Multivariable Calculus, 7th Edition

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

Chapter 11 - Multiple Integrals - 15.2 Exercises - Page 1011: 12

Answer

$\displaystyle \frac{4\sqrt{2}-2}{15}$

Work Step by Step

$\displaystyle \int_{0}^{1}\int_{0}^{1}xy\sqrt{x^{2}+y^{2}}dydx=\int_{0}^{1}[\int_{0}^{1}xy\sqrt{x^{2}+y^{2}}dy]dx=\left[\begin{array}{ll} t=x^{2}+y^{2} & \\ dt=2ydy & ydy=dt/2\\ y=0\Rightarrow t=x^{2}, & y=1\Rightarrow t=x^{2}+1 \end{array}\right]$ ... treat $x$ as a constant in the inner integral. $=\displaystyle \int_{0}^{1}[\frac{x}{2}\int_{x^{2}}^{x^{2}+1}t^{1/2}dt]dx$ $=\displaystyle \int_{0}^{1}\frac{x}{2}\cdot[\frac{t^{3/2}}{3/2}]_{x^{2}}^{x^{2}+1}dx$ $=\displaystyle \frac{1}{3}\int_{0}^{1}x[(x^{2}+1)^{3/2}-x^{3}]dx$ $=\displaystyle \frac{1}{3}\int_{0}^{1}[x(x^{2}+1)^{3/2}-x^{4}]dx=\frac{1}{3}\{\int_{0}^{1}x(x^{2}+1)^{3/2}dx-\int_{0}^{1}x^{4}dx\}$ ... for the first integral, $\left[\begin{array}{ll} t=x^{2}+1 & \\ dt=2xdx & \\ x=0\Rightarrow t=1, & x=1\Rightarrow t=2 \end{array}\right]$ $=\displaystyle \frac{1}{3}\{\int_{1}^{2}\frac{1}{2}t^{3/2}dt-[\frac{x^{5}}{5}]_{0}^{1}\}$ $=\displaystyle \frac{1}{3}\{[\frac{t^{5/2}}{2\cdot(5/2)}]_{1}^{2}-[\frac{1-0}{5}] \}$ $=\displaystyle \frac{1}{15}[2^{5/2}-1-1]$ $=\displaystyle \frac{2}{15}(2\sqrt{2}-1)$ $=\displaystyle \frac{4\sqrt{2}-2}{15}$
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.