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\begin{document}


\section{Ma 227 Homework 5 Solutions Fall 2007\newline
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Due 10/11/2007$\qquad \ $}

\vspace{1pt}{}

Pg. 842 Problems 5,7,21, 23

\vspace{1pt}

5)

$\int_{0}^{2}\int_{0}^{\frac{\pi }{2}}x\sin ydydx=\int_{0}^{2}xdx\int_{0}^{%
\frac{\pi }{2}}\sin ydy=\left[ \frac{x^{2}}{2}\right] _{0}^{2}\left[ -\cos y%
\right] _{0}^{\frac{\pi }{2}}=\left( 2-0\right) \left( 0+1\right) =2$

\vspace{1pt}

7)

$\int_{0}^{2}\int_{0}^{1}\left( 2x+y\right) ^{8}dxdy=\int_{0}^{2}\left[ 
\frac{1}{2}\frac{(2x+y)^{9}}{9}\right] _{x=0}^{x=1}dy$

\vspace{1pt}

Use substitituon: $u=2x+y\Longrightarrow dx=\frac{1}{2}du$

\vspace{1pt}

$\frac{1}{18}\int_{0}^{2}\left[ \left( 2+y\right) ^{9}-(0+y)^{9}\right] dy=%
\frac{1}{18}\left[ \frac{\left( 2+y\right) ^{10}}{10}-\frac{y^{10}}{10}%
\right] _{0}^{2}=\frac{1}{18}\left[ \left( 4^{10}-2^{10}\right) -\left(
2^{10}-0^{10}\right) \right] $

$=\frac{1,046,528}{180}=\allowbreak \frac{261,632}{45}$

\vspace{1pt}

\vspace{1pt}

21) Find the volume of the solid that lies under the plane $3x+2y+z=12$\ and
above the rectangle $R=\{(x,y)|0\leq x\leq 1,-2\leq y\leq 3\}$

\vspace{1pt}

$z=12-3x-2y$

\vspace{1pt}$V=\int
\int_{R}(12-3x-2y)dA=\int_{-2}^{3}\int_{0}^{1}(12-3x-2y)dxdy=$ $\int_{-2}^{3}%
\left[ 12x-\frac{3}{2}x^{2}-2xy\right] _{x=0}^{x=1}dy=\int_{-2}^{3}(\frac{21%
}{2}-2y)dy$

$\vspace{1pt}$

$=\left[ \frac{21}{2}y-2y^{2}\right] _{-2}^{3}=\frac{95}{2}$

\vspace{1pt}

\vspace{1pt}23) Find the volume of the solid that lies under the elliptic
paraboloid $\frac{x^{2}}{4}+\frac{y^{2}}{9}+z=1$\ and above the rectangle $%
R=[-1,1]\times \lbrack -2,2]$

$z=1-\frac{x^{2}}{4}-\frac{y^{2}}{9}$

$V=$\vspace{1pt}$\int \int_{R}\left( 1-\frac{x^{2}}{4}-\frac{y^{2}}{9}%
\right) dV=$\vspace{1pt}$\int_{-2}^{2}\int_{-1}^{1}\left( 1-\frac{x^{2}}{4}-%
\frac{y^{2}}{9}\right) dxdy$ $=4\int_{0}^{2}\int_{0}^{1}\left( 1-\frac{x^{2}%
}{4}-\frac{y^{2}}{9}\right) dxdy$

$=4\int_{0}^{2}\left[ x-\frac{1}{12}x^{3}-\frac{1}{9}y^{2}x\right]
_{x=0}^{x=1}dy=4\int_{0}^{2}(\frac{11}{12}-\frac{1}{9}y^{2})dy=4\left[ \frac{%
11}{12}y-\frac{1}{27}y^{3}\right] _{0}^{2}=4\cdot \frac{83}{54}=\frac{166}{27%
}$

\vspace{1pt}

Pg. 850 Problems 5, 7,11,15,19,33,37

\vspace{1pt}

\vspace{1pt}5)

\vspace{1pt}$\int_{0}^{\frac{\pi }{2}}\int_{0}^{\cos \theta }e^{\sin \theta
}drd\theta =\int_{0}^{\frac{\pi }{2}}re^{\sin \theta }]_{0}^{\cos \theta
}d\theta =\int_{0}^{\frac{\pi }{2}}\cos \theta e^{\sin \theta }d\theta
=e^{\sin \theta }]_{0}^{\frac{\pi }{2}}=e-1$

\vspace{1pt}

7)

\vspace{1pt}$\int
\int_{D}x^{3}y^{2}dA=\int_{0}^{2}\int_{-x}^{x}x^{3}y^{2}dydx=\int_{0}^{2}%
\left[ \allowbreak \frac{1}{3}x^{3}y^{3}\right] _{-x}^{x}=\int_{0}^{2}\frac{2%
}{3}x^{6}dx=\allowbreak \left[ \frac{2}{21}x^{7}\right] _{0}^{2}=\frac{256}{%
21}$

\vspace{1pt}

11)

\vspace{1pt}

$\int \int_{D}x\cos y$ $dA$\vspace{1pt}\qquad $D:y=0,y=x^{2},x=1$

\vspace{1pt}

$=\int_{0}^{1}\int_{0}^{x^{2}}(x\cos y)$ $dxdy=\int_{0}^{1}\left( x\sin
y\right) ]_{0}^{x^{2}}dx=\int_{0}^{1}(x\sin x^{2})dx=-\frac{1}{2}\cos
x^{2}]_{0}^{1}=-\frac{1}{2}\cos (1)+\frac{1}{2}=\frac{1}{2}(1-\cos (1))$

\vspace{1pt}

\vspace{1pt}

15)

$\int \int_{D}(2x-y)dA\qquad D:$\ $\ \ x^{2}+y^{2}=4$

The graph of $D$ is shown below. Graph 2D Polare was used to get it.

$r=2$\FRAME{dtbpFX}{4.4996in}{3in}{0pt}{}{}{Plot}{\special{language
"Scientific Word";type "MAPLEPLOT";width 4.4996in;height 3in;depth
0pt;display "USEDEF";plot_snapshots FALSE;mustRecompute FALSE;lastEngine
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"2";yviewmin "-2.00014832079422";yviewmax "2.00014832079422";plottype
8;axesFont "Times New Roman,12,0000000000,useDefault,normal";constrained
TRUE;numpoints 100;plotstyle "patch";axesstyle "normal";axestips
TRUE;gridLines TRUE;xis \TEXUX{v952};var1name \TEXUX{$\U{3b8} $};function
\TEXUX{$2$};linecolor "black";linestyle 1;pointstyle "point";linethickness
2;lineAttributes "Solid";coordinateSystem "polar";var1range
"-3.1416,3.1416";num-x-gridlines 100;curveColor
"[flat::RGB:0000000000]";curveStyle "Line";VCamFile 'JPN2MO0C.xvz';}}

$=\int_{-2}^{2}\int_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}}(2x-y)dydx=%
\int_{-2}^{2}(2xy-\frac{1}{2}y^{2})]_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}}$ $dx$

$=\int_{-2}^{2}2x\sqrt{4-x^{2}}-\frac{1}{2}(4-x^{2})-2x(-\sqrt{4-x^{2}})+%
\frac{1}{2}(4-x^{2})$ $dx=\int_{-2}^{2}(4x\sqrt{4-x^{2}})dx=\frac{4}{3}%
(4-x^{2})^{\frac{3}{2}}]_{-2}^{2}=0$

\vspace{1pt}

19)

$V=\int_{1}^{2}\int_{1}^{7-3y}xydxdy=\int_{1}^{2}[\frac{1}{2}%
x^{2}y]_{x=1}^{x=7-3y}dy=\frac{1}{2}\int_{1}^{2}(48y-42y^{2}+9y^{3})dy=\frac{%
1}{2}[24y^{2}-14y^{3}+\frac{9}{4}y^{4}]_{1}^{2}=\frac{31}{8}$

\vspace{1pt}

33)

\vspace{1pt}

$\sqrt{x}$\FRAME{dtbpFX}{3in}{2.0003in}{0pt}{}{}{Plot}{\special{language
"Scientific Word";type "MAPLEPLOT";width 3in;height 2.0003in;depth
0pt;display "USEDEF";plot_snapshots FALSE;mustRecompute FALSE;lastEngine
"MuPAD";xmin "0";xmax "4.2";xviewmin "0";xviewmax "4.2";yviewmin
"0";yviewmax "2";viewset"XY";rangeset"X";plottype 4;axesFont "Times New
Roman,12,0000000000,useDefault,normal";constrained TRUE;numpoints
49;plotstyle "patch";axesstyle "normal";axestips TRUE;gridLines TRUE;xis
\TEXUX{x};yis \TEXUX{y};var1name \TEXUX{$x$};var2name \TEXUX{$y$};function
\TEXUX{$\sqrt{x}$};linecolor "black";linestyle 1;pointstyle
"point";linethickness 2;lineAttributes "Solid";var1range
"0,4.2";num-x-gridlines 49;curveColor "[flat::RGB:0000000000]";curveStyle
"Line";VCamFile 'JPN2MO0B.xvz';}}

\vspace{1pt}Because the region of integration is

$D=\{(x,y)|0\leq y\leq \sqrt{x},0\leq x\leq 4\}=\{(x,y)|y^{2}\leq x\leq
4,0\leq y\leq 2\}$

we have

$\int_{0}^{4}\int_{0}^{\sqrt{x}}f(x,y)dydx=\int
\int_{D}f(x,y)dA=\int_{0}^{2}\int_{y^{2}}^{4}f(x,y)dxdy$

\vspace{1pt}

37)

\FRAME{dtbpFX}{3in}{2.0003in}{0pt}{}{}{Plot}{\special{language "Scientific
Word";type "MAPLEPLOT";width 3in;height 2.0003in;depth 0pt;display
"USEDEF";plot_snapshots FALSE;mustRecompute FALSE;lastEngine "MuPAD";xmin
"0";xmax "3";xviewmin "0";xviewmax "3";yviewmin "0";yviewmax
"2";viewset"XY";rangeset"X";plottype 4;axesFont "Times New
Roman,12,0000000000,useDefault,normal";numpoints 49;plotstyle
"patch";axesstyle "normal";axestips TRUE;gridLines TRUE;xis \TEXUX{x};yis
\TEXUX{y};var1name \TEXUX{$x$};var2name \TEXUX{$y$};function \TEXUX{$\ln
x$};linecolor "black";linestyle 1;pointstyle "point";linethickness
2;lineAttributes "Solid";var1range "0,3";num-x-gridlines 49;curveColor
"[flat::RGB:0000000000]";curveStyle "Line";VCamFile 'JPN2MO0A.xvz';}}

Because the region of integration is

$D=\{(x,y)|0\leq y\leq \ln x,1\leq x\leq 2\}=\{(x,y)|e^{y}\leq x\leq 2,0\leq
y\leq \ln 2\}$

we have

$\int_{1}^{2}\int_{0}^{\ln x}f(x,y)dydx=\int \int_{D}y\cos
x^{2}dA=\int_{0}^{\ln 2}\int_{e^{y}}^{2}f(x,y)dydx$

\end{document}
