Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 15 - Multiple Integrals - 15.3 Exercises - Page 1020: 30

Answer

$\dfrac{16}{3}$

Work Step by Step

Since, $z \gt 0$ because it is in the first quadrant, so we have: $z=f(x,y)=\sqrt {4-y^2}$ The volume under the surface $z=f(x,y)$ and above the region $D$ in the xy-plane can be defined as: $$\ Volume =\iint_{D} f(x,y) \ dA$$ $D=\left\{ (x, y) | 0 \leq x \leq 2, \ 0 \leq x \leq 2y \right\} $ Now, $V =\iint_{D} f(x,y) \ dA \\=\int_{0}^{2} \int_{0}^{2y} \sqrt{4-y^2} \ dx \ dy \\ = \int_{0}^{2} 2y \sqrt{4-y^2} \ dy $ Consider $4-y^2=a$ and $ -da= 2y dy$ Now, $V=\int_{4}^{0} -\sqrt a da \\=-[\dfrac{2a^{3/2}}{3}]_0^4 \\= \dfrac{2}{3} \times [ (4)^{3/2}] \\=\dfrac{16}{3}$
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