Calculus: Early Transcendentals 8th Edition

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

Chapter 6 - Section 6.2 - Volume - 6.2 Exercises - Page 447: 33

Answer

a.) $V=78.95684$ b.) $V=78.95684$

Work Step by Step

a.) $x^2+4y^2=4\\ x^2+4(0)^2=4\\ x^2=4\\ x=-2 \qquad x=2$ $\displaystyle{A(x)=\pi\left(2+\sqrt{\frac{4-x^2}{4}}\right)^2-\pi\left(2-\sqrt{\frac{4-x^2}{4}}\right)^2}\\ \displaystyle{A(x)=\pi\left(4\sqrt{4-x^2}\right)}$ $\begin{aligned} V &=\int_{-2}^{2} A(x) \ d x \\ V &=\int_{-2}^{2} \pi\left(4\sqrt{4-x^2}\right) \ d x \\ V &=4\pi \int_{-2}^{2} \sqrt{4-x^2}\ dx \\ V &=8\pi \int_{0}^{2} \sqrt{4-x^2}\ dx \\ V &=78.95684 \end{aligned}$ b.) $x^2+4y^2=4\\ (0)^2+4y^2=4\\ y^2=1\\ y=-1 \qquad y=1$ $\displaystyle{A(y)=\pi\left(2+\sqrt{4-4y^2}\right)^2-\pi\left(2-\sqrt{4-4y^2}\right)^2}\\ \displaystyle{A(y)=\pi\left(8\sqrt{4-4y^2}\right)}$ $\begin{aligned} V &=\int_{-1}^{1} A(y) \ d y \\ V &=\int_{-1}^{1} \pi\left(8\sqrt{4-4y^2}\right) \ d y \\ V &=8\pi \int_{-1}^{1} \sqrt{4-4y^2}\ dy \\ V &=16\pi \int_{0}^{1} \sqrt{4-4y^2}\ dy \\ V &=78.95684 \end{aligned}$
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