5.2 Double integrals over general regions  (Page 8/12)

The region $D$ bounded by $y=0,x=\mathrm{-10}+y,\text{and}x=10-y$ as given in the following figure.

Find the volume of the solid under the graph of the function $f\left(x,y\right)=x+y$ and above the region in the figure from the previous exercise.

The region $D$ bounded by $y=0,x=y-1,$ $x=\frac{\pi }{2}$ as given in the following figure.

The region $D$ bounded by $y=0$ and $y=x^2-1$ as given in the following figure.

Let $D$ be the region bounded by the curves of equations $y=x,y=\text{−}x,$ and $y=2-x^2.$ Explain why $D$ is neither of Type I nor II.

Let $D$ be the region bounded by the curves of equations $y=\text{cos}x$ and $y=4-x^2$ and the $x$ -axis. Explain why $D$ is neither of Type I nor II.

$f\left(x,y\right)=2x+5y$ and $D=\left\{\left(x,y\right)|0\le x\le 1,x^3\le y\le x^3+1\right\}$

$f\left(x,y\right)=1$ and $D=\left\{\left(x,y\right)|0\le x\le \frac{\pi }{2},\text{sin}x\le y\le 1+\text{sin}x\right\}$

$f\left(x,y\right)=2$ and $D=\left\{\left(x,y\right)|0\le y\le 1,y-1\le x\le \text{arccos}y\right\}$

$f\left(x,y\right)=xy$ and $D=\left\{\left(x,y\right)|-1\le y\le 1,y^2-1\le x\le \sqrt{1-y^2}\right\}$

$f\left(x,y\right)=\text{sin}y$ and $D$ is the triangular region with vertices $\left(0,0\right),\left(0,3\right),\text{and}\left(3,0\right)$

$f\left(x,y\right)=\text{−}x+1$ and $D$ is the triangular region with vertices $\left(0,0\right),\left(0,2\right),\text{and}\left(2,2\right)$

${\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \underset{2x}{\overset{3x}{\int }}\left(x+y^2\right)}}dydx$

${\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \underset{2\sqrt{x}}{\overset{2\sqrt{x}+1}{\int }}\left(xy+1\right)}}dydx$

${\displaystyle \underset{e}{\overset{e^2}{\int }}{\displaystyle \underset{\text{ln}u}{\overset{2}{\int }}\left(v+\text{ln}u\right)}}dvdu$

${\displaystyle \underset{1}{\overset{2}{\int }}{\displaystyle \underset{\text{−}{u}^2-1}{\overset{\text{−}u}{\int }}\left(8uv\right)}}dvdu$

${\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \underset{\text{−}\sqrt{1-y^2}}{\overset{\sqrt{1-y^2}}{\int }}\left(2x+4x^3\right)}}dxdy$

${\displaystyle \underset{0}{\overset{1\text{/}2}{\int }}{\displaystyle \underset{\text{−}\sqrt{1-4y^2}}{\overset{\sqrt{1-4y^2}}{\int }}4}}dxdy$

Let $D$ be the region bounded by $y=1-x^2,y=4-x^2,$ and the $x$ - and $y$ -axes.

1. Show that ${\displaystyle \underset{D}{\iint }xdA={\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \underset{1-x^2}{\overset{4-x^2}{\int }}xdydx}}}+{\displaystyle \underset{1}{\overset{2}{\int }}{\displaystyle \underset{0}{\overset{4-x^2}{\int }}xdydx}}$ by dividing the region $D$ into two regions of Type I.
2. Evaluate the integral ${\displaystyle \underset{D}{\iint }xdA}.$

Let $D$ be the region bounded by $y=1,$ $y=x,$ $y=\text{ln}x,$ and the $x$ -axis.

1. Show that ${\displaystyle \underset{D}{\iint }}ydA={\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \underset{0}{\overset{x}{\int }}ydydx}}+{\displaystyle \underset{1}{\overset{e}{\int }}{\displaystyle \underset{\text{ln}x}{\overset{1}{\int }}ydydx}}$ by dividing $D$ into two regions of Type I.
2. Evaluate the integral ${\displaystyle \underset{D}{\iint }ydA}.$

1. Show that ${\displaystyle \underset{D}{\iint }}{y}^{2}dA={\displaystyle \underset{\mathrm{-1}}{\overset{0}{\int }}{\displaystyle \underset{\text{−}x}{\overset{2-x^2}{\int }}{y}^{2}dydx}}+{\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \underset{x}{\overset{2-x^2}{\int }}{y}^{2}dydx}}$ by dividing the region $D$ into two regions of Type I, where $D=\left\{(x,y)|y\ge x,y\ge -x,y\le 2-x^2\right\}.$
2. Evaluate the integral ${\displaystyle \underset{D}{\iint }{y}^2}dA.$

Let $D$ be the region bounded by $y=x^2,y=x+2,$ and $y=\text{−}x.$

1. Show that ${\displaystyle \underset{D}{\iint }}xdA={\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \underset{\text{−}y}{\overset{\sqrt{y}}{\int }}xdxdy}}+{\displaystyle \underset{1}{\overset{2}{\int }}{\displaystyle \underset{y-2}{\overset{\sqrt{y}}{\int }}xdxdy}}$ by dividing the region $D$ into two regions of Type II, where $D=\left\{(x,y)|y\ge x^2,y\ge -x,y\le x+2\right\}.$
2. Evaluate the integral ${\displaystyle \underset{D}{\iint }x}dA.$

The region $D$ bounded by $x=0,y=x^5+1,$ and $y=3-x^2$ is shown in the following figure. Find the area $A\left(D\right)$ of the region $D.$

The region $D$ bounded by $y=\text{cos}x,y=4\text{cos}x,$ and $x=\pm \frac{\pi }{3}$ is shown in the following figure. Find the area $A\left(D\right)$ of the region $D.$

Find the area $A(D)$ of the region $D=\left\{(x,y)|y\ge 1-x^2,y\le 4-x^2,y\ge 0,x\ge 0\right\}.$

Let $D$ be the region bounded by $y=1,y=x,y=\text{ln}x,$ and the $x$ -axis. Find the area $A\left(D\right)$ of the region $D.$

Find the average value of the function $f\left(x,y\right)=\text{sin}y$ on the triangular region with vertices $\left(0,0\right),\left(0,3\right),$ and $\left(3,0\right).$

Find the average value of the function $f\left(x,y\right)=\text{−}x+1$ on the triangular region with vertices $\left(0,0\right),\left(0,2\right),$ and $\left(2,2\right).$

${\displaystyle \underset{\mathrm{-1}}{\overset{\pi \text{/}2}{\int }}{\displaystyle \underset{0}{\overset{x+1}{\int }}\text{sin}xdydx}}$

${\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \underset{x-1}{\overset{1-x}{\int }}xdydx}}$

${\displaystyle \underset{\mathrm{-1}}{\overset{0}{\int }}{\displaystyle \underset{\text{−}\sqrt{y+1}}{\overset{\sqrt{y+1}}{\int }}{y}^{2}dxdy}}$

${\displaystyle \underset{\text{−}1\text{/}2}{\overset{1\text{/}2}{\int }}{\displaystyle \underset{\text{−}\sqrt{y^2+1}}{\overset{\sqrt{y^2+1}}{\int }}ydxdy}}$