Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 16: Integrals and Vector Fields - Section 16.6 - Surface Integrals - Exercises 16.6 - Page 1000: 2



Work Step by Step

Consider $\vec{r} (x,y) =x i+y j+\sqrt {4-y^2} k$ So, $\vec{r_x} \times \vec{r_y}=\dfrac{y j }{\sqrt {4-y^2}} + k$ and $|\vec{r_x} \times \vec{r_y}| =\sqrt {\dfrac{y^2 }{4-y^2}+1}=\dfrac{2}{\sqrt {4-y^2}}$ Now, $\iint_{S} 6 (x,y,z) \ d \theta= \int_{1}^2 \int_{0}^3 6 (x,y,z) \ d \theta $ or, $=\int_{1}^2 \int_{-2}^{2} \sqrt {4-y^2} (\dfrac{2}{\sqrt {4-y^2}}) \ dy \ dx$ By using a calculator, we evaluate the integral as : $\int_{1}^2 \int_{-2}^{2} \sqrt {4-y^2} (\dfrac{2}{\sqrt {4-y^2}}) \ dy \ dx=24$
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