Calculating Volumes, Areas, and Average Values

Learning Objectives

Use double integrals to calculate the volume of a region between two surfaces or the area of a plane region.

We can use double integrals over general regions to compute volumes, areas, and average values. The methods are the same as those in Double Integrals over Rectangular Regions, but without the restriction to a rectangular region, we can now solve a wider variety of problems.

Example: finding the volume of a tetrahedron

Find the volume of the solid bounded by the planes $x = 0$ , $y = 0, \ z = 0$ , and $2x+3y+z=6$ .

Show Solution

The solid is a tetrahedron with the base on the $xy$ -plane and a height $z=6-2x-3y$ The base is the region $D$ bounded by the lines, $x=0$ , $y=0$ , and $2x+3y=6$ where $z=0$ (Figure 1). Note that we can consider the region $D$ as Type I or as Type II, and we can integrate in both ways.

This figure shows a tetrahedron bounded by x = 0, y = 0, z = 0, and 2x + 3y = 6 (or z = 6 minus 2x minus 3y).

Figure 1. A tetrahedron consisting of the three coordinate planes and the plane $\small{z=6-2x-3y}$ , with the base bound by $\small{x=0}$ , $\small{y=0}$ , and $\small{2x+3y=6}$ .

First, consider $D$ as a Type I region, and hence ${D} = {\left \{{(x,y)}{\mid}{0} \ {\leq} \ {x} \ {\leq} \ {3,0} \ {\leq} \ {y} \ {\leq} \ {2-{\frac{2}{3}}x} \right \}}$ .

Therefore, the volume is
$\hspace{3cm}\begin{align} V&=\displaystyle\int_{x=0}^{x=3}\displaystyle\int_{y=0}^{y=2-(2x/3)}(6-2x-3y)dydx=\displaystyle\int_{x=0}^{x=3}\left[\left(6y-2xy-\frac32y^2\right)\Bigg|_{y=0}^{y=2-(2x/3)}\right]dx \\ &=\displaystyle\int_{x=0}^{x=3}\left[\frac23(x-3)^2\right]dx=6 \end{align}$

Now consider $D$ as a Type II region, so ${D} = {\left \{{(x,y)}{\mid}{0} \ {\leq} \ {y} \ {\leq} \ {2,0} \ {\leq} \ {x} \ {\leq} \ {3-{\frac{3}{2}}y} \right \}}$ . In this calculation, the volume is
$\hspace{3cm}\begin{align} V&=\displaystyle\int_{y=0}^{y=2}\displaystyle\int_{x=0}^{x=3-(3y/2)}(6-2x-3y)dxdy=\displaystyle\int_{y=0}^{y=2}\left[(6x-x^2-3xy)\bigg|_{x=0}^{x=3-(3y/2)}\right]dy \\ &=\displaystyle\int_{y=0}^{y=2}\left[\frac94(y-2)^2\right]dy=6 \end{align}$

Therefore, the volume is $6$ cubic units.

try it

Find the volume of the solid bounded above by $f(x, y)=10-2x+y$ over the region enclosed by the curves $y=0$ and ${y} = {e^x}$ , where $x$ is in the interval $[0,1]$ .

Show Solution

$\frac{e^2}4+10e-\frac{49}4$ cubic units.

Finding the area of a rectangular region is easy, but finding the area of a nonrectangular region is not so easy. As we have seen, we can use double integrals to find a rectangular area. As a matter of fact, this comes in very handy for finding the area of a general nonrectangular region, as stated in the next definition.

definition

The area of a plane-bounded region $D$ is defined as the double integral $\underset{D}{\displaystyle\iint}{1dA}$ .

We have already seen how to find areas in terms of single integration. Here we are seeing another way of finding areas by using double integrals, which can be very useful, as we will see in the later sections of this chapter.

Example: finding the area of a region

Find the area of the region bounded below by the curve $y=x^{2}$ and above by the line $y=2x$ in the first quadrant (Figure 2).

The line y = 2 x (also marked x = y/2) is shown, as is y = x squared (also marked x = the square root of y). There are vertical and horizontal shadings giving for small stretch of this region, denoting that it can be treated as a Type I or Type II area.

Figure 2. The region bounded by $y=x^{2}$ and $y=2x$ .

Show Solution

We just have to integrate the constant function $f(x, y)=1$ over the region. Thus, the area $A$ of the bounded region is $\displaystyle\int_{x=0}^{x=2}\displaystyle\int_{y=x^2}^{y=2x}dydx$ or $\displaystyle\int_{y=0}^{x=4}\displaystyle\int_{x=y/2}^{x=\sqrt{y}}dxdy$ :

$A=\underset{D}{\displaystyle\iint}1dxdy=\displaystyle\int_{x=0}^{x=2}\displaystyle\int_{y=x^2}^{y=2x}1dydx=\displaystyle\int_{x=0}^{x=2}\left[y\bigg|_{y=x^2}^{y=2x}\right]dx=\displaystyle\int_{x=0}^{x=2}(2x-x^2)dx=x^2-\frac{x^3}3\bigg|_0^2=\frac43.$

try it

Find the area of a region bounded above by the curve $y=x^{3}$ and below by $y=0$ over the interval $[0,3]$ .

Show Solution

$\frac{81}4$ square units.

Watch the following video to see the worked solution to the above Try It

You can view the transcript for “CP 5.13” here (opens in new window).

We can also use a double integral to find the average value of a function over a general region. First, recall how we find the average value of a function using single-variable calculus.

Recall: Average Value of a Function (Single-variable version)

If $f(x)$ is continuous on $[a,b]$ , then the average value of $f(x)$ on $[a,b]$ is

$f_{ave}=\frac{1}{b-a} \int_a^b f(x) dx$

The following definition is a direct extension of the formula above.

definition

If $f(x, y)$ is integrable over a plane-bounded region $D$ with positive area $A(D)$ then the average value of the function is

$\large{{f_{ave}} = {\frac{1}{A(D)}}\underset{D}{\displaystyle\iint}{f(x,y)}{dA}}$

Note that the area is ${A(D)}=\underset{D}{\displaystyle\iint}{1dA}$ .

Example: finding an average value

Find the average value of the function $f(x, y)=7xy^{2}$ on the region bounded by the line $x=y$ and the curve $x=\sqrt{y}$ (Figure 3).

The lines x = y and x = the square root of y bound a shaded region. There are horizontal dashed lines marked throughout the region.

Figure 3. The region bounded by $\small{x=y}$ and $\small{x=\sqrt{y}}$

Show Solution

First find the area $A(D)$ where the region $D$ is given by the figure. We have

$A(d)=\underset{D}{\displaystyle\iint}1dA=\displaystyle\int_{y=0}^{y=1}\displaystyle\int_{x=y}^{x=\sqrt{y}}1dxdy \displaystyle\int_{y=0}^{y=1}\left[x\bigg|_{x=y}^{x=\sqrt{y}}\right]dy=\frac23y^{3/2}-\frac{y^2}2\Bigg|_0^1=\frac16.$

Then the average value of the given function over this region is

$\begin{align} f_{ave} &= \frac1{A(D)}\underset{D}{\displaystyle\iint}f(x,y)dA = \frac1{A(D)}\displaystyle\int_{y=0}^{y=1}\displaystyle\int_{x=y}^{x=\sqrt{y}}yxy^2dxdy=\frac1{1/6}\displaystyle\int_{y=0}^{y=1}\left[\frac72x^2y^s\Bigg|_{x=y}^{x=\sqrt{y}}\right]dy \\ &=6\displaystyle\int_{y=0}^{y=1}\left[\frac72y^2(y-y^2)\right]dy=6\displaystyle\int_{y=0}^{y=1}\left[\frac72(y^3-y^4)\right]dy=\frac{42}2(\frac{y^4}4-\frac{y^5}5)\bigg|_0^1=\frac{42}{40}=\frac{21}{20}. \end{align}$

try it

Find the average value of the function $f(x, y)=xy$ over the triangle with vertices $(0, 0)$ , $(1, 0)$ , and $(1, 3)$ .

Show Solution

$\frac34$

Module 5: Multiple Integration