Chapter 15.4, #23

\int^{0}_{1}\int^{y}_{\sqrt{2-y^{2}}}(x+y)dydx

From the limits of integration, we see that y is from 0 to 1. Moreover, x=y, which implies \pi/4, and x is from 0 to \sqrt{2}.

\int^{\pi/4}_{0}\int^{\sqrt{2}}_{0}(rcos\theta+rsin\theta)rdrd\theta=
\int^{\pi/4}_{0}\int^{\sqrt{2}}_{0}(r^{2}cos\theta+r^{2}sin\theta)drd\theta=
\int^{\pi/4}_{0}\frac{1}{3}r^{3}(cos\theta+sin\theta\left. \right|_{0}^{\sqrt{2}})d\theta=
\frac{1}{3}\int^{\pi/4}_{0}(2\sqrt{2}cos\theta+2\sqrt{2}sin\theta)d\theta=
\frac{1}{3}(2\sqrt{2}sin\theta-2\sqrt{2}cos\theta\left. \right|_{0}^{\pi/4})=
\frac{1}{3}[(2-2)-(0-2\sqrt{2})]=
\frac{2\sqrt{2}}{3}

This is getting troublesome, as I don't use LATEX enough to remember key markup bits.