I have to prove that $$\int_D xy\; dxdy=\frac{3}{8}$$ with $D=\{(x,y)\in \mathbb{R}^2: x^2+y^2 \geq 1,\ \frac{x^2}{4}+y^2 \leq 1,\ x \geq 0,\ y \geq 0 \}$. So I have the intersection of ellipse $\frac{x^2}{4}+y^2 \leq 1$ and the ball $x^2+y^2 \leq 1$. If I use the sets $$\begin{eqnarray} P&=&\{(x,y)\in\mathbb{R}^2: x^2+y^2 \leq 1,\ x \geq 0,\ y \geq 0 \}\\ E&=&\Big\{(x,y)\in\mathbb{R}^2: \frac{x^2}{4}+y^2 \leq 1,\ x \geq 0,\ y \geq 0 \Big\} \end{eqnarray}$$ I obtain that $\displaystyle \int_D xy\; dxdy=\int_E xy\; dxdy - \int_P xy\; dxdy$.
The integral $\int_E xy\; dxdy$ can be calculated whith the elliptic coordinates $x=2\rho\cos\theta$, $y=\rho \sin \theta$ and the set E became $$\Big\{\rho \leq 1,\ \theta \in \Big[0,\frac{\pi}{2}\Big]\Big\}.$$ So $\displaystyle\int_E xy\; dxdy =\int_0^1 2\rho^3\; d\rho \int_0^{\frac{\pi}{2}}\cos\theta\sin\theta\;d\theta =\left.2\frac{\rho^4}{4}\right|^1_0 \cdot \left.\Big(-\frac{1}{2}\cos^2\theta\Big)\right|_0^{\pi/2} =\frac{1}{4}\cdot$
In the same way we can use the coordinate $x=\rho \cos\theta$, $y=\rho \sin \theta$ and the set P became $$\Big\{\rho\leq 1,\ \theta\in\Big[0,\frac{\pi}{2}\Big]\Big\}.$$ So $\displaystyle\int_P xy\; dxdy=\int_0^1 \rho^\; d\rho \int_0^{\frac{\pi}{2}}\cos\theta\sin\theta\; d\theta =\left.\frac{\rho^4}{4}\right|^1_0 \cdot \left.\Big(-\frac{1}{2}\cos^2\theta\Big)\right|_0^{\pi/2} =\frac{1}{8}\cdot$
Then $\displaystyle\int_D xy\; dxdy=\frac{1}{4}-\frac{1}{8}=\frac{1}{8}\cdot$ Where am I doing wrong?
