Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 6 - Applications of the Integral - Chapter Review Exercises - Page 319: 18


$9 \pi cm^3/s$

Work Step by Step

The flow rate $(Q)$ through the pipe can be calculated as: $Q=2 \pi \int_{0}^{a} r \ v(r) \ dr$ Now, $Q= 2 \pi \int_{0}^{3} r (3-r) \ dr \\=2 \pi \int_0^3 (-r^2+3r) \ dr \\=2 \pi [-\dfrac{r^3}{3}+\dfrac{3r^2}{2}]_0^{3} \\=2 \pi \times [-\dfrac{(3)^3}{3}+\dfrac{3(3^2)}{2}] \\=2 \pi [-\dfrac{27}{2} -9] \\=9 \pi cm^3/s$
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.