Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 16 - Review - Exercises - Page 1150: 28


$32 \pi \sqrt 3$

Work Step by Step

The surface area is calculated as: $S=\iint_S dS=\iint_D\sqrt {1+z_x^2 +z_y^2} dA$ Given: $z=4+x+y$ Thus, $dS=\iint_D\sqrt {1+(1)^2 +(1)^2} dy dx=\iint_D\sqrt {3} dy dx$ Now, $\iint_S x dS=\iint_D x \sqrt {3} dy dx=\iint_D (x^2+y^2) \sqrt {4+x+y}\sqrt 3 dx dy$ Now, in polar co-ordinates we have $\iint_S x dS=\int_0^{2\pi}\int_0^{2}r^2(4+r \cos \theta+r \sin \theta)\sqrt 3r dr d\theta=\sqrt 3 \int_0^{2\pi}\int_0^2 4r^3+r^4\cos \theta+r^4 \sin \theta dr d \theta$ Then $\iint_S x dS=\sqrt 3 \int_0^{2 \pi} [r^4+\dfrac{r^5(\cos \theta+\sin \theta)}{5}]_0^2 d\theta$ or, $=\sqrt 3 [16 \theta+\dfrac{32(\sin \theta-\cos \theta)}{5}]_0^{2 \pi}$ Hence, $\iint_S x^2 z+y^2 z dS=32 \pi \sqrt 3$
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