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 1061: 53

Answer

$\dfrac{\pi}{2}$

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: $$\ Volume =\iint_{D} f(x,y) \ dA$$ $f(x,y) =1-x^2-y^2$ The plane and paraboloid intersect along the circle $x^2+y^2=1$ and $D$ is the projection of the surface on the xy-plane. Thus, we can express the region $D$ using the point of intersection as follows: $$D=\left\{ (x, y) | -1 \leq x \leq 1 , -\sqrt {1-x^2} \leq y \leq \sqrt {1-x^2}\right\}$$ Therefore, $$\ Volume =\iint_{D} f(x,y) \ dA\\=\iint_{D} (1-x^2-y^2) \ dA \\= \int_{-1}^{1} \int_{-\sqrt {1-x^2}}^{\sqrt {1-x^2}} (1-x^2-y^2) \ dy \ dx $$ Now, we will use a calculator to compute the integral: $\ Volume = \int_{-1}^{1} \int_{-\sqrt {1-x^2}}^{\sqrt {1-x^2}} (1-x^2-y^2) \ dy \ dx \approx \dfrac{\pi}{2}$
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