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

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

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

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

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

Therefore, the volume is $6$ cubic units.