Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 15 - Section 15.2 - Double Integrals over General Regions - 15.2 Exercise - Page 1060: 40

Answer

$$\dfrac{16r^3}{3}$$

Work Step by Step

The volume under the surface $z=f(x,y)$ and above the region $D$ in the xy-plane can be defined as: $$\ Volume =\iint_{D} f(x,y) \ dA$$ We have: $D$ is a circle having a radius $r$ with center at the origin. Thus, we can express the region $D$ as follows: $$D=\left\{ (x, y) | -r \leq x \leq r, \ -\sqrt {r^2-x^2} \leq y \leq \sqrt {r^2-x^2} \right\} $$ Therefore, $$\ Volume =\iint_{D} f(x,y) \ dA \\=2 \int_{-r}^{r} \int_{-\sqrt {r^2-x^2} }^{\sqrt {r^2-x^2} } \sqrt {r^2-y^2} dy \ dx \\ = 2\int_{-r}^{r} [\dfrac{r}{2} \sqrt {r^2-y^2}+\dfrac{r^2}{2} \arcsin (y/r) ]_{-\sqrt {r^2-x^2} }^{\sqrt {r^2-x^2} } dx \\ =4 \int_{0}^{r}x \sqrt {r^2-x^2} +r^2 \arccos (x/r) \ dx \\=\dfrac{4r^3}{3}+4r^3\\=\dfrac{16r^3}{3}$$
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