vr}{|\vr|}$ so that

\begin{align*} \int_{\partial V}\vF\cdot\hn\ \dee{S} &=\int_{|\vr|=1} \frac{\vr}{|\vr|^3}\cdot \frac{\vr}{|\vr|}\ \dee{S} =\int_{|\vr|=1} \frac{1}{|\vr|^2}\ \dee{S} =\int_{|\vr|=1}\dee{S}\\ &=4\pi\ne 0 \end{align*}

Subsection 4.2.1 Optional — An Application of the Divergence Theorem — the Heat Equation

Subsubsection 4.2.1.1 Derivation of the Heat Equation

Let $T(x,y,z,t)$ be the temperature at time $t$ at the point $(x,y,z)$ in some object $\cB\text{.}$ The heat equation

⁸

The heat equation was formulated by the French mathematician and physicist Jean-Baptiste Joseph Fourier in 1807. He lived from 1768 to 1830, a period which included both the French revolution and the reign of Napoleon. Indeed Fourier served on his local Revolutionary Committee, was imprisoned briefly during the Terror, and was Napoleon Bonaparte’s scientific advisor on his Egyptian expedition of 1798. Fourier series and the Fourier transform are named after him. Fourier is also credited with discovering the greenhouse effect.

is the partial differential equation that describes the flow of heat energy and consequently the behaviour of $T\text{.}$ We now use the divergence theorem to derive the heat equation from two physical “laws”, that we assume are valid:

The amount of heat energy required to raise the temperature of an object by $\De T$ degrees is $CM\,\De T$ where, $M$ is the mass of the object and $C$ is a positive physical constant determined by the material contained in the object. It is called the specific heat, or specific heat capacity
⁹
Heat is now understood to arise from the internal energy of the object. In an earlier theory, heat was viewed as measuring an invisible fluid, called the caloric. The amount of caloric that an object could hold was called its “heat capacity” by the Scottish physician and chemist Joseph Black (1728–1799).
, of the object.
Think of heat energy as a moving fluid. We will rig its velocity field so that heat flows in the direction opposite to the temperature gradient. Precisely, we choose its velocity field to be $-\ka\vnabla T(x,y,z,t)\text{.}$ Here $\ka$ is another positive physical constant called the thermal conductivity of the object. So the rate at which heat is conducted across an element of surface area $\dee{S}$ at $(x,y,z)$ in the direction of its unit normal $\hn$ is given by $-\ka\hn\cdot\vnabla T(x,y,z,t)\,\dee{S}$ at time $t\text{.}$ (See Lemma 3.4.1.) For example, in the figure

the temperature gradient, which points in the direction of increasing temperature, is opposite $\hn\text{.}$ Consequently the flow rate $-\ka\hn\cdot\vnabla T(x,y,z,t)\,\dee{S}$ is positive, indicating flow in the direction of $\hn\text{.}$ This is just what you would expect — heat flows from hot regions to cold regions. Also the rate of flow increases as the magnitude of the temperature gradient increases. This also makes sense (and is reminiscent of Newton’s law of cooling).

Let $V\subset\cB$ be any three dimensional region in the object and denote by $\partial V$ the surface of $V$ and by $\hn$ the outward normal to $\partial V\text{.}$ The amount of heat that enters $V$ across an infinitesimal piece $\dee{S}$ of $\partial V$ in an infinitesimal time interval $\dee{t}$ is $-\big(-\ka\hn\cdot\vnabla T(x,y,z,t)\,\dee{S}\big)\,\dee{t}\text{.}$ The amount of heat that enters $V$ across all of $\partial V$ in the time interval $\dee{t}$ is given by the integral

\begin{equation*} \dblInt_{\partial V} \ka\hn\cdot\vnabla T(x,y,z,t)\,\dee{S}\,\dee{t} \end{equation*}

In this same time interval, the temperature at a point $(x,y,z)$ in $V$ changes by $\frac{\partial T}{\partial t}(x,y,z,t)\,\dee{t}\text{.}$ If the density of the object at $(x,y,z)$ is $\rho(x,y,z)\text{,}$ the amount of heat energy required to increase the temperature of an infinitesimal volume $\dee{V}$ of the object centred at $(x,y,z)$ by $\frac{\partial T}{\partial t}(x,y,z,t)\,\dee{t}$ is $C(\rho\dee{V})\,\frac{\partial T}{\partial t}(x,y,z,t)\,\dee{t}\text{.}$ The amount of heat energy required to increase the temperature by $\frac{\partial T}{\partial t}(x,y,z,t)\,\dee{t}$ at all points $(x,y,z)$ in $V$ is then

\begin{equation*} \tripInt_V C\rho\frac{\partial T}{\partial t}(x,y,z,t)\,\dee{V}\,\dee{t} \end{equation*}

Assuming that the object is not generating or destroying

¹⁰

The caloric theory of heat was itself destroyed by the cannon boring experiment of 1798. In this experiment the American/British physicist Benjamin Thompson (1753–1814) boiled water just using the heat generated by friction during the boring of a cannon.

heat itself, this must be same as the amount of heat that entered $V$ in the time interval $\dee{t}\text{.}$ That is

\begin{equation*} \dblInt_{\partial V} \ka\hn\cdot\vnabla T\,\dee{S}\,\dee{t} =\tripInt_V C\rho\frac{\partial T}{\partial t}\,\dee{V}\,\dee{t} \end{equation*}

Now we cancel the common factor of $\dee{t}\text{.}$ We can then rewrite the left hand side as an integral over $V$ by applying the divergence theorem giving

\begin{equation*} \tripInt_{V} \ka\vnabla\cdot\vnabla T\,\dee{V} =\tripInt_V C\rho\frac{\partial T}{\partial t}\,\dee{V} \end{equation*}

As both integrals are over the same volume $V\text{,}$ we have

\begin{align*} &\tripInt_{V} \ka\vnabla\cdot\vnabla T\,\dee{V} -\tripInt_V C\rho\frac{\partial T}{\partial t}\,\dee{V} =0\\ &\hskip1in\implies \tripInt_{V} \left[\ka\vnabla^2 T -C\rho\frac{\partial T}{\partial t}\right]\,\dee{V}=0 \tag{H} \end{align*}

where $\vnabla^2=\vnabla\cdot\vnabla=\frac{\partial^2 }{\partial x^2}+ \frac{\partial^2 }{\partial y^2}+\frac{\partial^2 }{\partial z^2}$ is the Laplacian. This must be true for all volumes $V$ in the object and for all times $t\text{.}$ We claim that this forces

\begin{equation*} \ka\vnabla^2 T(x,y,z,t)-C\rho\frac{\partial T}{\partial t}(x,y,z,t)=0 \end{equation*}

for all $(x,y,z)$ in the object and all $t\text{.}$

Suppose that to the contrary there was a point $(x_0,y_0,z_0)$ in the object and a time $t_0$ with, for example, $\ka\vnabla^2 T(x_0,y_0,z_0,t_0) -C\rho\frac{\partial T}{\partial t}(x_0,y_0,z_0,t_0) \gt 0\text{.}$ By continuity, which we are assuming, $\ka\vnabla^2 T(x,y,z,t_0) -C\rho\frac{\partial T}{\partial t}(x,y,z,t_0)$ must remain close to $\ka\vnabla^2 T(x_0,y_0,z_0,t_0) -C\rho\frac{\partial T}{\partial t}(x_0,y_0,z_0,t_0)$ when $(x,y,z)$ is close to $(x_0,y_0,z_0)\text{.}$ So we would have

\begin{equation*} \ka\vnabla^2 T(x,y,z,t_0) -C\rho\frac{\partial T}{\partial t}(x,y,z,t_0) \gt 0 \end{equation*}

for all $(x,y,z)$ in some small ball $B$ centered on $(x_0,y_0,z_0)\text{.}$ Then, necessarily,

\begin{equation*} \tripInt_{B} \Big[\ka\vnabla\cdot\vnabla T(x,y,z,t_0) -C\rho\frac{\partial T}{\partial t}(x,y,z,t_0)\Big]\,\dee{V} \gt 0 \end{equation*}

which violates (H) for $V=B\text{.}$ This completes our derivation of the heat equation, which is

Equation 4.2.8.

\begin{equation*} \frac{\partial T}{\partial t}(x,y,z,t) =\alpha \vnabla^2 T(x,y,z,t) \end{equation*}

where $\alpha=\frac{\ka}{C\rho}$ is called the thermal diffusivity.

Subsubsection 4.2.1.2 An Application of the Heat Equation

As an application, we look at the temperature a short distance below the surface of the Earth. For simplicity, we make the Earth flat

¹¹

Insert sarcastic footnote here.

and we assume that the temperature, $T\text{,}$ depends only on time, $t\text{,}$ and the vertical coordinate, $z\text{.}$ Then the heat equation simplifies to

\begin{equation*} \frac{\partial T}{\partial t}(z,t) =\alpha \frac{\partial^2 T}{\partial z^2}(z,t) \tag{HE} \end{equation*}

We choose a coordinate system having the surface of the Earth at $z=0$ and having $z$ increase downward. We also assume that the temperature $T(0,t)$ at the surface of the Earth is primarily determined by solar heating and is given by

\begin{equation*} T(0,t)=T_0+T_A\cos(\sigma t)+T_D\cos(\delta t) \tag{BC} \end{equation*}

Here $T_0$ is the long term average of the temperature at the surface of the Earth, $T_A\cos(\sigma t)$ gives seasonal temperature variations and $T_D\cos(\delta t)$ gives daily temperature variations.

We measure time in days so that $\delta =2\pi$ and $\sigma =\frac{2\pi}{1\ {\rm year}}=\frac{2\pi}{365{\rm days}}\text{.}$ Then $T_A\cos(\sigma t)$ has period one year and $T_D\cos(\delta t)$ has period one day. The solution to the initial value problem (HE)+(BC) can be found by separation of variables, a standard topic in courses on partial differential equations. The solution is

\begin{align*} T(z,t)&=T_0 +T_Ae^{-\sqrt{\sigma \over 2\alpha}\ z} \cos\Big(\sigma t-\sqrt{\frac{\sigma }{2\alpha}}\ z\Big)\\ &\hskip1in+T_De^{-\sqrt{\delta \over 2\alpha}\ z} \cos\Big(\delta t-\sqrt{\frac{\delta }{2\alpha}}\ z\Big) \tag{SLN} \end{align*}

Whether or not you can find this solution, you can, and should, check that (SLN) satisfies both (HE) and (BC).

Now let’s see what we can learn from the solution (SLN). For any fixed $z\text{,}$ the time average of $T(z,t)$ is $T_0$ (just because the average value if cosine is zero), the same as the average temperature at the surface $z=0\text{.}$ That is, under the hypotheses that we have made, the long term average temperature at any depth $z$ is is the same as the long term average temperature at the surface.

The term $T_Ae^{-\sqrt{\sigma \over 2\alpha}\ z} \cos\Big(\sigma t-\sqrt{\frac{\sigma }{2\alpha}}\ z\Big)$

oscillates in time with a period of one year, just like $T_A\cos(\sigma t)$
has an amplitude $T_Ae^{-\sqrt{\sigma \over 2\alpha}\ z}$ which is $T_A$ at the surface and decreases exponentially as $z$ increases. Increasing the depth $z$ by a distance $\sqrt{\frac{2\alpha}{\sigma}}$ causes the amplitude of the oscillation to decrease by a factor of $\frac{1}{e}\text{.}$ Both of these first two bullet points are probably very consistent with your intuition. But this term also has a third property that you may find less obvious. It has
has a time lag of $\frac{z}{\sqrt{2\alpha\sigma}}$ with respect to $T_A\cos(\sigma t)\text{.}$ The surface term $T_A\cos(\sigma t)$ takes its maximum value when $t=0,\ \frac{2\pi}{\sigma},\ \frac{4\pi}{\sigma},\ \cdots\text{.}$ At depth $z\text{,}$ the corresponding term $T_Ae^{-\sqrt{\sigma \over 2\alpha}\ z} \cos\Big(\sigma t-\sqrt{\frac{\sigma }{2\alpha}}\ z\Big)$ takes its maximum value when $\ \sigma t-\sqrt{\frac{\sigma }{2\alpha}}\ z =0,\ 2\pi, 4\pi,\ \cdots$ so that $t=\frac{z}{\sqrt{2\alpha\sigma}},\ \frac{2\pi}{\sigma}+\frac{z}{\sqrt{2\alpha\sigma}}, \ \frac{4\pi}{\sigma}+\frac{z}{\sqrt{2\alpha\sigma}},\ \cdots\text{.}$

Similarly, the term $T_De^{-\sqrt{\delta \over 2\alpha}\ z} \cos\Big(\delta t-\sqrt{\frac{\delta }{2\alpha}}\ z\Big)$

oscillates in time with a period of one day, just like $T_D\cos(\delta t)$
has an amplitude which is $T_D$ at the surface and decreases by a factor of $\frac{1}{e}$ for each increase of $\sqrt{\frac{2\alpha}{\delta}}$ in depth.
has a time lag of $\frac{z}{\sqrt{2\alpha\delta}}$ with respect to $T_D\cos(\delta t)\text{.}$

For water $\alpha$ is approximately $0.012$ m$^2$/day. This $\alpha$ gives

\begin{equation*} \sqrt{\frac{2\alpha}{\sigma}}\approx 1.2\,{\rm m}\quad \sqrt{\frac{2\alpha}{\delta}}\approx 0.062\,{\rm m} \quad \frac{z}{\sqrt{2\alpha\sigma}}\approx 49\, z\,{\rm days} \quad \frac{z}{\sqrt{2\alpha\delta}}\approx 2.6\, z\,{\rm days} \end{equation*}

for $z$ measured in centimeters. So at a depth of a couple of meters, the temperature is pretty constant in time. What variation there is lags the surface variations by several months.

Subsection 4.2.2 Variations of the Divergence Theorem

Here are a couple useful variations of the divergence theorem.

Theorem 4.2.9. Variations on the divergence theorem.

If $V$ is a solid with surface $\partial V\text{,}$ then

\begin{align*} \dblInt_{\partial V} \vF\cdot\hn\,\dee{S} &=\tripInt_V\vnabla\cdot\vF\ \dee{V}\\ \dblInt_{\partial V} f \hn\,\dee{S} &=\tripInt_V\vnabla f\ \dee{V}\\ \dblInt_{\partial V} \hn\times\vF\,\dee{S} &=\tripInt_V\vnabla\times\vF\ \dee{V} \end{align*}

where $\hn$ is the outward unit normal of $\partial V\text{.}$

Memory Aid. All three formulae can be combined into

\begin{equation*} \dblInt_{\partial V} \hn *\tilde F\,\dee{S} =\tripInt_V\vnabla *\tilde F\ \dee{V} \end{equation*}

where $*$ can be either $\cdot\text{,}$ $\times$ or nothing. When $*=\cdot$ or $*=\times\text{,}$ then $\tilde F=\vF\text{.}$ When $*$ is nothing, $\tilde F=f\text{.}$

Proof.

The first formula is exactly the divergence theorem and was proven in Theorem 4.2.2.

To prove the second formula, set $\vF=f\va\text{,}$ where $\va$ is any constant vector, and apply the divergence theorem.

\begin{align*} \dblInt_{\partial V} f\va\cdot\hn\,\dee{S} &=\tripInt_V\vnabla\cdot(f\va)\ \dee{V}\\ &=\tripInt_V\big[(\vnabla f)\cdot\va +f\underbrace{\vnabla\cdot\va}_{=0}\big]\ \dee{V}\\ &=\tripInt_V(\vnabla f)\cdot\va\ \dee{V} \end{align*}

To get the second line, we used the vector identity Theorem 4.1.4.c. To get the third line, we just used that $\va$ is a constant, so that all of its derivatives are zero. Rewrite

\begin{equation*} \tripInt_V(\vnabla f)\cdot\va\ \dee{V} =\tripInt_V\va\cdot(\vnabla f)\ \dee{V} \end{equation*}

Since $\va$ is a constant, we can factor it out of both integrals, so

\begin{align*} &\va\cdot\dblInt_{\partial V} f\hn\,\dee{S} =\va\cdot\tripInt_V\vnabla f\ \dee{V}\\ \implies &\va\cdot\bigg\{\dblInt_{\partial V} f\hn\,\dee{S} -\tripInt_V\vnabla f\ \dee{V}\bigg\}=0 \end{align*}

In particular, choosing $\va=\hi\text{,}$ $\hj$ and $\hk\text{,}$ we see that all three components of the vector $\dblInt_{\partial V} f\hn\,\dee{S} -\tripInt_V\vnabla f\ \dee{V}$ are zero. So

\begin{equation*} \dblInt_{\partial V} f\hn\,\dee{S} -\tripInt_V\vnabla f\ \dee{V}=0 \end{equation*}

which is what we wanted show.

To prove the third formula, apply the divergence theorem, but with $\vF$ replaced by $\va\times\vF\text{,}$ where $\va$ is any constant vector.

\begin{align*} &\dblInt_{\partial V} (\va\times\vF)\cdot\hn\ \dee{S} =\tripInt_V\vnabla\cdot(\va\times\vF)\ \dee{V}\\ &\hskip0.5in=\tripInt_V\big[\vF\cdot\underbrace{(\vnabla \times \va)}_{=\vZero} -\va\cdot(\vnabla\times\vF)\big]\ \dee{V}\\ &\hskip0.5in=-\tripInt_V\va\cdot(\vnabla \times\vF)\ \dee{V} =-\va\cdot\tripInt_V\vnabla \times\vF\ \dee{V} \end{align*}

To get the second line, we used the vector identity Theorem 4.1.4.d. To get the third line, we again used that $\va$ is a constant, so that all of its derivatives are zero. For all vectors $(\va\times\vb)\cdot\vc=\va\cdot(\vb\times\vc)$ (in case you don’t remember this, it was Lemma 4.1.8.a) so that

\begin{equation*} (\va\times\vF)\cdot\hn =\va\cdot(\vF\times\hn) \end{equation*}

and

\begin{align*} &\va\cdot\dblInt_{\partial V} \vF\times\vn\ \dee{S} =-\va\cdot\tripInt_V\vnabla \times\vF\ \dee{V}\\ \implies &\va\cdot\bigg\{\dblInt_{\partial V} \vF\times\vn\ \dee{S} +\tripInt_V\vnabla\times\vF\ \dee{V}\bigg\}=0 \end{align*}

In particular, choosing $\va=\hi\text{,}$ $\hj$ and $\hk\text{,}$ we see that all three components of the vector $\dblInt_{\partial V} \vF\times\vn\ \dee{S} +\tripInt_V\vnabla\times\vF\ \dee{V}$ are zero. So

\begin{equation*} \tripInt_V\vnabla\times\vF\ \dee{V} =-\dblInt_{\partial V} \vF\times\vn\ \dee{S} =\dblInt_{\partial V} \hn\times\vF\ \dee{S} \end{equation*}

which is what we wanted show.

Subsection 4.2.3 An Application of the Divergence Theorem — Buoyancy

In this section, we use the divergence theorem to show that when you immerse an object in a fluid the net effect of fluid pressure acting on the surface of the object is a vertical force (called the buoyant force) whose magnitude equals the weight of fluid displaced by the object. This is known as Archimedes’ principle

¹²

The interested reader should do a net search for the story of Archimedes and the golden crown.

We shall also show that the buoyant force acts through the “centre of buoyancy” which is the centre of mass of the fluid displaced by the object. The design of self-righting

¹³

The first design of a self-righting boat was entered by William Wouldhave in a lifeboat design competition organised by South Shield’s Law House committee in 1789.

boats exploits the fact that the centre of buoyancy and the centre of gravity, where gravity acts, need not be the same.

We start by computing the total force due to the pressure of the fluid pushing on the object. Recall that pressure

is the force per unit surface area that the fluid exerts on the object
acts perpendicularly to the surface
pushes on the object

Thus the force due to pressure that acts on an infinitesimal piece of the object’s surface at $\vr=(x,y,z)$ with surface area $\dee{S}$ and outward normal $\hn$ is $-p(\vr)\,\hn \dee{S}\text{.}$ The minus sign is there because pressure is directed into the object. If the object fills the volume $V$ and has surface $\partial V\text{,}$ then the total force on the object due to fluid pressure, called the buoyant force, is

\begin{equation*} \vB=-\dblInt_{\partial V}p(\vr)\,\hn\,\dee{S} \end{equation*}

We now wish to apply a variant of the divergence theorem to rewrite $\vB=-\tripInt_V \vnabla p\ \dee{V}\text{.}$ But there is a problem with this: $p(\vr)$ is the fluid pressure at $\vr$ and is only defined where there is fluid. In particular, there is no fluid

¹⁴

A cup of tea in the galley doesn’t count.

inside the object, so $p(\vr)$ is not defined for any $\vr$ in the interior of $V\text{.}$

So we pretend that we remove the object from the fluid and we call $P(\vr)$ the fluid pressure at $\vr$ when there is no object in the fluid. We also make the assumption that at any point $\vr$ outside of the object, the pressure at $\vr$ does not depend on whether the object is in the fluid or not. In other words, we assume that

\begin{equation*} p(\vr)=\begin{cases} P(\vr)& \text{if $\vr$ is not in $V$}\\ \text{not defined} & \text{if $\vr$ is in the $V$} \end{cases} \end{equation*}

This assumption is only an approximation to reality, but, in practice, it is a very good approximation. So, by Theorem 4.2.9,

\begin{equation} \vB=-\dblInt_{\partial V}p(\vr)\,\hn\,\dee{S} =-\dblInt_{\partial V}P(\vr)\,\hn\,\dee{S} =-\tripInt_{V}\vnabla P(\vr)\,\dee{V}\tag{4.2.1} \end{equation}

Our next job is to compute $\vnabla P\text{.}$ Concentrate on an infinitesimal cube of fluid whose edges are parallel to the coordinate axes. Call the lengths of the edges $\dee{x}\text{,}$ $\dee{y}$ and $\dee{z}$ and the position of the centre of the cube $(x,y,z)\text{.}$ The forces applied to the various faces of the cube by the pressure of fluid outside the cube are illustrated in the figure

The total force due to the pressure acting on the cube is the sum

\begin{align*} &-P\left(x+\frac{\dee{x}}{2},y,z\right)\,dy\dee{z}\,\hi +P\left(x-\frac{\dee{x}}{2},y,z\right)\,dy\dee{z}\,\hi\\ &-P\left(x,y+\frac{dy}{2},z\right)\,\dee{x}\dee{z}\,\hj +P\left(x,y-\frac{dy}{2},z\right)\,\dee{x}\dee{z}\,\hj\\ &-P\left(x,y,z+\frac{\dee{z}}{2}\right)\,\dee{x}dy\,\hk +P\left(x,y,z-\frac{\dee{z}}{2}\right)\,\dee{x}dy\,\hk \end{align*}

of the forces acting on the six faces. Consider the $\hi$ component and rewrite it as

\begin{align*} &-P\left(x+\frac{\dee{x}}{2},y,z\right)\,dy\dee{z}\,\hi +P\left(x-\frac{\dee{x}}{2},y,z\right)\,dy\dee{z}\,\hi\\ &\hskip1in =-\frac{P(x+{\dee{x}\over2},y,z)-P(x-{\dee{x}\over2},y,z)}{\dee{x}}\,\hi\ \dee{x} \dee{y} \dee{z}\\ &\hskip1in = - \frac{\partial P}{\partial x}(x,y,z)\,\hi\ \dee{x} \dee{y} \dee{z} \end{align*}

Doing this for the other components as well, we see that the total force due to the pressure acting on the cube is

\begin{gather*} -\Big\{\frac{\partial P}{\partial x}(x,y,z)\,\hi \!+\!\frac{\partial P}{\partial y}(x,y,z)\,\hj \!+\!\frac{\partial P}{\partial z}(x,y,z)\,\hk \Big\}\dee{x} \dee{y} \dee{z} =-\vnabla P(x,y,z)\, \dee{x} \dee{y} \dee{z} \end{gather*}

We shall assume that the only other force acting on the cube is gravity and that the fluid is stationary (or at least not accelerating). Hence the total force acting on the cube is zero. If the fluid has density $\rhof\text{,}$ then the cube has mass $\rhof\, \dee{x}\dee{y}\dee{z}$ so that the force of gravity is $- g\rhof\, \dee{x}\dee{y}\dee{z}\,\hk\text{.}$ The vanishing of the total force now tells us that

\begin{equation*} -\vnabla P(\vr)\, \dee{x} \dee{y} \dee{z} - g\rhof\, \dee{x} \dee{y} \dee{z}\,\hk=0 \implies \vnabla P(\vr)= - g\rhof \,\hk \end{equation*}

Subbing this into (4.2.1) gives

\begin{equation*} \vB= g\,\hk\tripInt_{V}\rhof\,\dee{V} = gM_f\,\hk \end{equation*}

where $M_f =\tripInt_V \rhof\,\dee{V}$ is the mass of the fluid displaced by the object — not the mass of the object itself. Thus the buoyant force acts straight up and has magnitude equal to $gM_f \text{,}$ which is also the magnitude of the force of gravity acting on the fluid displaced by the object. In other words, it is the weight of the displaced fluid. This is exactly Archimedes’ principle.

We next consider the rotational motion of our submerged object. The physical law that determines the rotational motion of a rigid body about a point $\vr_0$ is analogous to the familiar Newton’s law, $m\frac{d\vv}{\dee{t}}=\vF\text{,}$ that determines the translational motion of the object. For the rotational law of motion,

the mass $m$ is replaced by a physical quantity, characteristic of the object, called the moment of inertia, and
the ordinary velocity $\vv$ is replaced by the angular velocity, which is a vector whose length is the rate of rotation (i.e. angle rotated per unit time) and whose direction is parallel to the axis of rotation (with the sign determined by a right hand rule), and
the force $\vF$ is replaced by a vector called the torque about $\vr_0\text{.}$ A force $\vF$ applied at $\vr=(x,y,z)$ produces the torque
¹⁵
This is what Archimedes was referring to when he said “Give me a lever and a place to stand and I will move the earth.”
$(\vr-\vr_0)\times\vF$ about $\vr_0\text{.}$

This is derived in the optional §4.2.4, entitled “Torque”, and is all that we need to know about rotational motion of rigid bodies in this discussion.

Fix any point $\vr_0\text{.}$ The total torque about $\vr_0$ produced by force of pressure acting on the surface of the submerged object is

\begin{equation*} \vT=\dblInt_{\partial V} (\vr-\vr_0)\times\big(-p(\vr)\hn\big)\,\dee{S} =\dblInt_{\partial V} \hn\times\big(P(\vr)\, (\vr-\vr_0)\big)\,\dee{S} \end{equation*}

Recall that in these integrals $\vr=(x,y,z)$ is the position of the infinitesimal piece $\dee{S}$ of the surface $S\text{.}$ Applying the cross product variant of the divergence theorem in Theorem 4.2.9, followed by the vector identity Theorem 4.1.5.c, gives

\begin{align*} \vT&=\tripInt_V \vnabla\times\big(P(\vr)\,(\vr-\vr_0)\big)\,\dee{V}\\ &=\tripInt_V \big\{\vnabla P(\vr)\times(\vr-\vr_0) +P(\vr)\underbrace{\vnabla\times(\vr-\vr_0)}_{=\vZero}\big\}\,\dee{V}\cr &=\tripInt_V \vnabla P(\vr)\times(\vr-\vr_0)\,\dee{V} \end{align*}

since $\vnabla\times\vr_0=0\text{,}$ because $\vr_0$ is a constant, and

\begin{equation*} \vnabla\times\vr=\det\left[\begin{matrix} \hi&\hj&\hk\cr \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ x&y&z \end{matrix}\right] =0 \end{equation*}

We have already found that $\vnabla P(\vr)=-g\rhof\hk\text{.}$ Substituting it in gives

\begin{align*} \vT&=-\tripInt_V g\rhof\hk\times(\vr-\vr_0)\,\dee{V}\\ &=-g\hk\times\tripInt_V \rhof (\vr-\vr_0)\,\dee{V}\\ &=-g\hk\times\bigg\{\tripInt_V\! \vr\rhof\,\dee{V} -\vr_0\!\tripInt_V\! \rhof\,\dee{V}\bigg\}\\ &=-g\bigg\{\tripInt_V\! \rhof\,\dee{V}\bigg\}\hk\times \bigg\{ \frac{\tripInt_V \vr\rhof\,\dee{V}}{\tripInt_V\rhof\, \dee{V}}-\vr_0\bigg\}\\ &=-\vB\times\bigg\{\frac{\tripInt_V \vr\rhof\,\dee{V}} {\tripInt_V\rhof\,\dee{V}}-\vr_0\bigg\}\\ &=\bigg\{\frac{\tripInt_V \vr\rhof\,\dee{V}} {\tripInt_V\rhof\,\dee{V}}-\vr_0\bigg\}\times\vB \end{align*}

So the torque generated at $\vr_0$ by pressure over the entire surface is the same the torque generated at $\vr_0$ by a force $\vB$ applied at the single point

\begin{equation*} \vC_\vB=\frac{\tripInt_V \vr\rhof\,\dee{V}}{\tripInt_V\rhof\,\dee{V}} \end{equation*}

This point is called the centre of buoyancy. It is the centre of mass of the displaced fluid.

The moral of the above discussion is that the buoyant force, $\vB\text{,}$ on a rigid body

acts straight upward,
has magnitude equal to the weight of the displaced fluid and
acts at the centre of buoyancy, which is the centre of mass of the displaced fluid.

As above, denoting by $\rhob$ the density of the object, the torque about $\vr_0$ due to gravity acting on the object is

\begin{equation*} \tripInt_V (\vr-\vr_0)\times(-g\rhob\hk)\,\dee{V} =\bigg\{\frac{\tripInt_V \vr\rhob\,\dee{V}} {\tripInt_V\rhob\,\dee{V}}-\vr_0\bigg\}\times \left(-g\bigg\{\tripInt_V \rhob\,\dee{V}\bigg\}\ \hk\right) \end{equation*}

So the gravitational force, $\vG\text{,}$

acts straight down,
has magnitude equal to the weight $g M_b=g\tripInt_V \vr\rhob\,\dee{V}$ (where $\rhob$ is the density of the object) of the object and
acts at the centre of mass, $\vC_\vG=\frac{\tripInt_V \vr\rhob\,\dee{V}}{\tripInt_V\rhob\,\dee{V}} \text{,}$ of the object.

Because the mass distribution of the object need not be the same as the mass distribution of the displaced fluid, buoyancy and gravity may act at two different points. This is exploited in the design of self-righting boats.

These boats are constructed with a heavy, often lead (which is cheap and dense), keel. As a result, the centre of gravity is lower in the boat than the center of buoyancy, which, because the displaced fluid has constant density, is at the geometric centre of the boat. As the figure below illustrates, a right side up configuration of such a boat is stable, while an upside down configuration is unstable. The boat rotates so as to keep the centre of gravity straight below the centre of buoyancy. To see this pretend that you are holding on to the boat with one hand holding the centre of buoyancy and the other hand holding the centre of gravity. Use your hands to apply forces in the directions of the arrows and think about how the boat will respond.

Subsection 4.2.4 Optional — Torque

In this section, we derive the properties of torque that we used in the last section. Newton’s law of motion says that the position $\vr(t)$ of a single particle moving under the influence of a force $\vF$ obeys $m\vr''(t)=\vF\text{.}$ Similarly, the positions $\vr_i(t)\text{,}$ $1\le i\le n\text{,}$ of a set of particles moving under the influence of forces $\vF_i$ obey $m\vr_i''(t)=\vF_i\text{,}$ $1\le i\le n\text{.}$ Very often systems of interest consist of some small number of rigid bodies. Suppose that we are interested in the motion of a single rigid body, say a piece of wood. The piece of wood is made up of a huge number

¹⁶

Just 12 grams of carbon contains about $6\times 10^{23}$ atoms.

of atoms. So the system of equations determining the motion of all of the individual atoms in the piece of wood is huge. On the other hand, we shall see that because the piece of wood is rigid, its configuration is completely determined by the position of, for example, its centre of mass and its orientation (we won’t get into what precisely is meant by “orientation”, but it is certainly determined by, for example, the positions of a few of the corners of the piece of wood). To be precise, we shall extract from the huge system of equations that determine the motion of all of the individual atoms, a small system of equations that determine the motion of the centre of mass and the orientation. We’ll do so now.

Imagine a piece of wood moving in $\bbbr^3\text{.}$

Furthermore, imagine that the piece of wood consists of a huge number of particles joined by a huge number of weightless but very strong

¹⁷

Mathematicians and their idealizations! Really the rods just represent the atomic/chemical forces that hold the wood together.

steel rods. The steel rod joining particle number one to particle number two just represents a force acting between particles number one and two. Suppose that

there are $n$ particles, with particle number $i$ having mass $m_i\text{,}$
at time $t\text{,}$ particle number $i$ has position $\vr_i(t)\text{,}$
at time $t\text{,}$ the external force (gravity and the like) acting on particle number $i$ is $\vF_i(t)\text{,}$ and
at time $t\text{,}$ the force acting on particle number $i\text{,}$ due to the steel rod joining particle number $i$ to particle number $j$ is $\vF_{i,j}(t)\text{.}$ If there is no steel rod joining particles number $i$ and $j\text{,}$ just set $\vF_{i,j}(t)=0\text{.}$ In particular, $\vF_{i,i}(t)=0\text{.}$

The only assumptions that we shall make about the steel rod forces are

(A1): for each $i\ne j\text{,}$ $\vF_{i,j}(t)=-\vF_{j,i}(t)\text{.}$ In words, the steel rod joining particles $i$ and $j$ applies equal and opposite forces to particles $i$ and $j\text{.}$
(A2): for each $i\ne j\text{,}$ there is a function $M_{i,j}(t)$ such that $\vF_{i,j}(t)=M_{i,j}(t)\big[\vr_i(t)-\vr_j(t)\big]\text{.}$ In words, the force due to the rod joining particles $i$ and $j$ acts parallel to the line joining particles $i$ and $j\text{.}$ For (A1) to be true, that is to have $M_{i,j}(t)\big[\vr_i(t)-\vr_j(t)\big] =-M_{j,i}(t)\big[\vr_j(t)-\vr_i(t)\big]\text{,}$ we need $M_{i,j}(t)=M_{j,i}(t)\text{.}$

Newton’s law of motion, applied to particle number $i\text{,}$ now tells us that

\begin{equation*} m_i \vr''_i(t) = \vF_i(t)+\sum_{j=1}^n \vF_{i,j}(t) \tag{$N_i$} \end{equation*}

Adding up all of the equations ($N_i$), for $i=1,\ 2,\ 3,\ \cdots,\ n$ gives

\begin{equation*} \sum_{i=1}^n m_i \vr''_i(t) = \sum_{i=1}^n \vF_i(t)+\sum_{1\le i,j\le n} \vF_{i,j}(t) \tag{$\Sigma N_i$} \end{equation*}

The sum $\sum\limits_{1\le i,j\le n} \vF_{i,j}(t)$ contains $\vF_{1,2}(t)$ exactly once and it also contains $\vF_{2,1}(t)$ exactly once and these two terms cancel exactly, by assumption (A1). In this way, all terms in $\sum\limits_{1\le i,j\le n} \vF_{i,j}(t)$ with $i\ne j$ exactly cancel. All terms with $i=j$ are assumed to be zero. So $\sum\limits_{1\le i,j\le n} \vF_{i,j}(t)=0$ and the equation ($\Sigma N_i$) simplifies to

\begin{equation*} \sum_{i=1}^n m_i \vr''_i(t) = \sum_{i=1}^n \vF_i(t) \tag{$ \Sigma N_i $} \end{equation*}

Phew! Denote by $M=\sum\limits_{i=1}^n m_i$ the total mass of the body, by $\vR(t)=\frac{1}{M}\sum\limits_{i=1}^n m_i\vr_i(t)$ the centre of mass

¹⁸

Note that this is just the weighted average (no pun intended) of the positions of the particles.

of the body and by $\vF(t)=\sum\limits_{i=1}^n \vF_i(t)$ the total external force acting on the system. In this notation, equation ($\Sigma N_i$) can be written as

Equation 4.2.10.

\begin{equation*} M\vR''(t)=\vF(t) \end{equation*}

The upshot is that the centre of mass of the system moves just like a single particle of mass $M$ subject to the total external force. This is why we can often replace an extended object by a point mass at its centre of mass.

Now take the cross product of $\vr_i(t)$ and equation ($N_i$) and sum over $i\text{.}$ This gives

\begin{align*} &\sum_{i=1}^n m_i\ \vr_i(t)\times\vr''_i(t)\\ &\hskip0.5in= \sum_{i=1}^n \vr_i(t)\times\vF_i(t) +\sum_{1\le i,j\le n} \vr_i(t)\times\vF_{i,j}(t) \tag{$\Sigma \vr_i\times N_i$} \end{align*}

By the assumption (A2)

\begin{align*} \vr_1(t)\times\vF_{1,2}(t) &=M_{1,2}(t)\ \vr_1(t)\times\big[\vr_1(t)-\vr_2(t)\big]\\ \vr_2(t)\times\vF_{2,1}(t) &=M_{2,1}(t)\ \vr_2(t)\times\big[\vr_2(t)-\vr_1(t)\big]\\ &=-M_{1,2}(t)\ \vr_2(t)\times\big[\vr_1(t)-\vr_2(t)\big]\\ \end{align*}

so that

\begin{align*} \vr_1(t)\times\vF_{1,2}(t) +\vr_2(t)\times\vF_{2,1}(t) &=M_{1,2}(t)\ \big[\vr_1(t)-\vr_2(t)\big]\times\big[\vr_1(t)-\vr_2(t)\big] =0 \end{align*}

because the cross product of any two parallel vectors is zero.

The last equation says that the $i=1\text{,}$ $j=2$ term in $\sum\limits_{1\le i,j\le n} \vr_i(t)\times\vF_{i,j}(t)$ exactly cancels the $i=2\text{,}$ $j=1$ term. In this way all of the terms in $\sum\limits_{1\le i,j\le n} \vr_i(t)\times\vF_{i,j}(t)$ with $i\ne j$ cancel. Each term with $i=j$ is exactly zero because $\vF_{ii}=0\text{.}$ So $\sum\limits_{1\le i,j\le n} \vr_i(t)\times\vF_{i,j}(t)=0$ and ($\Sigma \vr_i\times N_i$) simplifies to

\begin{equation*} \sum_{i=1}^n m_i\ \vr_i(t)\times\vr''_i(t) = \sum_{i=1}^n \vr_i(t)\times\vF_i(t) \tag{$\Sigma \vr_i\times N_i$} \end{equation*}

At this point it makes sense to define vectors

\begin{align*} \vL(t)&= \sum_{i=1}^n m_i\ \vr_i(t)\times\vr'_i(t)\\ \vT(t)&=\sum_{i=1}^n \vr_i(t)\times\vF_i(t) \end{align*}

because, in this notation, ($\Sigma \vr_i\times N_i$) becomes

Equation 4.2.11.

\begin{equation*} \diff{ }{t}\vL(t)=\vT(t) \end{equation*}

Equation 4.2.11 plays the role of Newton’s law of motion for rotational motion. $\vT(t)$ is called the torque and plays the role of “rotational force”. $\vL(t)$ is called the angular momentum (about the origin) and is a measure of the rate at which the piece of wood is rotating. For example, if a particle of mass $m$ is travelling in a circle of radius $\rho$ in the $xy$-plane at $\om$ radians per unit time, then $\vr(t)=\rho\cos(\om t)\hi+\rho\sin(\om t)\hj$ and

\begin{align*} m\vr(t)\times\vr'(t) &= m \big[\rho\cos(\om t)\hi+\rho\sin(\om t)\hj\big] \times\big[-\om\rho\sin(\om t)\hi+\om\rho\cos(\om t)\hj\big]\\ &=m \rho^2\ \om\ \hk \end{align*}

is proportional to $\om\text{,}$ which is the rate of rotation about the origin and is in the direction $\hk\text{,}$ which is normal to the plane containing the circle.

In any event, in order for the piece of wood to remain stationary, equations 4.2.10 and 4.2.11 force $\vF(t)=\vT(t)=0\text{.}$

Now suppose that the piece of wood is a seesaw

¹⁹

Or teeter-totter for those who speak a different English dialect.

that is supported on a fulcrum at $\vp\text{.}$ The forces consist of gravity, $-m_ig\hk\text{,}$ acting on particle number $i\text{,}$ for each $1\le i\le n\text{,}$ and the

force $\vPhi$ imposed by the fulcrum that is pushing up on the particle at $\vp\text{.}$ The total external force is $\vF=\vPhi-\sum\limits_{i=1}^n m_ig\hk =\vPhi-Mg\hk\text{.}$ If the seesaw is to remain stationary, this must be zero so that $\vPhi=Mg\hk\text{.}$

The total torque (about the origin) is

\begin{equation*} \vT=\vp\times\vPhi-\sum_{i=1}^n m_ig \vr_i\times\hk =g\Big(M\vp-\sum_{i=1}^n m_i \vr_i\Big)\times\hk \end{equation*}

If the seesaw is to remain stationary, this must also be zero. This will be the case if the fulcrum is placed at

\begin{equation*} \vp=\frac{1}{M}\sum_{i=1}^n m_i \vr_i \end{equation*}

which is just the centre of mass of the piece of wood.

More generally, suppose that the external forces acting on the piece of wood consist of $\vF_i\text{,}$ acting on particle number $i\text{,}$ for each $1\le i\le n\text{,}$ and a “fulcrum force” $\vPhi$ acting on a particle at $\vp\text{.}$ The total external force is $\vF=\vPhi+\sum\limits_{i=1}^n \vF_i\text{.}$ If the piece of wood is to remain stationary, this must be zero so that $\vPhi=-\sum\limits_{i=1}^n \vF_i\text{.}$ The total torque (about the origin) is

\begin{equation*} \vT=\vp\times\vPhi+\sum_{i=1}^n \vr_i\times\vF_i =\sum_{i=1}^n (\vr_i-\vp)\times\vF_i \end{equation*}

If the piece of wood is to remain stationary, this must also be zero. That is, the torque about point $\vp$ due to all of the forces $\vF_i\text{,}$ $1\le i\le n\text{,}$ must be zero.

Subsection 4.2.5 Optional — Solving Poisson’s Equation

In this section we shall use the divergence theorem to find a formula for the solution of Poisson’s equation

\begin{equation*} \vnabla^2\varphi = 4\pi\rho \end{equation*}

Here $\rho=\rho(\vr)$ is a given (continuous) function and $\varphi$ is the unknown function that we wish to find. This equation arises, for example, in electrostatics, where $\rho$ is the charge density and $\varphi$ is the electric potential.

The main step in finding this solution formula will be to consider an

arbitrary (smooth) function $\varphi$ and an
arbitrary (smooth) region $V$ in $\bbbr^3$ and an
arbitrary point $\vr_0$ in the interior of $V$

and to find an auxiliary formula which expresses $\varphi(\vr_0)$ in terms of

$\vnabla^2\varphi(\vr)\text{,}$ with $\vr$ running over $V$ and
$\vnabla\varphi(\vr)$ and $\varphi(\vr)\text{,}$ with $\vr$ running only over $\partial V\text{.}$

This auxiliary formula, which we shall derive below, is

\begin{align*} \varphi(\vr_0)&=-\frac{1}{4\pi}\bigg\{ \tripInt_V\frac{ \vnabla^2\varphi(\vr)}{|\vr-\vr_0|}\ d^3\vr -\dblInt_{\partial V}\varphi(\vr)\frac{\vr-\vr_0}{|\vr-\vr_0|^3}\cdot\hn\ \dee{S}\\ &\hskip1in-\dblInt_{\partial V}\frac{\vnabla\varphi(\vr)}{|\vr-\vr_0|}\cdot\hn\ \dee{S}\bigg\} \tag{$V$} \end{align*}

When we take the limit as $V$ expands to fill all of $\bbbr^3$ then, assuming that $\varphi$ and $\vnabla\varphi$ go to zero sufficiently quickly

²⁰

Suppose, for example, that, for large $|\vr-\vr_0|\text{,}$ $|\varphi(\vr)|$ is bounded by a constant times $\frac{1}{|\vr-\vr_0|}$ and $|\vnabla\varphi(\vr)|$ is bounded by a constant times $\frac{1}{|\vr-\vr_0|^2}\text{.}$ Then, if $\partial V$ is the sphere of radius $R$ centred on $\vr_0\text{,}$ $\partial V$ has surface area $4\pi R^2$ and the two integrals over $\partial V$ are bounded by a constant times $\frac{1}{R}\text{.}$

at $\infty\text{,}$ the two integrals over $\partial V$ will converge to zero and we will end up with the formula

\begin{equation*} \varphi(\vr_0)=-\frac{1}{4\pi} \tripInt_{\bbbr^3}\frac{ \vnabla^2\varphi(\vr)}{|\vr-\vr_0|}\ d^3\vr \end{equation*}

This expresses $\varphi$ evaluated at an arbitrary point, $\vr_0\text{,}$ of $\bbbr^3$ in terms of $\vnabla^2\varphi(\vr)\text{,}$ with $\vr$ running over $\bbbr^3\text{,}$ which is exactly what we want, since $\vnabla^2\varphi = 4\pi\rho$ for any solution of Poisson’s equation. So once we have proven (V) we will have proven

²¹

Note that the theorem does not claim that the $\varphi$ defined in the theorem obeys $\vnabla^2\varphi = 4\pi\rho\text{.}$ It does, but the proof is beyond our scope.

Theorem 4.2.12.

Assume that $\rho(\vr)$ is continuous and decays sufficiently quickly as $\vr\rightarrow\infty\text{.}$ If $\varphi$ obeys $\vnabla^2\varphi = 4\pi\rho$ on $\bbbr^3\text{,}$ and $\varphi$ and $\vnabla\varphi$ decay sufficiently quickly as $\vr\rightarrow\infty\text{,}$ then

\begin{equation*} \varphi(\vr_0)= -\tripInt_{\bbbr^3}\frac{ \rho(\vr)}{|\vr-\vr_0|}\ d^3\vr \end{equation*}

for all $\vr_0$ in $\bbbr^3\text{.}$

Let

\begin{align*} \vr(x,y,z)&=x\,\hi+y\,\hj+z\,\hk\\ \vr_0&=x_0\,\hi+y_0\,\hj+z_0\,\hk \end{align*}

We shall exploit three properties of the function $\frac{1}{|\vr-\vr_0|}\text{.}$ The first two properties are

\begin{align*} \vnabla \frac{1}{|\vr-\vr_0|} &=-\frac{\vr-\vr_0}{|\vr-\vr_0|^3}\tag{P1}\\ \vnabla^2 \frac{1}{|\vr-\vr_0|} &=-\vnabla\cdot\frac{\vr-\vr_0}{|\vr-\vr_0|^3} =0 \tag{P2} \end{align*}

and are valid for all $\vr\ne \vr_0\text{.}$ Verification of the first property is a simple one line computation. Verification of the second property is a simple three line computation. (See the solution to Question 6 in Section 4.1.)

The other property of $\frac{1}{|\vr-\vr_0|}$ that we shall use is the following. Let $S_\veps$ be the sphere of radius $\veps$ centered on $\vr_0\text{.}$ Then, for any continuous function $\psi(\vr)\text{,}$

\begin{align*} \lim_{\veps\rightarrow 0+}\dblInt_{S_\veps} \frac{\psi(\vr)}{|\vr-\vr_0|^p}\ \dee{S} &=\lim_{\veps\rightarrow 0+} \frac{1}{\veps^p} \dblInt_{S_\veps} \psi(\vr)\ \dee{S} =\lim_{\veps\rightarrow 0+}\frac{\psi(\vr_0)}{\veps^p} \dblInt_{S_\veps}\ \dee{S}\\ &=\lim_{\veps\rightarrow 0+}\frac{\psi(\vr_0)}{\veps^p}4\pi\veps^2 \notag\\ &=\begin{cases}4\pi\psi(\vr_0)& \text{if } p=2 \\ 0 & \text{if } p \lt 2\\ \text{undefined} & \text{if } p \gt 2 \end{cases} \tag{P3} \end{align*}

Derivation of (V):

Here is the derivation of ($V$). Let $V_\veps$ be the part of $V$ outside of $S_\veps\text{.}$

Note that the boundary $\partial V_\veps$ of $V_\veps$ consists of two parts — the boundary $\partial V$ of $V$ and the sphere $S_\veps$ — and that the unit outward normal to $\partial V_\veps$ on $S_\veps$ is $-\frac{\vr-\vr_0}{|\vr-\vr_0|}\text{,}$ because it points towards $\vr_0$ and hence outside of $V_\veps\text{.}$

Recall the vector identity Theorem 4.1.7.d, which says

\begin{equation*} \vnabla\cdot(f\vnabla g-g\vnabla f)=f\,\vnabla^2g-g\,\vnabla^2f \end{equation*}

Applying this identity with $f= \frac{1}{|\vr-\vr_0|}$ and $g=\varphi$ gives

\begin{align*} \vnabla\cdot\Big(\frac{1}{|\vr-\vr_0|}\vnabla\varphi -\varphi\vnabla\frac{1}{|\vr-\vr_0|}\Big) &=\frac{\vnabla^2\varphi}{|\vr-\vr_0|} -\varphi\,\overbrace{ \vnabla^2\frac{1}{|\vr-\vr_0|} }^{=0\text{ by (P2)}}\\ &=\frac{\vnabla^2\varphi}{|\vr-\vr_0|} \end{align*}

which is the integrand of the first integral on the right hand side of (V). So, by the divergence theorem

\begin{align*} &\tripInt_{V_\veps} \frac{\vnabla^2\varphi}{|\vr-\vr_0|} dV =\tripInt_{V_\veps}\vnabla\cdot\Big(\frac{1}{|\vr-\vr_0|}\vnabla\varphi -\varphi\vnabla\frac{1}{|\vr-\vr_0|}\Big) dV\\ &\hskip0.25in=\dblInt_{\partial V}\Big(\frac{1}{|\vr-\vr_0|}\vnabla\varphi -\varphi\vnabla\frac{1}{|\vr-\vr_0|}\Big)\cdot\hn\ \dee{S}\\ &\hskip1.0in+\dblInt_{S_\veps}\Big(\frac{1}{|\vr-\vr_0|}\vnabla\varphi -\varphi\vnabla\frac{1}{|\vr-\vr_0|}\Big)\cdot \Big(-\frac{\vr-\vr_0}{|\vr-\vr_0|}\Big) \dee{S} \tag{M} \end{align*}

To see the connection between (M) and the rest of (V), note that,

by (P1), the first term on the right hand side of (M) is
\begin{align*} &\dblInt_{\partial V}\Big(\frac{1}{|\vr-\vr_0|}\vnabla\varphi -\varphi\vnabla\frac{1}{|\vr-\vr_0|}\Big)\cdot\hn\ \dee{S}\\ &\hskip0.5in=\dblInt_{\partial V}\frac{\vnabla\varphi(\vr)}{|\vr-\vr_0|}\cdot\hn\ \dee{S} +\dblInt_{\partial V}\varphi(\vr)\frac{\vr-\vr_0}{|\vr-\vr_0|^3}\cdot\hn\ \dee{S} \tag{R1} \end{align*}
which is $4\pi$ times the second and third terms on the right hand side of (V),
and substituting in $\vnabla \frac{1}{|\vr-\vr_0|} =-\frac{\vr-\vr_0}{|\vr-\vr_0|^3}\text{,}$ from (P1), and applying (P3) with $p=2\text{,}$ the limit of the second term on the right hand side of (M) is
\begin{align*} &\lim_{\veps\rightarrow 0+} \dblInt_{S_\veps}\Big(\frac{1}{|\vr-\vr_0|}\vnabla\varphi -\varphi\vnabla\frac{1}{|\vr-\vr_0|}\Big)\cdot \Big(-\frac{\vr-\vr_0}{|\vr-\vr_0|}\Big) \dee{S}\\ &\hskip1in=-\lim_{\veps\rightarrow 0+} \dblInt_{B_\veps}\big[\vnabla\varphi\cdot(\vr-\vr_0)+\varphi\big] \frac{1}{|\vr-\vr_0|^2} \dee{S}\\ &\hskip1in=-4\pi\Big[\vnabla\varphi\cdot(\vr-\vr_0)+\varphi\Big]_{\vr=\vr_0}\\ &\hskip1in=-4\pi\varphi(\vr_0) \tag{R2} \end{align*}

So applying

²²

You might worry about the singularity in $\frac{\vnabla^2\varphi}{|\vr-\vr_0|}$ when applying $\lim_{\veps\rightarrow 0+}$ to $\tripInt_{V_\veps} \frac{\vnabla^2\varphi}{|\vr-\vr_0|} dV\text{.}$ That this singularity is harmless may be seen using spherical coordinates centred on $\vr_0\text{.}$ Then $\dee{V}$ contains a factor of $|\vr-\vr_0|^2$ (see §A.6.3), which completely eliminates the singularity.

$\lim_{\veps\rightarrow 0+}$ to (M) and substituting in (R1) and (R2) gives

\begin{gather*} \tripInt_{V}\frac{\vnabla^2\varphi}{|\vr-\vr_0|}\ dV =\dblInt_{\partial V}\frac{\vnabla\varphi(\vr)}{|\vr-\vr_0|}\cdot\hn\ \dee{S} +\dblInt_{\partial V}\varphi(\vr)\frac{\vr-\vr_0}{|\vr-\vr_0|^3}\cdot\hn\ \dee{S} -4\pi\varphi(\vr_0) \end{gather*}

which is exactly equation (V).

Exercises 4.2.6 Exercises

Exercise Group.

Exercises — Stage 1

1.

Let $V$ be the cube

\begin{gather*} V=\Set{(x,y,z)}{0\le x\le 1,\ 0\le y\le 1,\ 0\le z\le 1} \end{gather*}

and $R$ be the square

\begin{gather*} R=\Set{(x,y)}{0\le x\le 1,\ 0\le y\le 1} \end{gather*}

and let $f(x,y,z)$ have continuous first partial derivatives.

Use the fundamental theorem of calculus to show that
\begin{gather*} \tripInt_V\frac{\partial f}{\partial z}(x,y,z)\,\dee{x}\,\dee{y}\,\dee{z} =\dblInt_R\! f(x,y,1)\,\dee{x}\,\dee{y} -\! \dblInt_R\! f(x,y,0)\,\dee{x}\,\dee{y} \end{gather*}
Use the divergence theorem to show that
\begin{gather*} \tripInt_V\frac{\partial f}{\partial z}(x,y,z)\,\dee{x}\,\dee{y}\,\dee{z} =\dblInt_R\! f(x,y,1)\,\dee{x}\,\dee{y} -\! \dblInt_R\! f(x,y,0)\,\dee{x}\,\dee{y} \end{gather*}

2.

By applying the divergence theorem to $\vF=\phi\,\va\text{,}$ where $\va$ is an arbitrary constant vector, show that
\begin{equation*} \tripInt_V \vnabla\phi\,\dee{V}=\dblInt_{\partial V}\phi\,\hn\,\dee{S} \end{equation*}
Show that the centroid $(\bar x,\bar y,\bar z)$ of a solid $V$ with volume $|V|$ is given by
\begin{equation*} (\bar x,\bar y,\bar z)=\frac{1}{2|V|}\dblInt_{\partial V} (x^2+y^2+z^2)\,\hn\,\dee{S} \end{equation*}

Exercise Group.

Exercises — Stage 2

3.

Let $S$ be the unit sphere centered at the origin and oriented by the outward pointing normal. If

\begin{equation*} \vF(x,y,z)=\big(x,y,z^2\big) \end{equation*}

evaluate the flux of $\vF$ through $S$

directly and
by applying the divergence theorem.

4.

Evaluate, by two methods, the integral $\dblInt_S\vF\cdot \hn\,\dee{S}\text{,}$ where $\vF=z\,\hk\text{,}$ $S$ is the surface $x^2+y^2+z^2=a^2$ and $\hn$ is the outward pointing unit normal to $S\text{.}$

First, by direct computation of the surface integral.
Second, by using the divergence theorem.

5.

Let

$\ \vF=zy^3\,\hi+yx\, \hj+(2z+y^2)\hk\ $ and
$V$ be the solid in 3-space defined by
\begin{equation*} 0\le z\le \frac{9-x^2-y^2}{9+x^2+y^2} \end{equation*}
and
$D$ be the bottom surface of $V\text{.}$ Because $\frac{9-x^2-y^2}{9+x^2+y^2}$ is positive for $x^2+y^2 \lt 9$ and negative for $x^2+y^2 \gt 9\text{,}$ the bottom surface is $z=0\text{,}$ $x^2+y^2\le 9\text{.}$
Let $S$ be the curved portion of the boundary of $V\text{.}$ It is $z={9-x^2-y^2\over 9+x^2+y^2}\text{,}$ $x^2+y^2\le 9\text{.}$ Here is a sketch of the first octant part of $S$ and $D\text{.}$

Denote by $|V|$ the volume of $V$ and compute, in terms of $|V|\text{,}$

$\ds \dblInt_{D}\vF\cdot \hn\,\dee{S}$ with $\hn$ pointing downward
$\displaystyle \ds \tripInt_V\vnabla\cdot F\,\dee{V}$
$\ds \dblInt_S\vF\cdot\hn\,\dee{S}$ with $\hn$ pointing outward

Use the divergence theorem to answer at least one of parts (a), (b) and (c).

6.

Evaluate the integral $\dblInt_S\vF\cdot \hn\,\dee{S}\text{,}$ where $\vF=(x,y,1)\text{,}$ $S$ is the surface $z=1-x^2-y^2$ for $x^2+y^2\le 1\text{,}$ and $\hn$ is the upward pointing normal, by two methods.

First, by direct computation of the surface integral.
Second, by using the divergence theorem.

7. (✳).

Find the divergence of the vector field $\vF = (z + \sin y, zy, \sin x \cos y)\text{.}$
Find the flux of the vector field $\vF$ of (a) outward through the sphere of radius $3$ centred at the origin in $\bbbr^3$ .

8.

The sides of a grain silo are described by the portion of the cylinder $x^2 + y^2 = 1$ with $0 \le z\le 1\text{.}$ The top of the silo is given by the portion of the sphere $x^2+ y^2+ z^2 = 2$ lying within the cylinder and above the $xy$-plane. Find the flux of the vector field

\begin{equation*} \vV(x,y,z) = (x^2yz\,,\, yz+e^xz\,,\, x^2+y ) \end{equation*}

out of the silo.

9.

Let $B$ be the ball of volume $V$ centered at the point $(x_0, y_0, z_0)\text{,}$ and let $S$ be the sphere that is the boundary of $B\text{.}$ Find the flux of $\vF = x^2\hi + xy\hj+(3 z - yz)\hk$ outward (from $B$) through $S\text{.}$

10. (✳).

Let

\begin{equation*} \vF(x, y, z) = \big( 1 + z^{1+z^{1+z}}\,,\, 1 + z^{1+z^{1+z}}\,,\,1\big) \end{equation*}

Let $S$ be the portion of the surface

\begin{equation*} x^2 + y^2 = 1 - z^4 \end{equation*}

which is above the $xy$-plane. What is the flux of $\vF$ downward through $S\text{?}$

11. (✳).

Use the divergence theorem to find the flux of $x\hi+y\hj+2z\hk$ through the part of the ellipsoid

\begin{equation*} x^2+y^2+2z^2=2 \end{equation*}

with $z\ge 0\text{.}$ [Note: the ellipsoid $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$ has volume $\frac{4}{3}\pi abc\text{.}$]

12. (✳).

Let $\vF(x,y,z) = \vr/r^3$ where $\vr= x\,\hi + y\,\hj + z\,\hk$ and $r = |\vr|\text{.}$

Find $\vnabla \cdot \vF\text{.}$
Find the flux of $\vF$ outwards through the spherical surface $x^2 + y^2 + z^2 = a^2\text{.}$
Do the results of (a) and (b) contradict the divergence theorem? Explain your answer.
Let $E$ be the solid region bounded by the surfaces $z^2 - x^2 - y^2 + 1 = 0\text{,}$ $z = 1$ and $z = -1\text{.}$ Let $\sigma$ be the bounding surface of $E\text{.}$ Determine the flux of $\vF$ outwards through $\sigma\text{.}$
Let $R$ be the solid region bounded by the surfaces $z^2 - x^2 - y^2 + 4y - 3 = 0\text{,}$ $z = 1$ and $z = - 1\text{.}$ Let $\Sigma$ be the bounding surface of $R\text{.}$ Determine the flux of $\vF$ outwards through $\Sigma\text{.}$

13. (✳).

Consider the ellipsoid $S$ given by

\begin{equation*} x^2 + \frac{y^2}{4} +\frac{z^2}{4} = 1 \end{equation*}

with the unit normal pointing outward.

Parameterize $S\text{.}$
Compute the flux $\dblInt_S \vF\cdot\hn\, \dee{S}$ of the vector field
\begin{equation*} \vF(x,y,z) = (x, y, z) \end{equation*}
Verify your answer in (b) using the divergence theorem.

14. (✳).

Evaluate the flux integral $\dblInt_S \vF\cdot\hn\, \dee{S}\text{,}$ where

\begin{equation*} \vF(x,y,z) = \big(x^3 + \cos(y^2)\,,\, y^3 + ze^x\,,\, z^2 + \arctan(xy)\big) \end{equation*}

and $S$ is the surface of the solid region bounded by the cylinder $x^2 + y^2 = 2$ and the planes $z = 0$ and $z = 2x + 3\text{.}$ The surface is positively oriented (its unit normal points outward).

15. (✳).

Find the flux of the vector field $(x + y, x + z, y + z)$ through the cylindrical surface whose equation is $x^2 + z^2 = 4\text{,}$ and which extends from $y = 0$ to $y = 3\text{.}$ (Only the curved part of the cylinder is included, not the two disks bounding it on the left and right.) The orientation of the surface is outward, i.e., pointing away from the $y$-axis.

16. (✳).

The surface $S$ is the part above the $xy$-plane of the surface obtained by revolving the graph of $z = 1 - x^4$ around the $z$-axis. The surface $S$ is oriented such that the normal vector has positive $z$-component. The circle with radius $1$ and centre at the origin in the $xy$-plane is the boundary of $S\text{.}$

Find the flux of the divergenceless vector field $\vF(x, y, z) = (yz, x + z, x^2 + y^2)$ through $S\text{.}$

17. (✳).

Let $S$ be the part of the paraboloid $z = 2 - x^2 - y^2$ contained in the cone $z = \sqrt{x^2+y^2}$ and oriented in the upward direction. Let

\begin{equation*} \vF = (\tan \sqrt{z} + \sin(y^3))\,\hi + e^{-x^2}\,\hj + z\hk \end{equation*}

Evaluate the flux integral $\dblInt_S \vF \cdot\hn\,\dee{S}\text{.}$

18. (✳).

Evaluate the surface integral

\begin{equation*} \dblInt_S \vF\cdot\hn\,\dee{S} \end{equation*}

where $\vF(x, y, z) = \big(\cos z + xy^2\,,\, xe^{-z}\,,\, \sin y + x^2 z\big)$ and $S$ is the boundary of the solid region enclosed by the paraboloid $z = x^2 + y^2$ and the plane $z = 4\text{,}$ with outward pointing normal.

19. (✳).

Let $S$ be the part of the sphere $x^2 + y^2 + z^2 = 4$ between the planes $z = 1$ and $z = 0$ oriented away from the origin. Let

\begin{gather*} \vF = (e^y + xz)\,\hi + (zy + \sin(x))\,\hj + (z^2 - 1)\,\hk \end{gather*}

Compute the flux integral

\begin{gather*} \dblInt_S \vF\cdot\hn\,\dee{S}. \end{gather*}

20. (✳).

Let $B$ be the solid region lying between the planes $x=-1\text{,}$ $x=1\text{,}$ $y=0\text{,}$ $y=2$ and bounded below by the plane $z=0$ and above by the plane $z+y=3\text{.}$ Let $S$ be the surface of $B\text{.}$ Find the flux of the vector field

\begin{equation*} \vF(x,y,z) = \big(x^2 z +\cos \pi y\big)\,\hi +\big(yz +\sin \pi z\big)\,\hj +(x-y^2)\,\hk \end{equation*}

21. (✳).

Let $S$ be the hemisphere $x^2 + y^2 + z^2 = 1\text{,}$ $z\ge 0\text{,}$ oriented with $\hn$ pointing away from the origin. Evaluate the flux integral

\begin{equation*} \dblInt_S \vF\cdot\hn\, \dee{S} \end{equation*}

where

\begin{equation*} \vF = \big(x + \cos(z^2)\big)\,\hi + \big(y + \ln(x^2 + z^5)\big)\,\hj + \sqrt{x^2 + y^2}\,\hk \end{equation*}

22.

Let $E$ be the solid region between the plane $z = 4$ and the paraboloid $z = x^2 + y^2\text{.}$ Let

\begin{equation*} \vF = \Big(-\frac{1}{3}x^3 + e^{z^2}\Big)\hi + \Big(-\frac{1}{3}y^3 + x\sin z\Big)\hj + 4z\hk \end{equation*}

Compute the flux of $\vF$ outward through the boundary of $E\text{.}$
Let $S$ be the part of the paraboloid $z = x^2 + y^2$ lying below the $z = 4$ plane oriented with a normal vector $\hN$ that has a positive $\hk$ component. Compute the flux of $\vF$ through $S\text{.}$

23. (✳).

Consider the vector field

\begin{equation*} \vF(x,y,z) = \frac{x\,\hi+y\,\hj+z\,\hk} {{[x^2 + y^2 + z^2\big]}^{3/2}} \end{equation*}

Compute $\vnabla\cdot\vF\text{.}$
Let $S_1$ be the sphere given by
\begin{equation*} x^2 + (y - 2)^2 + z^2 = 9 \end{equation*}
oriented outwards. Compute $\ds \dblInt_{S_1} \vF\cdot\hn\,\dee{S}\text{.}$
Let $S_2$ be the sphere given by
\begin{equation*} x^2 + (y - 2)^2 + z^2 = 1 \end{equation*}
oriented outwards. Compute $\ds \dblInt_{S_2} \vF\cdot\hn\,\dee{S}\text{.}$
Are your answers to (b) and (c) the same or different? Give a mathematical explanation of your answer.

24. (✳).

Let $\vF$ be the vector field defined by

\begin{equation*} \vF(x,y, z) = \big(y^3 z + 2x\big)\,\hi +\big(3y - e^{\sin z}\big)\,\hj + \big(e^{x^2+y^2}+ z\big)\,\hk \end{equation*}

Calculate the flux integral $\dblInt_S \vF\cdot\hn\,\dee{S}$ where $S$ is the boundary surface of the solid region

\begin{equation*} E\ :\ 0 \le x \le 2,\quad 0 \le y \le 2,\quad 0 \le z \le 2+ y \end{equation*}

with outer normal.

25. (✳).

Consider the vector field

\begin{equation*} \vF(x, y, z) = \big(z \arctan(y^2)\,,\, z^3 \ln(x^2 + 1)\,,\, 3z\big) \end{equation*}

Let the surface $S$ be the part of the sphere $x^2 + y^2 + z^2 = 4$ that lies above the plane $z = 1$ and be oriented downwards.

Find the divergence of $\vF\text{.}$
Compute the flux integral $\dblInt_S \vF \cdot\hn\,\dee{S}\text{.}$

26. (✳).

Let $S$ be the sphere $x^2+y^2+z^2=3$ oriented inward. Compute the flux integral

\begin{equation*} \dblInt_S\vF\cdot\hn\,\dee{S} \end{equation*}

where

\begin{equation*} \vF = \big(xy^2 + y^4z^6\,,\, yz^2+x^4z\,,\,zx^2+xy^4\big) \end{equation*}

27. (✳).

Consider the vector field $\vF(x,y,z) =-2xy\,\hi+\big(y^2+\sin(xz)\big)\,\hj+(x^2+y^2)\,\hk\text{.}$

Calculate $\nabla\cdot \vF\text{.}$
Find the flux of $\vF$ through the surface $S$ defined by
\begin{equation*} x^2+y^2+(z-12)^2=13^2,\ z\ge 0 \end{equation*}
using the outward normal to $S\text{.}$

28. (✳).

Let $S$ be the portion of the hyperboloid $x^2 + y^2 -z^2 = 1$ between $z=-1$ and $z=1\text{.}$ Find the flux of $\vF = (x+e^{yz})\,\hi +\big(2yz+\sin(xz)\big)\,\hj +(xy-z-z^2)\,\hk$ out of $S$ (away from the origin).

29. (✳).

Let $\vF$ be the vector field $\vF(x,y,z)=(x^2-y-1)\,\hi+(e^{\cos y}+z^3)\,\hj+(2xz+z^5)\,\hk\text{.}$ Evaluate $\dblInt_S\vnabla\times \vF\cdot\hn\,\dee{S}$ where $S$ is the part of the ellipsoid $x^2+y^2+2z^2=1$ with $z\ge 0\text{.}$

30. (✳).

Let $S$ be the portion of the sphere $x^2+y^2+(z-1)^2=4$ that lies above the $xy$-plane. Find the flux of $\vF=(x^2+e^{y^2})\,\hi+(e^{x^2}+y^2)\,\hj +(4+5x)\,\hk$ outward across $S\text{.}$

31. (✳).

Find the flux of $\vF=xy^2\hi+x^2y\hj+\hk$ outward through the hemispherical surface

\begin{equation*} x^2+y^2+z^2=4,\qquad z\ge 0 \end{equation*}

32. (✳).

Let $D$ be the cylinder $x^2+y^2\le 1\text{,}$ $0\le z\le 5\text{.}$ Calculate the flux of the vector field

\begin{equation*} \vF=(x+xye^z)\,\hi+\half y^2ze^z\,\hj+(3z-yze^z)\,\hk \end{equation*}

outward through the curved part of the surface of $D\text{.}$

33.

Find the flux of $\vF=(y+xz)\hi+(y+yz)\hj-(2x+z^2)\hk$ upward through the first octant part of the sphere $x^2+y^2+z^2=a^2\text{.}$

34.

Let $\ \vF=(x-yz)\hi+(y+xz)\hj+(z+2xy)\hk\ $ and let

$S_1$ be the portion of the cylinder $\ x^2+y^2=2\ $ that lies inside the sphere $\ x^2+y^2+z^2=4$
$S_2$ be the portion of the sphere $\ x^2+y^2+z^2=4\ $ that lies outside the cylinder $\ x^2+y^2=2\ $
$V$ be the solid bounded by $S_1$ and $S_2$

Compute

$\dblInt_{S_1}\vF\cdot \hn\,\dee{S}$ with $\hn$ pointing inward
$\displaystyle \tripInt_V\vnabla\cdot F\,\dee{V}$
$\dblInt_{S_2}\vF\cdot \hn\,\dee{S}$ with $\hn$ pointing outward

Use the divergence theorem to answer at least one of parts (a), (b) and (c).

Exercise Group.

Exercises — Stage 3

35.

Let $\vE(\vr)$ be the electric field due to a charge configuration that has density $\rho(\vr)\text{.}$ Gauss’ law states that, if $V$ is any solid in $\bbbr^3$ with surface $\partial V\text{,}$ then the electric flux

\begin{equation*} \dblInt_{\partial V}\vE\cdot\hn\, \dee{S}=4\pi Q \qquad{\rm where}\qquad Q=\tripInt_V\rho\ \dee{V} \end{equation*}

is the total charge in $V\text{.}$ Here, as usual, $\hn$ is the outward pointing unit normal to $\partial V\text{.}$ Show that

\begin{equation*} \vnabla\cdot\vE(\vr)=4\pi \rho(\vr) \end{equation*}

for all $\vr$ in $\bbbr^3\text{.}$ This is one of Maxwell’s equations. Assume that $\vnabla\cdot\vE(\vr)$ and $\rho(\vr)$ are well--defined and continuous everywhere.

36.

Let $V$ be a solid in $\bbbr^3$ with surface $\partial V\text{.}$ Show that

\begin{equation*} \dblInt_{\partial V}\vr\cdot\hn\,\dee{S}=3\,\text{Volume}(V) \end{equation*}

where $\vr=x\,\hi+y\,\hj+z\,\hk$ and, as usual, $\hn$ is the outer normal to $\partial V\text{.}$ See if you can explain this result geometrically.

37. (✳).

Let $S$ be the sphere of radius $3\text{,}$ centered at the origin and with outward orientation. Given the vector field $\vF(x,y,z) = (0, 0, x + z)\text{:}$

Calculate (using the definition) the flux of $\vF$ through $S$
\begin{equation*} \dblInt_S \vF \cdot \hn\,\dee{S} \end{equation*}
That is, compute the flux by evaluating the surface integral directly.
Calculate the same flux using the divergence theorem.

38. (✳).

Consider the cube of side length $1$ that lies entirely in the first octant ($x \ge 0\text{,}$ $y \ge 0\text{,}$ $z \ge 0$) with one corner at the origin and another corner at point $(1, 1, 1)\text{.}$ As such, one face lies in the plane $x = 0\text{,}$ one lies in the plane $y = 0\text{,}$ and another lies in the plane $z = 0\text{.}$ The other three faces lie in the planes $x = 1\text{,}$ $y = 1\text{,}$ and $z = 1\text{.}$ Denote $S$ as the open surface that consists of the union of the $5$ faces of the cube that do not lie in the plane $z = 0\text{.}$ The surface $S$ is oriented in such a way that the unit normal vectors point outwards (that is, the orientation of $S$ is such that the unit normal vectors on the top face point towards positive $z$-directions). Determine the value of

\begin{equation*} I=\dblInt_S \vF \cdot\hn\, \dee{S} \end{equation*}

where $\vF$ is the vector field given by

\begin{equation*} \vF = \left(y \cos(y^2) + z - 1\,,\, \frac{z}{x+1}+1\,,\, xy e^{z^2}\right) \end{equation*}

39. (✳).

Find an upward pointing unit normal vector to the surface $z = xy$ at the point $(1, 1, 1)\text{.}$
Now consider the part of the surface $z = xy\text{,}$ which lies within the cylinder $x^2 + y^2 = 9$ and call it $S\text{.}$ Compute the upward flux of $\vF = (y, x, 3)$ through $S\text{.}$
Find the flux of $\vF = (y, x, 3)$ through the cylindrical surface $x^2 + y^2 = 9$ in between $z = xy$ and $z = 10\text{.}$ The orientation is outward, away from the z-axis.

40. (✳).

Find the divergence of the vector field $\vF = (x + \sin y, z + y, z^2)\text{.}$
Find the flux of $\vF$ through the upper hemisphere $x^2 + y^2 + z^2 = 25\text{,}$ $z \ge 0\text{,}$ oriented in the positive $z$-direction.
Specify an oriented closed surface $S\text{,}$ such that the flux $\dblInt_S\vF \cdot\hn\,\dee{S}$ is equal to $-9\text{.}$

41. (✳).

Evaluate the surface integrals. (Use any method you like.)

$\dblInt_S z^2\,\dee{S}\text{,}$ if $S$ is the part of the cone $x^2 + y^2 = 4z^2$ where $0 \le x \le y$ and $0 \le z \le 1\text{.}$
$\dblInt_S \vF \cdot \hn\,\dee{S}\text{,}$ if $\vF = z\hk$ and $S$ is the rectangle with vertices $(0, 2, 0)\text{,}$ $(0, 0, 4)\text{,}$ $(5, 2, 0)\text{,}$ $(5, 0, 4)\text{,}$ oriented so that the normal vector points upward.
$\dblInt_S \vF \cdot \hn\,\dee{S}\text{,}$ where $\vF = (y - z^2 )\hi + (z - x^2)\hj + z^2\hk$ and $S$ is the boundary surface of the box $0 \le x \le 1\text{,}$ $0 \le y \le 2\text{,}$ $0 \le z \le 3\text{,}$ with the normal vector pointing outward.

42. (✳).

Let $\sigma_1$ be the open surface given by $z = 1 - x^2 - y^2\text{,}$ $z \ge 0\text{.}$ Let $\sigma_2$ be the open surface given by $z = x^2 + y^2 - 1\text{,}$ $z \le 0\text{.}$ Let $\sigma_3$ be the planar surface given by $z = 0\text{,}$ $x^2 + y^2 \le 1\text{.}$ Let $\vF = [ a ( y^2 + z^2 ) + bxz ]\,\hi + [ c ( x^2 + z^2 ) + dyz ]\,\hj + x^2\,\hk$ where $a\text{,}$ $b\text{,}$ $c\text{,}$ and $d$ are constants.

Find the flux of $\vF$ upwards across $\sigma_1\text{.}$
Find all values of the constants $a\text{,}$ $b\text{,}$ $c\text{,}$ and $d$ so that the flux of $\vF$ outwards across the closed surface $\sigma_1 \cup \sigma_3$ is zero.
Find all values of the constants $a\text{,}$ $b\text{,}$ $c\text{,}$ and $d$ so that the flux of $\vF$ outwards across the closed surface $\sigma_1 \cup \sigma_2$ is zero.

43. (✳).

Let $S$ be the ellipsoid $x^2+2y^2+3z^2=16$ and $\hn$ its outward unit normal.

Find $\dblInt_S \vF\cdot\hn\,\dee{S}$ if $\vF(x,y,z)=\dfrac{(x,y,z)-(2,1,1)}{{\big[(x-2)^2+(y-1)^2+(z-1)^2\big]}^{3/2}}\text{.}$
Find $\dblInt_S \vG\cdot\hn\,\dee{S}$ if $\vG(x,y,z)=\dfrac{(x,y,z)-(3,2,2)}{{\big[(x-3)^2+(y-2)^2+(z-2)^2\big]}^{3/2}}\text{.}$

44. (✳).

Let $\Om\subset\bbbr^3$ be a smoothly bounded domain, with boundary $\partial\Om$ and outer unit normal $\hn\text{.}$ Prove that for any vector field $\vF$ which is continuously differentiable in $\Om\cup\partial\Om\text{,}$

\begin{equation*} \tripInt_\Om \vnabla\times\vF\,\dee{V} = -\dblInt_{\partial\Om} \vF\times\hn\,\dee{S} \end{equation*}

45. (✳).

Recall that if $S$ is a smooth closed surface with outer normal field $\hn\text{,}$ then for any smooth function $p(x,y,z)$ on $\bbbr^3\text{,}$ we have

\begin{equation*} \dblInt_S p\hn\ \dee{s}=\tripInt_E \nabla p\ \dee{V} \end{equation*}

where $E$ is the solid bounded by $S\text{.}$ Show that as a consequence, the total force exerted on the surface of a solid body contained in a gas of constant pressure is zero. (Recall that the pressure acts in the direction normal to the surface.)

46. (✳).

Let $\vF$ be a smooth 3-dimensional vector field such that the flux of $\vF$ out of the sphere $x^2+y^2+z^2=a^2$ is equal to $\pi(a^3+2a^4)$ for every $a \gt 0\text{.}$ Calculate $\vnabla\cdot \vF(0,0,0)\text{.}$

47. (✳).

Let $\ \vF= (x^2+y^2+z^2)\,\hi+(e^{x^2}+y^2)\,\hj+(3+x+z)\,\hk\ $ and let $S$ be the part of the surface $\ x^2+y^2+z^2=2az+3a^2\ $ having $\ z\ge 0\text{,}$ oriented with normal pointing away from the origin. Here $a \gt 0$ is a constant. Compute the flux of $\vF$ through $S\text{.}$

48. (✳).

Let $u=u(x,y,z)$ be a solution of Laplace’s Equation,

\begin{equation*} \frac{\partial^2 u}{\partial x^2} +\frac{\partial^2 u}{\partial y^2} +\frac{\partial^2 u}{\partial z^2} = 0, \end{equation*}

in $\bbbr^3\text{.}$ Let $\cR$ be a smooth solid in $\bbbr^3\text{.}$

Prove that the total flux of $\vF = \nabla u$ out through the boundary of $\cR$ is zero.
Prove that the total flux of $\vG = u\nabla u$ out through the boundary of $\cR$ equals
\begin{equation*} \tripInt_\cR \Big[\Big(\frac{\partial u}{\partial x}\Big)^2 + \Big(\frac{\partial u}{\partial y}\Big)^2 + \Big(\frac{\partial u}{\partial z}\Big)^2\Big]\,\dee{V} \end{equation*}

49. (✳).

Let $\cR$ be the part of the solid cylinder $x^2 + (y-1)^2 \le 1$ satisfying $0\le z \le y^2\text{;}$ let $\cS$ be the boundary of $\cR\text{.}$ Given $\vF = x^2\,\hi + 2y\,\hj - 2z\,\hk\text{,}$

Find the total flux of $\vF$ outward through $\cS\text{.}$
Find the total flux of $\vF$ outward through the (vertical) cylindrical sides of $\cS\text{.}$

Hint: $\ds \int_0^\pi \sin^{n}\theta\,\dee{\theta} = \frac{n-1}{ n}\int_0^\pi \sin^{n-2}\theta\,\dee{\theta}$ for $n=2,3,4,\ldots\text{.}$

50. (✳).

A smooth surface $\cS$ lies above the plane $z=0$ and has as its boundary the circle $x^2+y^2=4y$ in the plane $z=0\text{.}$ This circle also bounds a disk $D$ in that plane. The volume of the 3-dimensional region $R$ bounded by $\cS$ and $D$ is 10 cubic units. Find the flux of

\begin{equation*} \vF(x,y,z)=(x+x^2y)\hi+(y-xy^2)\hj+(z+2x+3y)\hk \end{equation*}

through $\cS$ in the direction outward from $R\text{.}$

51. (✳).

Let $E$ be the solid region between the plane $z = 4$ and the paraboloid $z = x^2 + y^2\text{.}$ Let

\begin{equation*} \vF = \Big(-\frac{1}{3}x^3 + e^{z^2}\Big)\hi + \Big(-\frac{1}{3}y^3 + x\root{3}\of{\tan z}\Big)\hj + 4z\hk \end{equation*}

Compute the flux of $\vF$ outward through the boundary of $E\text{.}$
Let $S$ be the part of the paraboloid $z = x^2 + y^2$ lying below the $z = 4$ plane oriented with a normal vector $\hN$ that has a positive $\hk$ component. Compute the flux of $\vF$ through $S\text{.}$

Prev Top Next