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CLP-4 Vector Calculus

Joel Feldman, Andrew Rechnitzer, Elyse Yeager

Section 3.4 Interpretation of Flux Integrals

We defined, in §3.3, two types of integrals over surfaces. We have seen, in §3.3.4, some applications that lead to integrals of the type \(\dblInt_S \rho\,\dee{S}\text{.}\) We now look at one application that leads to integrals of the type \(\dblInt_S \vF\cdot\hn\,\dee{S}\text{.}\) Recall that integrals of this type are called flux integrals. Imagine a fluid with
  • the density of the fluid (say in kilograms per cubic meter) at position \((x,y,z)\) and time \(t\) being \(\rho(x,y,z,t)\) and with
  • the velocity of the fluid (say in meters per second) at position \((x,y,z)\) and time \(t\) being \(\vv(x,y,z,t)\text{.}\)
We are going to determine the rate (say in kilograms per second) at which the fluid is flowing through a tiny piece \(\dee{S}\) of surface at \((x,y,z)\text{.}\) During a tiny time interval of length \(\dee{t}\) about time \(t\text{,}\) fluid near \(\dee{S}\) moves \(\vv(x,y,z,t)\dee{t}\text{.}\) The green line in the figure below is a side view of \(\dee{S}\) and \(\hn=\hn(x,y,z)\) is a unit normal vector to \(\dee{S}\text{.}\)
So during that tiny time interval
  • the red line moves to the green line and
  • the green line moves to the blue line so that
  • the fluid filling the dark grey region below the green line crosses through \(\dee{S}\) and moves to light grey region above the green line.
If we denote by \(\theta\) the angle between \(\hn\) and \(\vv\dee{t}\text{,}\)
  • the volume of fluid that crosses through \(\dee{S}\) during the time interval \(\dee{t}\) is the volume whose side view is the dark grey region below the green line. This region has base \(\dee{S}\) and height \(|\vv\dee{t}|\cos\theta\) and so has volume
    \begin{equation*} |\vv(x,y,z,t)\dee{t}|\cos\theta\ \dee{S} =\vv(x,y,z,t)\cdot\hn(x,y,z)\,\dee{t}\,\dee{S} \end{equation*}
    because \(\hn(x,y,z)\) has length one.
  • The mass of fluid that crosses \(\dee{S}\) during the time interval \(\dee{t}\) is then
    \begin{equation*} \rho(x,y,z,t)\vv(x,y,z,t)\cdot\hn(x,y,z)\,\dee{t}\,\dee{S} \end{equation*}
  • and the rate at which fluid is crossing through \(\dee{S}\) is
    \begin{equation*} \rho(x,y,z,t)\vv(x,y,z,t)\cdot\hn(x,y,z)\,\dee{S} \end{equation*}
Integrating \(\dee{S}\) over a surface \(S\text{,}\) we conclude that

Subsection 3.4.1 Examples of Flux Integrals

Example 3.4.2. Point Source.

In Example 2.1.2, we found that the vector field of a point source
 1 
You can imagine that a very small pipe pumps water to the origin.
(in three dimensions) that creates \(4\pi m\) liters per second is
\begin{equation*} \vv(x,y,z) = \frac{m}{r(x,y,z)^2}\,\hat\vr(x,y,z) \end{equation*}
where
\begin{equation*} r(x,y,z) = \sqrt{x^2+y^2+z^2}\qquad \hat\vr(x,y,z) = \frac{x\hi + y\hj + z\hk}{\sqrt{x^2+y^2+z^2}} \end{equation*}
We sketched it in Figure 2.1.3. We’ll now compute the flux of this vector field across a sphere centred on the origin. Suppose that the sphere has radius \(R\text{.}\)
Then the outward
 2 
It doesn’t really matter which unit normal we pick here. We just have to be clear which one we’re using. With the outward normal, the flux gives the rate at which fluid crosses the sphere in the outward direction. If we were to use the inward pointing normal, the flux would give the rate at which fluid crosses the sphere in the inward direction.
pointing normal at a point \((x,y,z)\) on the sphere is
\begin{equation*} \hn(x,y,z) = \hat\vr(x,y,z) = \frac{x\hi + y\hj + z\hk}{\sqrt{x^2+y^2+z^2}} = \frac{x\hi + y\hj + z\hk}{R} \end{equation*}
Note that \(\hat\vr(x,y,z)\cdot \hat\vr(x,y,z)=1\) and that, on the sphere, \(r(x,y,z)=R\text{.}\) So the flux of \(\vv\) outward through the sphere is
\begin{align*} \dblInt_S\vv\cdot\hn\ \dee{S} &= \dblInt_S \frac{m}{r(x,y,z)^2}\,\hat\vr(x,y,z) \cdot \hat\vr(x,y,z)\ \dee{S}\\ &= \dblInt_S \frac{m}{R^2}\ \dee{S} =\frac{m}{R^2} 4\pi R^2\\ &=4\pi m \end{align*}
This is the rate at which volume of fluid is exiting the sphere. In our derivation of the vector field we assumed that the fluid is incompressible, so it is also the rate at which the point source is creating fluid.

Example 3.4.3. Vortex.

In Figure 2.1.6, we sketched the vector field (in two dimensions)
\begin{equation*} \vv(x,y) = \Om\big(-y\hi +x\hj\big) \end{equation*}
We’ll now compute the flux of this vector field across a circle \(C\) centred on the origin. Suppose that the circle has radius \(R\text{.}\)
By definition, in two dimensions, the flux of a vector field across a curve \(C\) is \(\int_C\vv\cdot\hn\ \dee{s}\text{.}\)
This is the natural analog of the flux in three dimensions — the surface \(S\) has been replaced by the curve \(C\text{,}\) and the surface area \(\dee{S}\) of a tiny piece of \(S\) has been replaced by the arc length \(ds\) of a tiny piece of \(C\text{.}\)
The outward pointing unit normal at a point \((x,y)\) on our circle \(C\) is
\begin{equation*} \hn(x,y) = \frac{x\hi + y\hj}{\sqrt{x^2+y^2}} = \frac{x\hi + y\hj}{R} \end{equation*}
So
\begin{equation*} \vv(x,y)\cdot\hn(x,y) =\frac{\Om}{R}\big(-y\hi + x\hj\big)\cdot\big(x\hi + y\hj\big) =0 \end{equation*}
and the flux across \(C\) is
\begin{equation*} \int_C\vv\cdot\hn\ \dee{s}=0 \end{equation*}
This should not be a surprise — no fluid is crossing \(C\) at all. This is exactly what we would expect from looking at the arrows in Figure 2.1.6 or at the stream lines in Example 2.2.6.

Example 3.4.4.

Evaluate \(\ \dblInt_S\vF\cdot\hn\ dS\ \) where
\begin{equation*} \vF(x,y,z) = (x+y)\,\hi + (y+z)\,\hj + (x+z)\,\hk \end{equation*}
and \(S\) is the boundary of \(V=\Set{(x,y,z)}{0\le x^2+y^2\le 9,\ 0\le z\le 5}\text{,}\) and \(\hn\) is the outward normal
 3 
It is necessary that the problem specify, one way or another, whether \(\hn\) is the inward pointing normal or the outward pointing normal. Without this, the meaning of \(\ \dblInt_S\vF\cdot\hn\ dS\ \) is ambiguous. Think about where the orientation of the normal vector gets used in your solution.
to \(S\text{.}\)
Solution.
The volume \(V\) looks like a tin can of radius \(3\) and height \(5\text{.}\)
It is natural to decompose its surface \(S\) into three parts
\begin{align*} S_t &= \Set{(x,y,z)}{0\le x^2+y^2\le 9,\ z= 5} = \text{the top}\\ S_b &= \Set{(x,y,z)}{0\le x^2+y^2\le 9,\ z= 0} = \text{the bottom}\\ S_s &= \Set{(x,y,z)}{x^2+y^2 = 9,\ 0\le z\le 5} = \text{the side} \end{align*}
We’ll compute the flux through each of the three parts separately and then add them together.
The Top: On the top, the outward pointing normal to \(S\) is \(\hn=\hk\) and \(\dee{S} = \dee{x}\dee{y}\text{.}\) This is probably intuitively obvious. But if it isn’t, you can always derive it by parametrizing the top by \(\vr(x,y) = x\,\hi +y\,\hj + 5\,\hk\) with \(x^2+y^2\le 9\text{.}\) So the flux through the top is
\begin{align*} \dblInt_{S_t}\vF\cdot\hn\ \dee{S} &= \dblInt_{\atp{x^2+y^2\le 9}{z=5}} (x+z)\ \dee{x}\dee{y} = \dblInt_{x^2+y^2\le 9} (x+5)\ \dee{x}\dee{y} \end{align*}
The integral \(\dblInt_{x^2+y^2\le 9} x\ \dee{x}\dee{y}=0\) since \(x\) is odd and the domain of integration is symmetric about \(x=0\text{.}\) So
\begin{gather*} \dblInt_{S_t}\vF\cdot\hn\ \dee{S} = \dblInt_{x^2+y^2\le 9} 5\ \dee{x}\dee{y} =5\pi(3)^2 =45\pi \end{gather*}
The Bottom: On the bottom, the outward pointing normal to \(S\) is \(\hn=-\hk\) and \(\dee{S} = \dee{x}\dee{y}\text{.}\) So the flux through the bottom is
\begin{align*} \dblInt_{S_b}\vF\cdot\hn\ \dee{S} &= -\dblInt_{\atp{x^2+y^2\le 9}{z=0}} (x+z)\ \dee{x}\dee{y} = -\dblInt_{x^2+y^2\le 9} x\ \dee{x}\dee{y} =0 \end{align*}
again since \(x\) is odd and the domain of integration is symmetric about \(x=0\text{.}\)
The Side: We can parametrize the side by using cylindrical coordinates.
\begin{equation*} \vr(\theta,z) = \big(3\cos\theta\,,\,3\sin\theta\,,\,z\big)\qquad 0\le\theta \lt 2\pi,\ 0\le z\le 5 \end{equation*}
Then, using 3.3.1,
\begin{align*} \frac{\partial\vr}{\partial\theta} &=(-3\sin\theta\,,\,3\cos\theta\,,\,0)\\ \frac{\partial\vr}{\partial z} &=(0\,,\,0\,,\,1)\\ \hn\,\dee{S}&= \frac{\partial\vr}{\partial\theta} \times\frac{\partial\vr}{\partial z}\ \dee{\theta}\,\dee{z}\\ &=(3\cos\theta\,,\,3\sin\theta\,,\,0)\ \dee{\theta}\,\dee{z} \end{align*}
Note that \(\hn = (\cos\theta\,,\,\sin\theta\,,\,0)\) is outward pointing
 4 
To check, draw, in your head, a sketch of the top view of the can. “Top view” just means “ignore the \(z\)-coordinate”. The top view of the can is a circle of radius \(3\text{.}\) Then, at a generic point, \(\vr=(\cos\theta,\sin\theta)\text{,}\) on the can, draw the unit normal \(\hn = (\cos\theta\,,\,\sin\theta)\) with its tail at \(\vr\text{.}\) It is pointing away from the origin, just like \(\vr\) is. That is, \(\hn\) is pointing outward.
, as desired. Continuing,
\begin{align*} \vF\big(x(\theta,z),y(\theta,z),z(\theta,z)\big) &=3(\cos\theta\!+\!\sin\theta)\,\hi +(3\sin\theta\!+\!z)\,\hj+(3\cos\theta\!+\!z)\,\hk\\ \vF\cdot\hn\,\dee{S} &=\big\{9\cos^2\theta\!+\!3\sin\theta\cos\theta\!+\!9\sin^2\theta\!+\!3z\sin\theta\big\}\,\dee{\theta}\,\dee{z}\\ &=\big\{9 +\frac{3}{2}\,\sin(2\theta)+3z\sin\theta\big\}\ \dee{\theta}\,\dee{z} \end{align*}
So the flux through the side is
\begin{align*} \dblInt_{S_s}\vF\cdot\hn\ \dee{S} &=\int_0^{2\pi}\dee{\theta}\int_0^5\dee{z}\ \big\{9 +\frac{3}{2}\,\sin(2\theta)+3z\sin\theta\big\}\\ &=9\int_0^{2\pi}\dee{\theta}\int_0^5\dee{z} \quad\text{since }\int_0^{2\pi}\sin\theta\,\dee{\theta} =\int_0^{2\pi}\sin(2\theta)\,\dee{\theta} =0\\ &=9\times 2\pi\times 5 =90\pi \end{align*}
and the total flux is
\begin{align*} \dblInt_{S}\vF\cdot\hn\ \dee{S} &=\dblInt_{S_t}\vF\cdot\hn\ \dee{S} +\dblInt_{S_b}\vF\cdot\hn\ \dee{S} +\dblInt_{S_s}\vF\cdot\hn\ \dee{S}\\ &=45\pi+0+90\pi =135\pi \end{align*}

Example 3.4.5.

Evaluate \(\ \dblInt_S\vF\cdot\hn\ dS\ \) where \(\ \vF(x,y,z)=x^4\hi+2y^2\hj+z\hk,\ \) \(S\) is the half of the surface \(\ \frac{1}{4}x^2+\frac{1}{9}y^2+z^2=1\ \) with \(z\ge 0\text{,}\) and \(\hn\) is the upward pointing unit normal.
Solution 1.
We start by parametrizing the surface, which is half of an ellipsoid. By way of motivation for the parametrization, recall that spherical coordinates, with \(\rho=1\text{,}\) provide a natural way to parametrize the sphere \(x^2+y^2+z^2=1\text{.}\) Namely \(x=\cos\theta\sin\varphi\text{,}\) \(y=\sin\theta\sin\varphi\text{,}\) \(z= \cos\varphi\text{.}\) The reason that these spherical coordinates work is that the trig identity \(\cos^2\alpha+\sin^2\alpha=1\) implies
\begin{equation*} x^2+y^2 = \cos^2\theta\sin^2\varphi + \sin^2\theta\sin^2\varphi =\sin^2\varphi \end{equation*}
and then
\begin{equation*} \big(x^2+y^2\big) + z^2 = \sin^2\varphi +\cos^2\varphi = 1 \end{equation*}
The equation of our ellipsoid is
\begin{equation*} \Big(\frac{x}{2}\Big)^2 + \Big(\frac{y}{3}\Big)^2 + z^2 =1 \end{equation*}
so we can parametrize the ellipsoid by replacing \(x\) with \(\frac{x}{2}\) and \(y\) with \(\frac{y}{3}\) in our parametrization of the sphere. That is, we choose the parametrization
\begin{align*} x(\theta,\varphi)&=2\cos\theta\sin\varphi\\ y(\theta,\varphi)&=3\sin\theta\sin\varphi\\ z(\theta,\varphi)&=\cos\varphi \end{align*}
with \((\theta,\varphi)\) running over \(0\le\theta\le 2\pi,\ 0\le\varphi\le\pi/2\text{.}\) Note that
\begin{equation*} \frac{1}{4}x(\theta,\varphi)^2+\frac{1}{9}y(\theta,\varphi)^2 +z(\theta,\varphi)^2=1 \end{equation*}
as desired.
Then, using 3.3.1,
\begin{align*} \Big(\frac{\partial x}{\partial\theta}\,,\,\frac{\partial y}{\partial\theta} \,,\, \frac{\partial z}{\partial\theta}\Big) &=(-2\sin\theta\sin\varphi\,,\,3\cos\theta\sin\varphi\,,\,0)\\ \Big(\frac{\partial x}{\partial\varphi}\,,\,\frac{\partial y}{\partial\varphi}\,,\, \frac{\partial z}{\partial\varphi}\Big) &=(2\cos\theta\cos\varphi\,,\,3\sin\theta\cos\varphi\,,\,-\sin\varphi)\\ \hn\,\dee{S}&= -\Big(\frac{\partial x}{\partial\theta}\,,\,\frac{\partial y}{\partial\theta} \,,\, \frac{\partial z}{\partial\theta}\Big) \times\Big(\frac{\partial x}{\partial\varphi}\,,\,\frac{\partial y}{\partial\varphi} \,,\,\frac{\partial z}{\partial\varphi}\Big)\ \dee{\theta} \dee{\varphi}\\ &=-(-3\cos\theta\sin^2\varphi,-2\sin\theta\sin^2\varphi,-6\sin\varphi\cos\varphi)\dee{\theta} \dee{\varphi} \end{align*}
The extra minus sign in \(\hn\,\dee{S}\) was put there to make the \(z\) component of \(\hn\) positive. (The problem specified that \(\hn\) is to be upward unit normal.) As
\begin{align*} &\vF\big(x(\theta,\varphi)\,,\,y(\theta,\varphi)\,,\,z(\theta,\varphi)\big)\\ &\hskip0.5in=2^4\cos^4\theta\sin^4\varphi\ \hi +2\times 3^2\sin^2\theta\sin^2\varphi\ \hj+\cos\varphi\ \hk \end{align*}
we have
\begin{align*} \vF\cdot\hn\,\dee{S} &=\Big[3\times 2^4\cos^5\theta\sin^6\varphi\!+\!2\times 2 \times 3^2 \sin^3\theta\sin^4\varphi\!+\!6\sin\varphi\cos^2\varphi\Big]\,\dee{\theta}\dee{\varphi} \end{align*}
and the desired integral
\begin{align*} \dblInt_S\vF\cdot\hn\ \dee{S} &=\int_0^{\frac{\pi}{2}}\!\!\dee{\varphi}\int_0^{2\pi}\dee{\theta}\ \Big[3\times 2^4\cos^5\theta\sin^6\varphi+2\times 2\times 3^2 \sin^3\theta\sin^4\varphi\\ &\hskip3in+6\sin\varphi\cos^2\varphi\Big] \end{align*}
Since \(\ \int_0^{2\pi} \cos^m\theta\,\dee{\theta} =\int_0^{2\pi} \sin^m\theta\,\dee{\theta}=0\ \) for all odd
 5 
Look at the graphs of \(\cos^m\varphi\) and \(\sin^m\varphi\text{.}\)
natural numbers \(m\text{,}\)
\begin{align*} \dblInt_S \vF\cdot\hn\, \dee{S} &=\int_0^{\pi/2}\hskip-8pt \dee{\varphi}\int_0^{2\pi}\hskip-6pt\dee{\theta}\ 6\sin\varphi\cos^2\varphi =12\pi\int_0^{\pi/2}\hskip-8pt \dee{\varphi}\ \sin\varphi\cos^2\varphi\\ &=12\pi\Big[-\frac{1}{3}\cos^3\varphi\Big]_0^{\pi/2} =4\pi \end{align*}
The integral was evaluated by guessing (and checking) that \(-\frac{1}{3}\cos^3\varphi\) is an antiderivative of \(\sin\varphi\cos^2\varphi\text{.}\) It can also be done by substituting \(u=\cos\varphi\text{,}\) \(\dee{u}=-\sin\varphi\,\dee{\varphi}\text{.}\)
Solution 2.
This time we’ll parametrize the half-ellipsoid using a variant of cylindrical coordinates.
\begin{align*} x(r,\theta)&=2r\cos\theta\\ y(r,\theta)&=3r\sin\theta\\ z(r,\theta)&=\sqrt{1-r^2} \end{align*}
with \((r,\theta)\) running over \(0\le\theta\le 2\pi,\ 0\le r\le1\text{.}\) Because we built the factors of \(2\) and \(3\) into \(x(r,\theta)\) and \(y(r,\theta)\text{,}\) we have
\begin{align*} &\frac{x(r,\theta)^2}{4} + \frac{y(r,\theta)^2}{9} =r^2\cos^2\theta+r^2\sin^2\theta = r^2\\ \implies &\frac{x(r,\theta)^2}{4} + \frac{y(r,\theta)^2}{9} +z(r,\theta)^2 = r^2 + \left(\sqrt{1-r^2}\right)^2=1 \end{align*}
as desired. Further \(z(r,\theta)\ge 0\) by our choice of square root in the definition of \(z(r,\theta)\text{.}\)
So, using 3.3.1,
\begin{align*} \Big(\frac{\partial x}{\partial \theta},\frac{\partial y}{\partial \theta}, \frac{\partial z}{\partial \theta}\Big) &=(-2r\sin\theta,3r\cos\theta,0)\\ \Big(\frac{\partial x}{\partial r},\frac{\partial y}{\partial r}, \frac{\partial z}{\partial r}\Big) &=\Big(2\cos\theta,3\sin\theta,-\frac{r}{\sqrt{1-r^2}}\Big)\\ \hn \dee{S}&= -\Big(\frac{\partial x}{\partial\theta},\frac{\partial y}{\partial\theta}, \frac{\partial z}{\partial\theta}\Big) \times\Big(\frac{\partial x}{\partial r},\frac{\partial y}{\partial r}, \frac{\partial z}{\partial r}\Big)dr\,\dee{\theta}\\ &=-\Big(-\frac{3r^2\cos\theta}{\sqrt{1-r^2}}, -\frac{2r^2\sin\theta}{\sqrt{1-r^2}},-6r\Big)\dee{r}\,\dee{\theta} \end{align*}
Once again, the extra minus sign in \(\hn \dee{S}\) was put there to make the \(z\) component of \(\hn\) positive. Continuing,
\begin{align*} \vF\big(x(r,\theta)\,,\,y(r,\theta)\,,\,z(r,\theta)\big) &=2^4r^4\cos^4\theta\,\hi+2\times 3^2r^2\sin^2\theta\,\hj+\sqrt{1-r^2}\,\hk\\ \vF\cdot\hn\, \dee{S}&=\Big[3\times2^4\frac{r^6}{\sqrt{1-r^2}}\cos^5\theta +2^2 3^2\frac{r^4}{\sqrt{1-r^2}}\sin^3\theta\\ &\hskip2.3in+6r\sqrt{1-r^2}\Big]\,\dee{r}\,\dee{\theta} \end{align*}
Again using that \(\ \int_0^{2\pi} \cos^m\theta\,\dee{\theta} =\int_0^{2\pi} \sin^m\theta\,\dee{\theta}=0\ \) for all odd natural numbers \(m\text{,}\)
\begin{align*} \int_S \vF\cdot\hn\, \dee{S} &=\int_0^1 dr\int_0^{2\pi}\hskip-6pt\dee{\theta}\ 6r\sqrt{1-r^2}\\ &=12\pi\int_0^1 dr\ r\sqrt{1-r^2} =12\pi\Big[-\frac{1}{3}{(1-r^2)}^{3/2}\Big]_0^1\\ &=4\pi \end{align*}
The integral was evaluated by guessing (and checking) that \(-\frac{1}{3}{(1-r^2)}^{3/2}\) is an antiderivative of \(r\sqrt{1-r^2}\text{.}\) It can also be done by substituting \(u=1-r^2\text{,}\) \(\dee{u}=-2r\,\dee{r}\text{.}\)
Solution 3.
The surface is of the form \(G(x,y,z)=0\) with \(G(x,y,z)=\frac{1}{4}x^2+\frac{1}{9}y^2+z^2-1\text{.}\) Hence, using 3.3.3,
\begin{align*} \hn \dee{S}&=\frac{\vnabla G}{\vnabla G\cdot\hk}\dee{x}\,\dee{y} =\frac{\frac{x}{2}\hi+\frac{2y}{9}\hj+2z\hk}{2z}\dee{x}\,\dee{y}\\ &=\Big(\frac{x}{4z}\hi+\frac{y}{9z}\hj+\hk\Big)\dee{x}\,\dee{y}\\ \implies \vF\cdot\hn\,\dee{S} &=\Big(\frac{x^5}{4z}+\frac{2y^3}{9z}+z\Big)\dee{x}\,\dee{y} \end{align*}
It is true that \(\hn\dee{S}\text{,}\) and consequently \(\vF\cdot\hn\,\dee{S}\) become infinite
 6 
That’s because the ellipsoid is becoming vertical as \(z\rightarrow 0\text{,}\) so that \(x\) and \(y\) are not really good parameters there.
as \(z\rightarrow 0\text{.}\) So we should really treat the integral as an improper integral, first integrating over \(z\ge \veps\) and then taking the limit \(\veps\rightarrow 0^+\text{.}\) But, as we shall see, the singularity is harmless. So it is standard to gloss over this point. On \(S\text{,}\) \(z=z(x,y)=\sqrt{1-\frac{x^2}{4}-\frac{y^2}{9}}\) and \(\frac{x^2}{4}+\frac{y^2}{9}\le 1\text{,}\) so
\begin{align*} \int_S \vF\cdot\hn\, \dee{S} &=\dblInt_{\frac{x^2}{4}+\frac{y^2}{9}\le 1} \Big(\frac{x^5}{4z(x,y)}+\frac{2y^3}{9z(x,y)}+z(x,y)\Big)\ \dee{x}\,\dee{y} \end{align*}
Both \(\frac{x^5}{4z(x,y)}\) and \(\frac{2y^3}{9z(x,y)}\) are odd under \(x\rightarrow-x,\ y\rightarrow -y\) and the domain of integration is even under \(x\rightarrow-x,\ y\rightarrow -y\text{,}\) so their integrals are zero and
\begin{align*} \int_S \vF\cdot\hn\, \dee{S} &=\dblInt_{\frac{x^2}{4}+\frac{y^2}{9}\le 1}z(x,y)\ \dee{x}\,\dee{y}\\ &=\dblInt_{\frac{x^2}{4}+\frac{y^2}{9}\le 1} \sqrt{1-\frac{x^2}{4}-\frac{y^2}{9}}\ \dee{x}\,\dee{y} \end{align*}
To evaluate this integral, first make the change of variables
 7 
The reader interested in general changes of variables in multidimensional integrals should look up “Jacobian determinant”.
\(x=2X\text{,}\) \(\dee{x}=2\dee{X}\text{,}\) \(y=3Y\text{,}\) \(\dee{y}=3\dee{Y}\) to give
\begin{equation*} \int_S \vF\cdot\hn\, \dee{S} =\dblInt_{X^2+Y^2\le 1} \sqrt{1-X^2-Y^2}\ 6\,\dee{X}\,\dee{Y} \end{equation*}
Then switch to polar coordinates, \(X=r\cos\theta\text{,}\) \(Y=r\sin\theta\text{,}\) \(\dee{X}\dee{Y} = r\,\dee{r}\dee{\theta}\) to give
\begin{align*} \int_S \vF\cdot\hn\, \dee{S} &=\int_0^1 dr\int_0^{2\pi}\hskip-6pt\dee{\theta}\ 6r\sqrt{1-r^2} =12\pi\int_0^1 dr\ r\sqrt{1-r^2}\\ &=12\pi\Big[-\frac{1}{3}{(1-r^2)}^{3/2}\Big]_0^1 =4\pi \end{align*}
Solution 4.
The surface is of the form \(z=f(x,y)\) with \(f(x,y)=\sqrt{1-\frac{x^2}{4}-\frac{y^2}{9}}\text{.}\) Hence, using 3.3.2,
\begin{align*} \hn \dee{S} &=\Big[-\frac{\partial f}{\partial x}\hi\!-\!\frac{\partial f}{\partial y}\hj \!+\!\hk\Big]\,\dee{x}\,\dee{y} =\left[\frac{\frac{x}{4}\hi+\frac{y}{9}\hj} {\sqrt{1-\frac{x^2}{4}-\frac{y^2}{9}}}+\hk\right]\dee{x}\,\dee{y}\cr \implies\! \vF\cdot\hn\, \dee{S}&=\left[\frac{\frac{x^5}{4}+\frac{2y^3}{9}} {\sqrt{1-\frac{x^2}{4}-\frac{y^2}{9}}} +\sqrt{1-\frac{x^2}{4}-\frac{y^2}{9}}\right]\dee{x}\,\dee{y} \end{align*}
Note that our unit normal is upward pointing, as required. As in Solution 3, by the oddness of the \(x^5\) and \(y^3\) terms in the integrand,
\begin{align*} \int_S \vF\cdot\hn\, \dee{S} &=\dblInt_{\frac{x^2}{4}+\frac{y^2}{9}\le 1} \left[\frac{\frac{x^5}{4}+\frac{2y^3}{9}}{\sqrt{\ \cdots\ }} +\sqrt{1-\frac{x^2}{4}-\frac{y^2}{9}}\right]\dee{x}\,\dee{y}\\ &=\dblInt_{\frac{x^2}{4}+\frac{y^2}{9}\le 1} \sqrt{1-\frac{x^2}{4}-\frac{y^2}{9}}\ \dee{x}\,\dee{y} \end{align*}
Now continue as in Solution 3.
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