Thomas' Calculus 13th Edition

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

Chapter 15: Multiple Integrals - Section 15.5 - Triple Integrals in Rectangular Coordinates - Exercises 15.5 - Page 902: 31

Answer

$8\pi-\frac{32}{3}$

Work Step by Step

$ V=\int^4_0 \int^{\sqrt{16-y^2}/2}_0 \int^{4-y}_0$ dx dz dy =$\int^4_0 \int^{\sqrt{16-y^2}/2}_0(4-y)dz $ dy =$\int^4_0 \frac{\sqrt{16-y^2}}{2}(4-y)$ dy =$\int^4_0 2\sqrt{16-y^2}$ dy $-\frac{1}{2}\int^4_0 y\sqrt{16-y^2}$ dy =$[y\sqrt{16-y^2}+16sin^{-1}\frac{y}{4}]^4_0+[\frac{1}{6}(16-y^2)^{3/2}]^4_0$ =$16(\frac{\pi}{2}-\frac{1}{6}(16)^{3/2})$ =$8\pi -\frac{32}{3}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.