 Solution

For a one-to-one differentiable change of variables $(x,y)\mapsto (u,v)$, the change of variables formula iswhere

Here $u=x+y$, $v=x^2+y^2$, andAt fixed $v$, write $x=\sqrt{v}\cos \theta$, $y=\sqrt{v}\sin \theta$ with $0\le \theta \le \pi$. Then $u=\sqrt{v}(\cos \theta +\sin \theta )$, soThe constant-$v$ curves are semicircles and the constant-$u$ curves are lines $x+y=u$.

The map is two-to-one where $\sqrt{v}<u<\sqrt{2v}$, because both $(x,y)$ and $(y,x)$ lie in the upper half-disc, and one-to-one elsewhere. Since $|y-x|=\sqrt{2v-u^2}$, accounting for this multiplicity givesDirectly, symmetry makes ${\iint }_{D}xdA=0$, while polar coordinates give ${\iint }_{D}ydA={\int }_0^1{r}^{2}dr{\int }_0^{\pi }\sin \theta d\theta =2/3$.