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CLP-3 Multivariable Calculus

Joel Feldman, Andrew Rechnitzer, Elyse Yeager

Section 3.3 Applications of Double Integrals

Double integrals are useful for more than just computing areas and volumes. Here are a few other applications that lead to double integrals.

Subsection 3.3.1 Averages

In Section 2.2 of the CLP-2 text, we defined the average value of a function of one variable. We'll now extend that discussion to functions of two variables. First, we recall the definition of the average of a finite set of numbers.

Definition 3.3.1.

The average (mean) of a set of \(n\) numbers \(f_1\text{,}\) \(f_2\text{,}\) \(\cdots\text{,}\) \(f_n\) is
\begin{gather*} \bar f = \llt f\rgt =\frac{f_1+f_2+\cdots+f_n}{n} \end{gather*}
The notations \(\bar f\) and \(\llt f\rgt\) are both commonly used to represent the average.
Now suppose that we want to take the average of a function \(f(x,y)\) with \((x,y)\) running continuously over some region \(\cR\) in the \(xy\)-plane. A natural approach to defining what we mean by the average value of \(f\) over \(\cR\) is to
  • First fix any natural number \(n\text{.}\)
  • Subdivide the region \(\cR\) into tiny (approximate) squares each of width \(\De x=\frac{1}{n}\) and height \(\De y =\frac{1}{n}\text{.}\) This can be done by, for example, subdividing vertical strips into tiny squares, like in Example 3.1.11.
  • Name the squares (in any fixed order) \(R_1\text{,}\) \(R_2\text{,}\) \(\cdots\text{,}\) \(R_N\text{,}\) where \(N\) is the total number of squares.
  • Select, for each \(1\le i\le N\text{,}\) one point in square number \(i\) and call it \((x_i^*,y_i^*)\text{.}\) So \((x_i^*,y_i^*)\in R_i\text{.}\)
  • The average value of \(f\) at the selected points is
    \begin{gather*} \frac{1}{N}\sum_{i=1}^N f(x_i^*,y_i^*) =\frac{\sum_{i=1}^N f(x_i^*,y_i^*)}{{\sum_{i=1}^N 1 }} =\frac{\sum_{i=1}^N f(x_i^*,y_i^*)\, \De x\,\De y}{{\sum_{i=1}^N \De x\,\De y}} \end{gather*}
    We have transformed the average into a ratio of Riemann sums.
Once we have the Riemann sums it is clear what to do next. Taking the limit \(n\rightarrow\infty\text{,}\) we get exactly \(\frac{\dblInt_\cR f(x,y)\,\dee{x}\,\dee{y}} {\dblInt_\cR \dee{x}\,\dee{y}}\text{.}\) That's why we define

Definition 3.3.2.

Let \(f(x,y)\) be an integrable function defined on region \(\cR\) in the \(xy\)-plane. The average value of \(f\) on \(\cR\) is
\begin{gather*} \bar f=\llt f\rgt =\frac{\displaystyle\dblInt_\cR f(x,y)\,\dee{x}\,\dee{y}} {\displaystyle\dblInt_\cR \dee{x}\,\dee{y}} \end{gather*}
Let \(a \gt 0\text{.}\) A mountain, call it Half Dome 1 , has height \(z(x,y)=\sqrt{a^2-x^2-y^2}\) above each point \((x,y)\) in the base region \(\cR=\Set{(x,y)}{x^2+y^2\le a^2, x\le 0}\text{.}\) Find its average height.
Solution.
By Definition 3.3.2 the average height is
\begin{equation*} \bar z =\frac{\dblInt_\cR z(x,y)\,\dee{x}\,\dee{y}} {\dblInt_\cR \dee{x}\,\dee{y}} =\frac{\dblInt_\cR \sqrt{a^2-x^2-y^2}\,\dee{x}\,\dee{y}} {\dblInt_\cR \dee{x}\,\dee{y}} \end{equation*}
The integrals in both the numerator and denominator are easily evaluated by interpreting them geometrically.
  • The numerator \(\dblInt_\cR z(x,y)\,\dee{x}\,\dee{y} =\dblInt_\cR \sqrt{a^2-x^2-y^2}\,\dee{x}\,\dee{y}\) can be interpreted as the volume of
    \begin{align*} &\Big\{\ (x,y,z)\ \Big|\ x^2+y^2\le a^2,\ x\le 0,\ 0\le z\le \sqrt{a^2-x^2-y^2}\ \Big\}\\ &=\Set{(x,y,z)}{x^2+y^2+z^2\le a^2,\ x\le 0,\ z\ge 0} \end{align*}
    which is one quarter of the interior of a sphere of radius \(a\text{.}\) So the numerator is \(\frac{1}{3}\pi a^3\text{.}\)
  • The denominator \(\dblInt_\cR \dee{x}\,\dee{y}\) is the area of one half of a circular disk of radius \(a\text{.}\) So the denominator is \(\frac{1}{2} \pi a^2\text{.}\)
All together, the average height is
\begin{equation*} \bar z = \frac{\frac{1}{3}\pi a^3}{\frac{1}{2} \pi a^2} =\frac{2}{3}\, a \end{equation*}
Notice this this number is bigger than zero and less than the maximum height, which is \(a\text{.}\) That makes sense.
This last example was relatively easy because we could reinterpret the integrals as geometric quantities. For practice, let's go back and evaluate the numerator \(\dblInt_\cR \sqrt{a^2-x^2-y^2}\,\dee{x}\,\dee{y}\) of Example 3.3.3 as an iterated integral.
Here is a sketch of the top view of the base region \(\cR\text{.}\)
Using the slicing in the figure
\begin{align*} \dblInt_\cR \sqrt{a^2-x^2-y^2}\,\dee{x}\,\dee{y} &= \int_{-a}^a \dee{y} \int_{-\sqrt{a^2-y^2}}^0\dee{x}\ \sqrt{a^2-x^2-y^2} \end{align*}
Note that, in the inside integral \(\displaystyle \int_{-\sqrt{a^2-y^2}}^0\dee{x}\ \sqrt{a^2-x^2-y^2}\text{,}\) the variable \(y\) is treated as a constant, so that the integrand \(\sqrt{a^2-y^2-x^2} = \sqrt{C^2-x^2}\) with \(C\) being the constant \(\sqrt{a^2-y^2}\text{.}\) The standard protocol for evaluating this integral uses the trigonometric substitution
\begin{align*} x &= C\sin\theta\qquad\text{with } -\frac{\pi}{2}\le\theta\le\frac{\pi}{2}\\ \dee{x} &= C\cos\theta\,\dee{\theta} \end{align*}
Trigonometric substitution was discussed in detail in Section 1.9 in the CLP-2 text. Since
\begin{align*} x&=0 & \implies C\sin\theta&=0 &\implies \theta&=0\\ x&=-\sqrt{a^2-y^2}=-C &\implies C\sin\theta&=-C &\implies \theta&=-\frac{\pi}{2} \end{align*}
and
\begin{align*} \sqrt{a^2-x^2-y^2}&=\sqrt{C^2-C^2\sin^2\theta}=C\cos\theta \end{align*}
the inner integral
\begin{align*} &\int_{-\sqrt{a^2-y^2}}^0\dee{x}\ \sqrt{a^2-x^2-y^2} =\int_{-\pi/2}^0 C^2\cos^2\theta\ \dee{\theta}\\ &\hskip1in=C^2\int_{-\pi/2}^0\frac{1+\cos(2\theta)}{2}\ \dee{\theta} =C^2\left[\frac{\theta+\frac{\sin(2\theta)}{2}}{2}\right]_{-\pi/2}^0\\ &\hskip1in=\frac{\pi C^2}{4} =\frac{\pi}{4}\big(a^2-y^2\big) \end{align*}
and the full integral
\begin{align*} \dblInt_\cR \sqrt{a^2-x^2-y^2}\,\dee{x}\,\dee{y} &=\frac{\pi}{4} \int_{-a}^a \big(a^2-y^2\big)\ \dee{y} =\frac{\pi}{2} \int_0^a \big(a^2-y^2\big)\ \dee{y}\\ &=\frac{\pi}{2}\left[a^3-\frac{a^3}{3}\right]\\ &=\frac{1}{3}\pi a^3 \end{align*}
just as we saw in Example 3.3.3.

Remark 3.3.5.

We remark that there is an efficient, sneaky, way to evaluate definite integrals like \(\int_{-\pi/2}^0 \cos^2\theta\ \dee{\theta}\text{.}\) Looking at the figures
we see that
\begin{equation*} \int_{-\pi/2}^0 \cos^2\theta\ \dee{\theta} =\int_{-\pi/2}^0 \sin^2\theta\ \dee{\theta} \end{equation*}
Thus
\begin{align*} \int_{-\pi/2}^0 \cos^2\theta\ \dee{\theta} &=\int_{-\pi/2}^0 \sin^2\theta\ \dee{\theta} =\int_{-\pi/2}^0 \frac{1}{2}\big[\sin^2\theta+\cos^2\theta\big]\,\dee{\theta} =\frac{1}{2}\int_{-\pi/2}^0 \dee{\theta}\\ &=\frac{\pi}{4} \end{align*}
It is not at all unusual to want to find the average value of some function \(f(x,y)\) with \((x,y)\) running over some region \(\cR\text{,}\) but to also want some \((x,y)\)'s to play a greater role in determining the average than other \((x,y)\)'s. One common way to do so is to create a “weight function” \(w(x,y) \gt 0\) with \(\frac{w(x_1,y_1)}{w(x_2,y_2)}\) giving the relative importance of \((x_1,y_1)\) and \((x_2,y_2)\text{.}\) That is, \((x_1,y_1)\) is \(\frac{w(x_1,y_1)}{w(x_2,y_2)}\) times as important as \((x_2,y_2)\text{.}\) This leads to the definition

Definition 3.3.6.

\begin{gather*} \frac{\dblInt_\cR f(x,y)\,w(x,y)\,\dee{x}\,\dee{y}} {\dblInt_\cR w(x,y)\,\dee{x}\,\dee{y}} \end{gather*}
is called the weighted average of \(f\) over \(\cR\) with weight \(w(x,y)\text{.}\)
Note that if \(f(x,y)=F\text{,}\) a constant, then the weighted average of \(f\) is just \(F\text{,}\) just as you would want.

Subsection 3.3.2 Centre of Mass

One important example of a weighted average is the centre of mass. If you support a body at its centre of mass (in a uniform gravitational field) it balances perfectly. That's the definition of the centre of mass of the body. In Section 2.3 of the CLP-2 text, we found that the centre of mass of a body that consists of mass distributed continuously along a straight line, with mass density \(\rho(x)\)kg/m and with \(x\) running from \(a\) to \(b\text{,}\) is at
\begin{equation*} \bar x = \frac{\int_a^b x\ \rho(x)\,\dee{x}}{\int_a^b \rho(x)\,\dee{x}} \end{equation*}
That is, the centre of mass is at the average of the \(x\)-coordinate weighted by the mass density.
In two dimensions, the centre of mass of a plate that covers the region \(\cR\) in the \(xy\)-plane and that has mass density \(\rho(x,y)\) is the point \((\bar x, \bar y)\) where
If the mass density is a constant, the centre of mass is also called the centroid, and is the geometric centre of \(\cR\text{.}\) In this case
In Section 2.3 of the CLP-2 text, we did not have access to multivariable integrals, so we used some physical intuition to derive that the centroid of a body that fills the region
\begin{equation*} \cR=\big\{\ (x,y)\ \big|\ a\le x\le b,\ B(x)\le y\le T(x)\ \big\} \end{equation*}
in the \(xy\)-plane is \((\bar x,\bar y)\) where
\begin{align*} \bar x &= \frac{\int_a^b x [T(x)-B(x)]\,\dee{x}}{A}\\ \bar y &= \frac{\int_a^b\, [T(x)^2-B(x)^2]\,\dee{x}}{2A} \end{align*}
and \(A=\int_a^b [T(x)-B(x)]\,\dee{x}\) is the area of \(\cR\text{.}\) Now that we do have access to multivariable integrals, we can derive these formulae directly from 3.3.8. Using vertical slices, as in this figure,
we see that the area of \(\cR\) is
\begin{gather*} A= \dblInt_\cR\dee{x}\,\dee{y} =\int_a^b\dee{x}\int_{B(x)}^{T(x)}\dee{y} =\int_a^b\dee{x}\ \big[T(x)-B(x)\big] \end{gather*}
and that 3.3.8 gives
\begin{align*} \bar x&= \frac{1}{A} \dblInt_\cR x\ \dee{x}\,\dee{y} = \frac{1}{A} \int_a^b\dee{x}\int_{B(x)}^{T(x)}\dee{y}\ x =\frac{1}{A}\int_a^b\dee{x}\ x\big[T(x)-B(x)\big]\\ \bar y&= \frac{1}{A} \dblInt_\cR y\ \dee{x}\,\dee{y} = \frac{1}{A} \int_a^b\dee{x}\int_{B(x)}^{T(x)}\dee{y}\ y =\frac{1}{A}\int_a^b\dee{x}\ \left[\frac{T(x)^2}{2}-\frac{B(x)^2}{2}\right] \end{align*}
just as desired.
We'll start with a simple mechanical example.
In Example 2.3.4 of the CLP-2 text, we found the centroid of the quarter circular disk
\begin{equation*} D=\Set{(x,y)}{x\ge 0,\ y\ge 0,\ x^2+y^2\le r^2} \end{equation*}
by using the formulae of the last example. We'll now find it again using 3.3.8.
Since the area of \(D\) is \(\frac{1}{4}\pi r^2\text{,}\) we have
\begin{align*} \bar x =\frac{\dblInt_D x\ \dee{x}\,\dee{y}} {\frac{1}{4}\pi r^2} \qquad \bar y &=\frac{\dblInt_D y\ \dee{x}\,\dee{y}} {\frac{1}{4}\pi r^2} \end{align*}
We'll evaluate \(\dblInt_D x\ \dee{x}\,\dee{y}\) by using horizontal slices, as in the figure on the left below.
Looking at that figure, we see that
  • \(y\) runs from \(0\) to \(r\) and
  • for each \(y\) in that range, \(x\) runs from \(0\) to \(\sqrt{r^2-y^2}\text{.}\)
So
\begin{align*} \dblInt_D x\ \dee{x}\,\dee{y} &= \int_0^r\dee{y}\int_0^{\sqrt{r^2-y^2}}\dee{x} \ x = \int_0^r\dee{y}\ \left[\frac{x^2}{2}\right]_0^{\sqrt{r^2-y^2}}\\ &=\frac{1}{2} \int_0^r\dee{y}\ \big[r^2-y^2\big] =\frac{1}{2}\left[r^3-\frac{r^3}{3}\right]\\ &= \frac{r^3}{3} \end{align*}
and
\begin{equation*} \bar x = \frac{4}{\pi r^2}\left[\frac{r^3}{3}\right] =\frac{4r}{3\pi} \end{equation*}
This is the same answer as we got in Example 2.3.4 of the CLP-2 text. But because we were able to use horizontal slices, the integral in this example was a little easier to evaluate than the integral in CLP-2. Had we used vertical slices, we would have ended up with exactly the integral of CLP-2.
By symmetry, we should have \(\bar y=\bar x\text{.}\) We'll check that by evaluating \(\dblInt_D y\ \dee{x}\,\dee{y}\) by using vertical slices slices, as in the figure on the right above. From that figure, we see that
  • \(x\) runs from \(0\) to \(r\) and
  • for each \(x\) in that range, \(y\) runs from \(0\) to \(\sqrt{r^2-x^2}\text{.}\)
So
\begin{align*} \dblInt_D y\ \dee{x}\,\dee{y} &= \int_0^r\dee{x}\int_0^{\sqrt{r^2-x^2}}\dee{y} \ y = \frac{1}{2}\int_0^r\dee{x}\ \big[r^2-x^2\big] \end{align*}
This is exactly the integral \(\frac{1}{2} \int_0^r\dee{y}\ \big[r^2-y^2\big]\) that we evaluated above, with \(y\) renamed to \(x\text{.}\) So \(\dblInt_D y\ \dee{x}\,\dee{y}=\frac{r^3}{3}\) too and
\begin{equation*} \bar y = \frac{4}{\pi r^2}\left[\frac{r^3}{3}\right] =\frac{4r}{3\pi} =\bar x \end{equation*}
as expected.
Find the centroid of the region that is inside the circle \(r=4\cos\theta\) and to the left of the line \(x=1\text{.}\)
Solution.
Recall that we saw in Example 3.2.14 that \(r=4\cos\theta\) was indeed a circle, and in fact is the circle \((x-2)^2+y^2=4\text{.}\) Here is a sketch of that circle and of the region of interest, \(\cR\text{.}\)
From the sketch, we see that \(\cR\) is symmetric about the \(x\)-axis. So we expect that its centroid, \((\bar x,\bar y)\text{,}\) has \(\bar y=0\text{.}\) To see this from the integral definition, note that the integral \(\dblInt_\cR y\ \dee{x}\,\dee{y}\)
  • has domain of integration, namely \(\cR\text{,}\) invariant under \(y\rightarrow -y\) (i.e. under reflection in the \(x\)-axis), and
  • has integrand, namely \(y\text{,}\) that is odd under \(y\rightarrow -y\text{.}\)
So \(\dblInt_\cR y\ \dee{x}\,\dee{y}=0\) and consequently \(\bar y=0\text{.}\)
We now just have to find \(\bar x\text{:}\)
\begin{gather*} \bar x =\frac{\dblInt_\cR x\ \dee{x}\,\dee{y}} {\dblInt_\cR \dee{x}\,\dee{y}} \end{gather*}
We have already found, in Example 3.2.14, that
\begin{gather*} \dblInt_\cR \dee{x}\,\dee{y} = \frac{4\pi}{3}-\sqrt{3} \end{gather*}
So we just have to compute \(\dblInt_\cR x\ \dee{x}\,\dee{y}\text{.}\) Using \(\cR_1\) to denote the top half of \(\cR\text{,}\) and using polar coordinates, like we did in Example 3.2.14,
\begin{align*} \dblInt_{\cR_1} x\ \dee{x}\,\dee{y} &=\int_0^{\pi/3}\dee{\theta}\int_0^{1/\cos\theta}\dee{r}\ r \overbrace{(r\cos\theta)}^{x} +\int_{\pi/3}^{\pi/2}\dee{\theta}\int_0^{4\cos\theta}\dee{r}\ r \overbrace{(r\cos\theta)}^{x}\\ &=\int_0^{\pi/3}\dee{\theta}\ \cos\theta\int_0^{1/\cos\theta}\dee{r}\ r^2 +\int_{\pi/3}^{\pi/2}\dee{\theta}\ \cos\theta \int_0^{4\cos\theta}\dee{r}\ r^2\\ &=\int_0^{\pi/3}\dee{\theta}\ \frac{\sec^2\theta}{3} +\int_{\pi/3}^{\pi/2}\dee{\theta}\ \frac{64}{3}\cos^4\theta \end{align*}
The first integral is easy, provided we remember that \(\tan\theta\) is an antiderivative for \(\sec^2\theta\text{.}\) For the second integral, we'll need the double angle formula \(\cos^2\theta=\frac{1+\cos(2\theta)}{2}\text{:}\)
\begin{align*} \cos^4\theta =\big(\cos^2\theta\big)^2 &=\left[\frac{1+\cos(2\theta)}{2}\right]^2 =\frac{1}{4}\left[1+2\cos(2\theta)+\cos^2(2\theta)\right]\\ &=\frac{1}{4}\left[1+2\cos(2\theta)+\frac{1+\cos(4\theta)}{2}\right]\\ &=\frac{3}{8} +\frac{\cos(2\theta)}{2} + \frac{\cos(4\theta)}{8} \end{align*}
so
\begin{align*} \dblInt_{\cR_1} x\ \dee{x}\,\dee{y} &=\frac{1}{3}\tan\theta\Big|_0^{\pi/3} +\frac{64}{3}\left[\frac{3\theta}{8} +\frac{\sin(2\theta)}{4}+\frac{\sin(4\theta)}{32}\right]_{\pi/3}^{\pi/2}\\ &=\frac{1}{3}\times\sqrt{3} +\frac{64}{3}\left[\frac{3}{8}\times\frac{\pi}{6} -\frac{\sqrt{3}}{4\times 2} +\frac{\sqrt{3}}{32\times 2}\right]\\ &=\frac{4\pi}{3}-2\sqrt{3} \end{align*}
The integral we want, namely \(\dblInt_\cR x\ \dee{x}\,\dee{y}\text{,}\)
  • has domain of integration, namely \(\cR\text{,}\) invariant under \(y\rightarrow -y\) (i.e. under reflection in the \(x\)-axis), and
  • has integrand, namely \(x\text{,}\) that is even under \(y\rightarrow -y\text{.}\)
So \(\dblInt_\cR x\ \dee{x}\,\dee{y} =2\dblInt_{\cR_1} x\ \dee{x}\,\dee{y}\) and, all together,
\begin{align*} \bar x &= \frac{2\left(\frac{4\pi}{3}-2\sqrt{3}\right)}{\frac{4\pi}{3}-\sqrt{3}} = \frac{\frac{8\pi}{3}-4\sqrt{3}}{\frac{4\pi}{3}-\sqrt{3}} \approx 0.59 \end{align*}
As a check, note that \(0\le x\le 1\) on \(\cR\) and more of \(\cR\) is closer to \(x=1\) than to \(x=0\text{.}\) So it makes sense that \(\bar x\) is between \(\frac{1}{2}\) and \(1\text{.}\)
Evaluate \(\displaystyle\int_0^2\int_{-\sqrt{2x-x^2}}^{\sqrt{2x-x^2}} \big(2x+3y\big)\dee{y}\,\dee{x}\text{.}\)
Solution.
This is another integral that can be evaluated without using any calculus at all. This time by relating it to a centre of mass. By 3.3.8,
\begin{align*} \dblInt_\cR x\ \dee{x}\,\dee{y} &=\bar x\ \text{Area}(\cR)\\ \dblInt_\cR y\ \dee{x}\,\dee{y} &=\bar y\ \text{Area}(\cR) \end{align*}
so that we can easily evaluate \(\dblInt_\cR x\ \dee{x}\,\dee{y}\) and \(\dblInt_\cR y\ \dee{x}\,\dee{y}\) provided \(\cR\) is sufficiently simple and symmetric that we can easily determine its area and its centroid.
That is the case for the integral in this example. Rewrite
\begin{align*} \int_0^2\int_{-\sqrt{2x-x^2}}^{\sqrt{2x-x^2}} \big(2x+3y\big)\dee{y}\,\dee{x} &=2\int_0^2\dee{x}\left[\int_{-\sqrt{2x-x^2}}^{\sqrt{2x-x^2}}\dee{y}\ x\right]\\ &\hskip0.5in+3\int_0^2\dee{x}\left[\int_{-\sqrt{2x-x^2}}^{\sqrt{2x-x^2}}\dee{y}\ y\right] \end{align*}
On the domain of integration
  • \(x\) runs from \(0\) to \(2\) and
  • for each fixed \(0\le x\le 2\text{,}\) \(y\) runs from \(-\sqrt{2x-x^2}\) to \(+\sqrt{2x-x^2}\)
Observe that \(y=\pm \sqrt{2x-x^2}\) is equivalent to
\begin{equation*} y^2 = 2x-x^2 = 1-(x-1)^2 \iff (x-1)^2 + y^2 =1 \end{equation*}
Our domain of integration is exactly the disk
\begin{equation*} \cR = \Set{(x,y)}{(x-1)^2 + y^2 \le 1} \end{equation*}
of radius \(1\) centred on \((1,0)\text{.}\)
So \(\cR\) has area \(\pi\) and centre of mass \((\bar x,\bar y)=(1,0)\) and
\begin{align*} \int_0^2\int_{-\sqrt{2x-x^2}}^{\sqrt{2x-x^2}} \big(2x+3y\big)\dee{y}\,\dee{x} &=2\dblInt_\cR x\ \dee{x}\,\dee{y} +3\dblInt_\cR y\ \dee{x}\,\dee{y}\\ &=2\,\bar x\, \text{Area}(\cR) +3\,\bar y\, \text{Area}(\cR) =2\pi \end{align*}
In Section 2.3.1 of the CLP-2 text, we saw that it was possible to think of a body as being made up of a number of component parts, and then it was possible to compute the centre of mass of the body as a whole by using the centres of mass of the component parts. In that discussion, we could only consider bodies consisting of a finite number of point particles arrayed along a straight line. We will now see that the same procedure can be used with continuous mass distributions. Suppose that we have a dumbbell which consists of
  • one end contained in a region \(\cR_1\) and having mass density \(\rho_1(x,y)\) and
  • a second end contained in a region \(\cR_2\) and having mass density \(\rho_2(x,y)\) and
  • an infinitely strong, weightless (idealized) rod joining the two ends.
Suppose that both \(\cR_1\) and \(\cR_2\) are contained in the larger region \(\cR\text{,}\) that \(\rho_1(x,y)=0\) for all \((x,y)\) not in \(\cR_1\) and that \(\rho_2(x,y)=0\) for all \((x,y)\) not in \(R_2\text{.}\)
Then the mass and centre of mass of the \(R_1\) end are
\begin{equation*} M_1=\dblInt_{\cR_1} \rho_1(x,y)\,\dee{x}\,\dee{y} \end{equation*}
and
\begin{equation*} (\bar X_1,\bar Y_1)\quad\text{with}\quad \bar X_1 = \frac{\dblInt_{\cR_1} x\,\rho_1(x,y)\,\dee{x}\,\dee{y}}{M_1},\ \bar Y_1 = \frac{\dblInt_{\cR_1} y\,\rho_1(x,y)\,\dee{x}\,\dee{y}}{M_1} \end{equation*}
and the mass and centre of mass of the \(R_2\) end are
\begin{equation*} M_2=\dblInt_{\cR_2} \rho_2(x,y)\,\dee{x}\,\dee{y} \end{equation*}
and
\begin{equation*} (\bar X_2,\bar Y_2)\quad\text{with}\quad \bar X_2 = \frac{\dblInt_{\cR_2} x\,\rho_2(x,y)\,\dee{x}\,\dee{y}}{M_2},\ \bar Y_2 = \frac{\dblInt_{\cR_2} y\,\rho_2(x,y)\,\dee{x}\,\dee{y}}{M_2} \end{equation*}
The mass and centre of mass of the entire dumbbell are
\begin{align*} M&= \dblInt_{\cR} \{\rho_1(x,y)+\rho_2(x,y)\}\,\dee{x}\,\dee{y}\\ &= \dblInt_{\cR} \rho_1(x,y)\,\dee{x}\,\dee{y} +\dblInt_{\cR} \rho_2(x,y)\,\dee{x}\,\dee{y} \\ &= \dblInt_{\cR_1} \rho_1(x,y)\,\dee{x}\,\dee{y} +\dblInt_{\cR_2} \rho_2(x,y)\,\dee{x}\,\dee{y}\\ &=M_1+M_2 \end{align*}
and \((\bar X,\bar Y)\) with
\begin{align*} \bar X &=\frac{\dblInt_{\cR} x\{\rho_1(x,y)+\rho_2(x,y)\}\,\dee{x}\,\dee{y}}{M}\\ &=\frac{\dblInt_{\cR_1} x\,\rho_1(x,y)\,\dee{x}\,\dee{y} +\dblInt_{\cR_2} x\,\rho_2(x,y)\,\dee{x}\,\dee{y}}{M}\\ &=\frac{M_1 \bar X_1 + M_2 \bar X_2}{M_1+M_2}\\ \bar Y &=\frac{\dblInt_{\cR} y\{\rho_1(x,y)+\rho_2(x,y)\}\,\dee{x}\,\dee{y}}{M}\\ &=\frac{\dblInt_{\cR_1} y\,\rho_1(x,y)\,\dee{x}\,\dee{y} +\dblInt_{\cR_2} y\,\rho_2(x,y)\,\dee{x}\,\dee{y}}{M}\\ &=\frac{M_1 \bar Y_1 + M_2 \bar Y_2}{M_1+M_2} \end{align*}
So we can compute the centre of mass of the entire dumbbell by treating it as being made up of two point particles, one of mass \(M_1\) located at the centre of mass \((\bar X_1,\bar Y_1)\) of the \(\cR_1\) end, and one of mass \(M_2\) located at the centre of mass \((X_2,Y_2)\) of the \(\cR_2\) end.
In Example 2.3.6 of the CLP-2 text, we found the centroid, i.e. the centre of mass with density \(1\text{,}\) of the region \(\cR_1\) in the sketch
We now again compute the centroid of \(\cR_1\text{,}\) but this time we use the ideas of Example 3.3.13 above. We think of \(\cR_1\) as one end of the “dumbbell” whose other end, \(\cR_2\text{,}\) is the quarter circular disk which fills out \(\cR_1\) to the \(2\times 2\) square in the sketch below. That is, the full dumbbell is
\begin{equation*} \cR=\cR_1\cup\cR_2=\Set{(x,y)}{0\le x\le 2,\ 0\le y\le 2} \end{equation*}
with
\begin{align*} \cR_1&=\Set{(x,y)}{0\le x\le 2,\ 0\le y\le 2,\ x^2+y^2\ge 1} \\ \cR_2&=\Set{(x,y)}{x\ge 0,\ y\ge 0,\ x^2+y^2\le 1} \end{align*}
  • The full dumbbell \(\cR\) is a \(2\times 2\) square and so has area, and hence mass, \(M=2\times 2=4\text{.}\) Just by symmetry, the centre of mass of \(\cR\) is \((\bar X,\bar Y)=(1,1)\text{.}\)
  • The end \(\cR_2\) is one quarter of a circular disk of radius \(1\) and so has area, and hence mass, \(M_2=\tfrac{\pi}{4}\text{.}\) In Example 3.3.10 above with \(r=1\text{,}\) we found that the centre of mass of \(\cR_2\) is \((\bar X_2,\bar Y_2)=\big(\tfrac{4}{3\pi},\tfrac{4}{3\pi})\text{.}\)
So \(M_1=M-M_2=4-\tfrac{\pi}{4}\) and just filling all of this data into the formulae at the end of Example 3.3.13 , above, gives
\begin{alignat*}{3} 1&= \bar X &= \frac{M_1 \bar X_1 + M_2 \bar X_2}{M_1+M_2} &= \frac{ (4-\frac{\pi}{4})\bar X_1 + (\frac{\pi}{4})(\tfrac{4}{3\pi})}{4} &= \left(1-\frac{\pi}{16}\right)\bar X_1 + \frac{1}{12}\\ 1&= \bar Y &= \frac{M_1 \bar Y_1 + M_2 \bar Y_2}{M_1+M_2} &= \frac{ (4-\frac{\pi}{4})\bar Y_1 + (\frac{\pi}{4})(\tfrac{4}{3\pi})}{4} &= \left(1-\frac{\pi}{16}\right)\bar Y_1 + \frac{1}{12} \end{alignat*}
Solving gives
\begin{equation*} \bar X_1=\bar Y_1=\frac{\frac{11}{12}}{1-\frac{\pi}{16}}=\frac{44}{48-3\pi} \end{equation*}
which is exactly the answer that we found in Example 2.3.6 of the CLP-2 text.

Subsection 3.3.3 Moment of Inertia

Consider a plate that fills the region \(\cR\) in the \(xy\)-plane, that has mass density \(\rho(x,y)\,\text{kg}/\text{m}^2\text{,}\) and that is rotating at \(\om\,\text{rad}/\text{s}\) about some axis. Let's call the axis of rotation \(\cA\text{.}\) We are now going to determine the kinetic energy of that plate. Recall 2  that, by definition, the kinetic energy of a point particle of mass \(m\) that is moving with speed \(v\) is \(\frac{1}{2}mv^2\text{.}\)
To get the kinetic energy of the entire plate, cut it up into tiny rectangles 3 , say of size \(\dee{x}\times\dee{y}\text{.}\) Think of each rectangle as being (essentially) a point particle. If the point \((x,y)\) on the plate is a distance \(D(x,y)\) from the axis of rotation \(\cA\text{,}\) then as the plate rotates, the point \((x,y)\) sweeps out a circle of radius \(D(x,y)\text{.}\) The figure on the right below shows that circle as seen from high up on the axis of rotation.
The circular arc that the point \((x,y)\) sweeps out in one second subtends the angle \(\om\) radians, which is the fraction \(\frac{\om}{2\pi}\) of a full circle and so has length \(\frac{\om}{2\pi}\big(2\pi D(x,y)\big)=\om\,D(x,y)\text{.}\) Consequently the rectangle that contains the point \((x,y)\)
  • has speed \(\om\,D(x,y)\text{,}\) and
  • has area \(\dee{x}\,\dee{y}\text{,}\) and so
  • has mass \(\rho(x,y)\ \dee{x}\,\dee{y}\text{,}\) and
  • has kinetic energy
    \begin{equation*} \frac{1}{2}\overbrace{\big(\rho(x,y)\,\dee{x}\,\dee{y}\big)}^{m} \overbrace{(\om\,D(x,y))^2}^{v^2} =\frac{1}{2}\om^2\ D(x,y)\,^2\rho(x,y)\ \dee{x}\,\dee{y} \end{equation*}
So (via our usual Riemann sum limit procedure) the kinetic energy of \(\cR\) is
\begin{equation*} \dblInt_\cR \frac{1}{2}\om^2\ D(x,y)^2\,\rho(x,y)\,\dee{x}\,\dee{y} =\frac{1}{2}\om^2 \dblInt_\cR \ D(x,y)^2\,\rho(x,y)\,\dee{x}\,\dee{y} =\frac{1}{2} I_\cA\,\om^2 \end{equation*}
where

Definition 3.3.15. Moment of Inertia.

\begin{equation*} I_\cA=\dblInt_\cR D(x,y)^2\rho(x,y)\ \dee{x}\,\dee{y} \end{equation*}
is called the moment of inertial of \(\cR\) about the axis \(\cA\text{.}\) In particular the moment of inertia of \(\cR\) about the \(y\)-axis is
\begin{equation*} I_y=\dblInt_\cR x^2\,\rho(x,y)\ \dee{x}\,\dee{y} \end{equation*}
and the moment of inertia of \(\cR\) about the \(x\)-axis is
\begin{equation*} I_x=\dblInt_\cR y^2\,\rho(x,y)\ \dee{x}\,\dee{y} \end{equation*}
Notice that the expression \(\frac{1}{2} I_\cA\,\om^2\) for the kinetic energy has a very similar form to \(\frac{1}{2} m v^2\text{,}\) just with the velocity \(v\) replaced by the angular velocity \(\om\text{,}\) and with the mass \(m\) replaced by \(I_\cA\text{,}\) which can be thought of as being a bit like a mass.
So far, we have been assuming that the rotation was taking place in the \(xy\)-plane — a two dimensional world. Our analysis extends naturally to three dimensions, though the resulting integral formulae for the moment of inertia will then be triple integrals, which we have not yet dealt with. We shall soon do so, but let's first do an example in two dimensions.
Find the moment of inertia of the interior, \(\cR\text{,}\) of the circle \(x^2+y^2=a^2\) about the \(x\)-axis. Assume that it has density one.
Solution.
The distance from any point \((x,y)\) inside the disk to the axis of
rotation (i.e. the \(x\)-axis) is \(|y|\text{.}\) So the moment of inertia of the interior of the disk about the \(x\)-axis is
\begin{equation*} I_x = \dblInt_\cR y^2\ \dee{x}\dee{y} \end{equation*}
Switching to polar coordinates 4 ,
\begin{align*} I_x &= \int_0^{2\pi}\dee{\theta}\int_0^{a}\dee{r}\ r \overbrace{(r\sin\theta)^2}^{y^2} = \int_0^{2\pi}\dee{\theta}\ \sin^2\theta\int_0^a\dee{r}\ r^3\\ &=\frac{a^4}{4}\int_0^{2\pi}\dee{\theta}\ \sin^2\theta =\frac{a^4}{4} \int_0^{2\pi}\dee{\theta}\ \frac{1-\cos(2\theta)}{2}\\ &=\frac{a^4}{8} \left[\theta-\frac{\sin(2\theta)}{2}\right]_0^{2\pi}\\ &=\frac{1}{4}\pi a^4 \end{align*}
For an efficient, sneaky, way to evaluate \(\int_0^{2\pi} \sin^2\theta\ \dee{\theta}\text{,}\) see Remark 3.3.5.
Find the moment of inertia of the interior, \(\cR\text{,}\) of the cardiod \(r=a(1+\cos\theta)\) about the \(z\)-axis. Assume that the cardiod lies in the \(xy\)-plane and has density one.
Solution.
We sketched the cardiod (with \(a=1\)) in Example 3.2.3.
As we said above, the formula for \(I_\cA\) in Definition 3.3.15 is valid even when the axis of rotation is not contained in the \(xy\)-plane. We just have to be sure that our \(D(x,y)\) really is the distance from \((x,y)\) to the axis of rotation. In this example the axis of rotation is the \(z\)-axis so that \(D(x,y)=\sqrt{x^2+y^2}\) and that the moment of inertia is
\begin{equation*} I_\cA = \dblInt_\cR (x^2+y^2)\ \dee{x}\dee{y} \end{equation*}
Switching to polar coordinates, using \(\dee{x}\dee{y}=r\,\dee{r}\dee{\theta}\) and \(x^2+y^2=r^2\text{,}\)
\begin{align*} I_\cA &= \int_0^{2\pi}\dee{\theta}\int_0^{a(1+\cos\theta)}\dee{r}\ r\times r^2 = \int_0^{2\pi}\dee{\theta}\ \int_0^{a(1+\cos\theta)}\dee{r}\ r^3\\ &=\frac{a^4}{4}\int_0^{2\pi}\dee{\theta}\ \big(1+\cos\theta\big)^4\\ &=\frac{a^4}{4}\int_0^{2\pi}\dee{\theta}\ \big(1+4\cos\theta+6\cos^2\theta+4\cos^3\theta+\cos^4\theta\big) \end{align*}
Now
\begin{align*} \int_0^{2\pi}\dee{\theta}\ \cos\theta &= \sin\theta\Big|_0^{2\pi} =0\\ \int_0^{2\pi}\dee{\theta}\ \cos^2\theta &= \int_0^{2\pi}\dee{\theta}\ \frac{1+\cos(2\theta)}{2} = \frac{1}{2}\left[\theta+\frac{\sin(2\theta)}{2}\right]_0^{2\pi} =\pi\\ \int_0^{2\pi}\dee{\theta}\ \cos^3\theta &= \int_0^{2\pi}\dee{\theta}\ \cos\theta\big[1-\sin^2\theta\big] \eqover{u=\sin\theta} \int_0^0\dee{u}\ (1-u^2) =0 \end{align*}
To integrate \(\cos^4\theta\text{,}\) we use the double angle formula
\begin{align*} \cos^2\theta &=\frac{\cos(2\theta)+1}{2}\\ \implies \cos^4\theta &= \frac{\big(\cos(2\theta)+1\big)^2}{4} =\frac{\cos^2(2\theta)+2\cos(2\theta)+1}{4}\\ &=\frac{\frac{\cos(4\theta)+1}{2}+2\cos(2\theta)+1}{4}\\ &=\frac{3}{8} +\frac{1}{2}\cos(2\theta) +\frac{1}{8}\cos(4\theta) \end{align*}
to give
\begin{align*} \int_0^{2\pi}\dee{\theta}\ \cos^4\theta &=\int_0^{2\pi}\dee{\theta}\ \left[\frac{3}{8} +\frac{1}{2}\cos(2\theta) +\frac{1}{8}\cos(4\theta)\right]\\ &=\frac{3}{8} \times 2\pi +\frac{1}{2}\times 0 +\frac{1}{8}\times 0 =\frac{3}{4}\pi \end{align*}
All together
\begin{align*} I_\cA&= \frac{a^4}{4}\left[2\pi + 4\times 0 +6\times\pi+4\times 0 +\frac{3}{4}\pi\right]\\ &=\frac{35}{16}\pi a^4 \end{align*}

Exercises 3.3.4 Exercises

Exercise Group.

Exercises — Stage 1
1.
For each of the following, evaluate the given double integral without using iteration. Instead, interpret the integral in terms of, for example, areas or average values.
  1. \(\dblInt_D(x+3)\ \dee{x}\,\dee{y}\text{,}\) where \(D\) is the half disc \(0\le y\le \sqrt{4-x^2}\)
  2. \(\dblInt_R (x+y)\ \dee{x}\,\dee{y}\) where \(R\) is the rectangle \(0\le x\le a,\ 0\le y\le b\)

Exercise Group.

Exercises — Stage 2
2. (✳).
Find the centre of mass of the region \(D\) in the \(xy\)--plane defined by the inequalities \(x^2 \le y \le 1\text{,}\) assuming that the mass density function is given by \(\rho(x,y) = y\text{.}\)
3. (✳).
Let R be the region bounded on the left by \(x = 1\) and on the right by \(x^2 + y^2 = 4\text{.}\) The density in \(R\) is
\begin{equation*} \rho(x,y) =\frac{1}{\sqrt{x^2+y^2}} \end{equation*}
  1. Sketch the region \(R\text{.}\)
  2. Find the mass of \(R\text{.}\)
  3. Find the centre-of-mass of \(R\text{.}\)
Note: You may use the result \(\int \sec(\theta)\ \dee{\theta} = \ln |\sec \theta + \tan \theta| + C\text{.}\)
4. (✳).
A thin plate of uniform density \(1\) is bounded by the positive \(x\) and \(y\) axes and the cardioid \(\sqrt{x^2+y^2}=r=1+\sin\theta\text{,}\) which is given in polar coordinates. Find the \(x\)--coordinate of its centre of mass.
5. (✳).
A thin plate of uniform density \(k\) is bounded by the positive \(x\) and \(y\) axes and the circle \(x^2 + y^2 = 1\text{.}\) Find its centre of mass.
6. (✳).
Let \(R\) be the triangle with vertices \((0, 2)\text{,}\) \((1, 0)\text{,}\) and \((2, 0)\text{.}\) Let \(R\) have density \(\rho(x, y) = y^2\text{.}\) Find \(\bar y\text{,}\) the \(y\)--coordinate of the center of mass of \(R\text{.}\) You do not need to find \(\bar x\text{.}\)
7. (✳).
The average distance of a point in a plane region \(D\) to a point \((a, b)\) is defined by
\begin{equation*} \frac{1}{A(D)}\dblInt_D \sqrt{(x-a)^2+(y-b)^2}\ \dee{x}\,\dee{y} \end{equation*}
where \(A(D)\) is the area of the plane region \(D\text{.}\) Let \(D\) be the unit disk \(1 \ge x^2 + y^2\text{.}\) Find the average distance of a point in \(D\) to the center of \(D\text{.}\)
8. (✳).
A metal crescent is obtained by removing the interior of the circle defined by the equation \(x^2 + y^2 = x\) from the metal plate of constant density 1 occupying the unit disc \(x^2 + y^2 \le 1\text{.}\)
  1. Find the total mass of the crescent.
  2. Find the \(x\)-coordinate of its center of mass.
You may use the fact that \(\int_{-\pi/2}^{\pi/2}\cos^4(\theta)\ \dee{\theta}=\frac{3\pi}{8}\text{.}\)
9. (✳).
Let \(D\) be the region in the \(xy\)--plane which is inside the circle \(x^2+(y-1)^2=1\) but outside the circle \(x^2+y^2=2\text{.}\) Determine the mass of this region if the density is given by
\begin{equation*} \rho(x,y)=\frac{2}{\sqrt{x^2+y^2}} \end{equation*}

Exercise Group.

Exercises — Stage 3
10. (✳).
Let \(a\text{,}\) \(b\) and \(c\) be positive numbers, and let \(T\) be the triangle whose vertices are \((-a,0)\text{,}\) \((b,0)\) and \((0,c)\text{.}\)
  1. Assuming that the density is constant on \(T\text{,}\) find the center of mass of \(T\text{.}\)
  2. The medians of \(T\) are the line segments which join a vertex of \(T\) to the midpoint of the opposite side. It is a well known fact that the three medians of any triangle meet at a point, which is known as the centroid of \(T\text{.}\) Show that the centroid of \(T\) is its centre of mass.
There is a real Half-Dome mountain in Yosemite National Park. It has \(a=1445\,\text{m} \text{.}\)
If you don't recall, don't worry. We wouldn't lie to you. Or check it on Wikipedia. They wouldn't lie to you either.
The relatively small number of “rectangles” around the boundary of \(\cR\) won't actually be rectangles. But, as we have seen in the optional §3.2.4, one can still make things rigorous despite the rectangles being a bit squishy around the edges.
See how handy they are!
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