Lecture 14 Applications of Multiple Integrals

Text References: Course notes pp. 53-65 & Rogawski 11.3, 15.4-15.6

14.1 Recap

Last time, we rounded out our toolkit for evaluating double integrals with the change of variables formula.

Exercise 14.1 Consider the change of variables \(x=3u+v\) and \(y=u-2v\). Determine the area of the rectangle \(R=[2,5]\times[1,7\) under this transformation.

Solution. In order to determine the area scaling factor, we need to calculate the Jacobian of the transformation. We have

\[\begin{align*} \dfrac{\partial(x,y)}{\partial(u,v)}&=\det\begin{bmatrix}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u}& \frac{\partial y}{\partial v}\end{bmatrix} \\ &=\det \begin{bmatrix}3& 1\\ 1&-2\end{bmatrix}\\ &= -7 \end{align*}\]

Taking the absolute value, we get a scaling factor of \(7\). The area of the rectangle is \(3\cdot 6=18\) square units; multiplying by the scaling factor, we get \(126\) square units.

14.2 Learning Objectives

Explain several applications of integrals.

14.3 Applications of Double Integrals

Until now, we have mainly focused on interpreting integrals as either areas under curves (for single integrals) or volumes under surfaces (for double integrals). Today, we’re going to broaden our interpretation of integrals to include other types of applications.

14.3.1 Definite Integrals

Let’s take a look at the formal definition of a definite integral: \[\int_a^b f(x) dx = \lim_{n\to\infty}\sum_{i=1}^nf(x_i^*)\Delta x\] where \(x_i^*\in[x_{i-1}, x_i]\) and \([a,b]\) is divided into \(n\) sub-intervals of equal width \(\Delta x\).

The interpretation of this integral hinges on our interpretation of \(f(x)\) and \(x\). So far, we’ve been interpreting \(\Delta x\) and \(f(x_i^*)\) as the width and height of a small rectangle, which means that the result of the integral is an area.

Let’s take a look at other possible interpretations.

Example 14.1 If \(t\) measures time (s) and \(f(t)\) measures velocity (m/s) in a straight line, then \(\displaystyle \int_a^b f(t) dt\) is a sum of infinitesimal displacements \(ds=f(t) dt\). The integral therefore represents the total displacement of a particle moving with velocity \(f(t)\) between \(t=a\) and \(t=b\).

Example 14.2 Consider a charged rod of length \(L\). If \(x\) measures the distance (meters) from one end of the rod and \(\rho (x)\) is the linear charge density in the rod (coulombs/meter), then \(\displaystyle \int_0^L \rho(x) dx\) gives the total charge of the rod.

Example 14.3 If \(f(x,y)\) is a population density (organisms/km\(^2\)) and \(R\) is a region, then \(\displaystyle \iint_R f(x,y)dA\) gives the total population within the region \(R\).

Example 14.4 The average value of a function \(f(x,y)\) over a two-dimensional region \(R\) can be calculated as \(\displaystyle f_{avg}=\frac{1}{\mbox{Area}(R)}=\iint_R f(x,y)dA\).

Exercise 14.2 The population in a rural area has density \(\delta(x,y)=40xe^{0.1y}\) people per km\(^2\). How many people live in the region \(R: 2\leq x \leq 6, 1\leq y\leq 3\)?

Solution. The population in the region \(R\) is given by the integral of the population density:

\[\begin{align*} \displaystyle \iint_R 40xe^{0.1y} dA & = \int_1^3 \int_2^6 40xe^{0.1y} dxdy\\ &= \int_1^3 \left (20x^2e^{0.1y}\right |_{x=2}^6)dy \\ &= \int_1^3 640 e^{0.1y}dy\\ &= \left . 6400 e^{0.1y}\right |_{y=1}^3\\ & \approx 1566 \, \mbox{people} \end{align*}\]

14.3.2 Integrals with no Integrand

Thinking back to single-variable calculus, the integral \(\displaystyle \int_a^b dx\) is the length of the interval of integration \([a,b]\): \(\displaystyle \int_a^b dx=b-a\). A similar interpretation holds for double integrals: \(\displaystyle \iint_R dA\) is the area of the domain \(R\).

Exercise 14.3 Find the area of the region \(R\) bounded by the curves \(y=x\) and \(y=x^2\).

Solution. Note that the curves intersect at \((0,0)\) and \((1,1)\). One way to describe the region is \(0\leq x \leq 1\) and \(x^2\leq y \leq x\). The area of the region is therefore given by

\[\begin{align*} \displaystyle \iint_R dA & = \int_0^1 \int_{x^2}^x dy dx \\ &= \int_0^1 (x-x^2) dx\\ &= \left. \frac{x^2}{2}-\frac{x^3}{3}\right |_{x=0}^1\\ & = \frac{1}{6} \end{align*}\]

14.4 Applications of Triple Integrals

All of the applications discussed above can be extended to functions of three variables:

\(\displaystyle \iiint_R dV\) gives the volume of the three-dimensional region \(R\).
\(\displaystyle \iiint_R f(x,y,z)dV\) is interpreted as the sum of the infinitesimal quantities \(f dV\) over all the points in \(R\).
\(\displaystyle \frac{1}{\mbox{Volume}(R)} \iiint_R f(x,y,z) dV\) gives the mean value of the function \(f\) on the region \(R\).