# 5.2 Double integrals over general regions  (Page 4/12)

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Consider the region bounded by the curves $y=\text{ln}\phantom{\rule{0.2em}{0ex}}x$ and $y={e}^{x}$ in the interval $\left[1,2\right].$ Decompose the region into smaller regions of Type II.

$\left\{\left(x,y\right)|0\le y\le 1,1\le x\le {e}^{y}\right\}\cup \left\{\left(x,y\right)|1\le y\le e,1\le x\le 2\right\}\cup \left\{\left(x,y\right)|e\le y\le {e}^{2},\text{ln}\phantom{\rule{0.2em}{0ex}}y\le x\le 2\right\}$

Redo [link] using a union of two Type II regions.

Same as in the example shown.

## Changing the order of integration

As we have already seen when we evaluate an iterated integral, sometimes one order of integration leads to a computation that is significantly simpler than the other order of integration. Sometimes the order of integration does not matter, but it is important to learn to recognize when a change in order will simplify our work.

## Changing the order of integration

Reverse the order of integration in the iterated integral $\underset{x=0}{\overset{x=\sqrt{2}}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{y=0}{\overset{y=2-{x}^{2}}{\int }}x{e}^{{x}^{2}}dy\phantom{\rule{0.2em}{0ex}}dx.$ Then evaluate the new iterated integral.

The region as presented is of Type I. To reverse the order of integration, we must first express the region as Type II. Refer to [link] .

We can see from the limits of integration that the region is bounded above by $y=2-{x}^{2}$ and below by $y=0,$ where $x$ is in the interval $\left[0,\sqrt{2}\right].$ By reversing the order, we have the region bounded on the left by $x=0$ and on the right by $x=\sqrt{2-y}$ where $y$ is in the interval $\left[0,2\right].$ We solved $y=2-{x}^{2}$ in terms of $x$ to obtain $x=\sqrt{2-y}.$

Hence

$\begin{array}{ccccc}\hfill \underset{0}{\overset{\sqrt{2}}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{0}{\overset{2-{x}^{2}}{\int }}x{e}^{{x}^{2}}dy\phantom{\rule{0.2em}{0ex}}dx& =\underset{0}{\overset{2}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{0}{\overset{\sqrt{2-y}}{\int }}x{e}^{{x}^{2}}dx\phantom{\rule{0.2em}{0ex}}dy\hfill & & & \begin{array}{c}\text{Reverse the order of}\hfill \\ \text{integration then use}\hfill \\ \text{substitution.}\hfill \end{array}\hfill \\ & =\underset{0}{\overset{2}{\int }}\left[\frac{1}{2}{e{x}^{2}|}_{0}^{\sqrt{2-y}}\right]dy=\underset{0}{\overset{2}{\int }}\frac{1}{2}\left({e}^{2-y}-1\right)dy={-\frac{1}{2}\left({e}^{2-y}+y\right)|}_{0}^{2}\hfill & & & \\ & =\frac{1}{2}\left({e}^{2}-3\right).\hfill & & & \end{array}$

## Evaluating an iterated integral by reversing the order of integration

Consider the iterated integral $\underset{R}{\iint }f\left(x,y\right)dx\phantom{\rule{0.2em}{0ex}}dy$ where $z=f\left(x,y\right)=x-2y$ over a triangular region $R$ that has sides on $x=0,y=0,$ and the line $x+y=1.$ Sketch the region, and then evaluate the iterated integral by

1. integrating first with respect to $y$ and then
2. integrating first with respect to $x.$

A sketch of the region appears in [link] .

We can complete this integration in two different ways.

1. One way to look at it is by first integrating $y$ from $y=0\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}y=1-x$ vertically and then integrating $x$ from $x=0\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}x=1\text{:}$
$\begin{array}{cc}\hfill \underset{R}{\iint }f\left(x,y\right)dx\phantom{\rule{0.2em}{0ex}}dy& =\underset{x=0}{\overset{x=1}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{y=0}{\overset{y=1-x}{\int }}\left(x-2y\right)dy\phantom{\rule{0.2em}{0ex}}dx=\underset{x=0}{\overset{x=1}{\int }}{\left[xy-2{y}^{2}\right]}_{y=0}^{y=1-x}dx\hfill \\ & =\underset{x=0}{\overset{x=1}{\int }}\left[x\left(1-x\right)-{\left(1-x\right)}^{2}\right]dx=\underset{x=0}{\overset{x=1}{\int }}\left[-1+3x-2{x}^{2}\right]dx={\left[\text{−}x+\frac{3}{2}{x}^{2}-\frac{2}{3}{x}^{3}\right]}_{x=0}^{x=1}=-\frac{1}{6}.\hfill \end{array}$
2. The other way to do this problem is by first integrating $x$ from $x=0\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}x=1-y$ horizontally and then integrating $y$ from $y=0\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}y=1\text{:}$
$\begin{array}{cc}\hfill \underset{R}{\iint }f\left(x,y\right)dx\phantom{\rule{0.2em}{0ex}}dy& =\underset{y=0}{\overset{y=1}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{x=0}{\overset{x=1-y}{\int }}\left(x-2y\right)dx\phantom{\rule{0.2em}{0ex}}dy=\underset{y=0}{\overset{y=1}{\int }}{\left[\frac{1}{2}{x}^{2}-2xy\right]}_{x=0}^{x=1-y}dy\hfill \\ & =\underset{y=0}{\overset{y=1}{\int }}\left[\frac{1}{2}{\left(1-y\right)}^{2}-2y\left(1-y\right)\right]dy=\underset{y=0}{\overset{y=1}{\int }}\left[\frac{1}{2}-3y+\frac{5}{2}{y}^{2}\right]dy\hfill \\ & ={\left[\frac{1}{2}y-\frac{3}{2}{y}^{2}+\frac{5}{6}{y}^{3}\right]}_{y=0}^{y=1}=-\frac{1}{6}.\hfill \end{array}$

Evaluate the iterated integral $\underset{D}{\iint }\left({x}^{2}+{y}^{2}\right)dA$ over the region $D$ in the first quadrant between the functions $y=2x$ and $y={x}^{2}.$ Evaluate the iterated integral by integrating first with respect to $y$ and then integrating first with resect to $x.$

$\frac{216}{35}$

## Calculating volumes, areas, and average values

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.

## Finding the volume of a tetrahedron

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

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 $z=0$ ( [link] ). Note that we can consider the region $D$ as Type I or as Type II, and we can integrate in both ways.

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

Therefore, the volume is

$\begin{array}{cc}\hfill V& =\underset{x=0}{\overset{x=3}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{y=0}{\overset{y=2-\left(2x\text{/}3\right)}{\int }}\left(6-2x-3y\right)dy\phantom{\rule{0.2em}{0ex}}dx=\underset{x=0}{\overset{x=3}{\int }}\left[{\left(6y-2xy-\frac{3}{2}{y}^{2}\right)|}_{y=0}^{y=2-\left(2x\text{/}3\right)}\right]dx\hfill \\ & =\underset{x=0}{\overset{x=3}{\int }}\left[\frac{2}{3}{\left(x-3\right)}^{2}\right]dx=6.\hfill \end{array}$

Now consider $D$ as a Type II region, so $D=\left\{\left(x,y\right)|0\le y\le 2,0\le x\le 3-\frac{3}{2}y\right\}.$ In this calculation, the volume is

$\begin{array}{cc}\hfill V& =\underset{y=0}{\overset{y=2}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{x=0}{\overset{x=3-\left(3y\text{/}2\right)}{\int }}\left(6-2x-3y\right)dx\phantom{\rule{0.2em}{0ex}}dy=\underset{y=0}{\overset{y=2}{\int }}\left[{\left(6x-{x}^{2}-3xy\right)|}_{x=0}^{x=3-\left(3y\text{/}2\right)}\right]dy\hfill \\ & =\underset{y=0}{\overset{y=2}{\int }}\left[\frac{9}{4}{\left(y-2\right)}^{2}\right]dy=6.\hfill \end{array}$

Therefore, the volume is $6$ cubic units.

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