Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 17 - Line and Surface Integrals - 17.4 Parametrized Surfaces and Surface Integrals - Exercises - Page 959: 39

Answer

$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_G^{} {x^2}z{\rm{d}}S = 48\pi $

Work Step by Step

Write $f\left( {x,y,z} \right) = {x^2}z$. There are three faces that build up the total surface of the cylinder. These are the top, the bottom and the side of the cylinder. So, the integral over the cylinder is equal to $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_G^{} {x^2}z{\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_{bottom}}}^{} f\left( {x,y,z} \right)||{\bf{N}}\left( {x,y} \right)||{\rm{d}}x{\rm{d}}y$ $ + \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_{top}}}^{} f\left( {x,y,z} \right)||{\bf{N}}\left( {x,y} \right)||{\rm{d}}x{\rm{d}}y$ $ + \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_{side}}}^{} f\left( {x,y,z} \right)||{\bf{N}}\left( {x,y} \right)||{\rm{d}}x{\rm{d}}y$ 1. The bottom: $z=0$ (the $xy$-plane), $||{\bf{N}}\left( {x,y} \right)|| = || - {\bf{k}}|| = 1$ Since $z=0$, so $f\left( {x,y,z} \right) = 0$. $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_{bottom}}}^{} f\left( {x,y,z} \right)||{\bf{N}}\left( {x,y} \right)||{\rm{d}}x{\rm{d}}y = 0$ 2. The top: $z=3$, $f\left( {x,y,z} \right) = 3{x^2}$, $||{\bf{N}}\left( {x,y} \right)|| = ||{\bf{k}}|| = 1$ $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_{top}}}^{} f\left( {x,y,z} \right)||{\bf{N}}\left( {x,y} \right)||{\rm{d}}x{\rm{d}}y = 3\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_{top}}}^{} {x^2}{\rm{d}}x{\rm{d}}y$ Using $x = r\cos \theta $ and $y = r\sin \theta $, we evaluate the integral using polar coordinates: $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_{top}}}^{} f\left( {x,y,z} \right)||{\bf{N}}\left( {x,y} \right)||{\rm{d}}x{\rm{d}}y = 3\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_{top}}}^{} {x^2}{\rm{d}}x{\rm{d}}y$ $ = 3\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^2 \left( {{r^2}{{\cos }^2}\theta } \right)r{\rm{d}}r{\rm{d}}\theta = 3\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } {{\cos }^2}\theta {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^2 {r^3}{\rm{d}}r} \right)$ Using the identity $\cos 2\theta = 2{\cos ^2}\theta - 1$, we get $ = \frac{3}{2}\left[ {\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {1 + \cos 2\theta } \right){\rm{d}}\theta } \right]\left( {\mathop \smallint \limits_{r = 0}^2 {r^3}{\rm{d}}r} \right)$ $ = \frac{3}{8}\left[ {\left( {\theta + \frac{1}{2}\sin 2\theta } \right)|_0^{2\pi }} \right]\left( {{r^4}|_0^2} \right) = \frac{3}{8}\left( {2\pi } \right)\left( {16} \right) = 12\pi $ So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_{top}}}^{} f\left( {x,y,z} \right)||{\bf{N}}\left( {x,y} \right)||{\rm{d}}x{\rm{d}}y = 12\pi $. 3. The side: $r=2$, $0 \le \theta \le 2\pi $, $0 \le z \le 3$, $f\left( {x,y,z} \right) = {x^2}z$ We evaluate the integral using cylindrical coordinates. We know from Section 14.6 on page 749, in polar coordinates the unit normal vector is given by ${{\bf{e}}_r} = \left( {\cos \theta ,\sin \theta } \right)$. So, the normal vector to the side of the cylinder is parallel to the $xy$-plane given by ${\bf{N}}\left( {\theta ,z} \right) = \left( {2\cos \theta ,2\sin \theta ,0} \right)$ So, we have $||{\bf{N}}\left( {\theta ,z} \right)|| = \sqrt {4{{\cos }^2}\theta + 4{{\sin }^2}\theta } = 2$. $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_{side}}}^{} f\left( {x,y,z} \right)||{\bf{N}}\left( {x,y} \right)||{\rm{d}}x{\rm{d}}y = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_{side}}}^{} f\left( {\theta ,z} \right)||{\bf{N}}\left( {\theta ,z} \right)||{\rm{d}}\theta {\rm{d}}z$ $ = 2\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{z = 0}^3 \left( {4{{\cos }^2}\theta } \right)z{\rm{d}}z{\rm{d}}\theta = 8\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } {{\cos }^2}\theta {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{z = 0}^3 z{\rm{d}}z} \right)$ $ = 4\left[ {\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {1 + \cos 2\theta } \right){\rm{d}}\theta } \right]\left( {\mathop \smallint \limits_{z = 0}^3 z{\rm{d}}z} \right)$ $ = 4\left[ {\left( {\theta + \frac{1}{2}\sin 2\theta } \right)|_0^{2\pi }} \right]\left( {\frac{1}{2}{z^2}|_0^3} \right) = 4\left( {2\pi } \right)\left( {\frac{9}{2}} \right) = 36\pi $ So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_{side}}}^{} f\left( {x,y,z} \right)||{\bf{N}}\left( {x,y} \right)||{\rm{d}}x{\rm{d}}y = 36\pi $. Finally, we obtain the integral over the cylinder: $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_G^{} {x^2}z{\rm{d}}S = 0 + 12\pi + 36\pi = 48\pi $
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