<s>[INST] Integral in two variables 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?
 [/INST] It's better do the integral in Cartesian anyway
$$\int_0^1 \int_{\sqrt{1-y^2}}^{2\sqrt{1-y^2}}xy\:dxdy = \int_0^1\frac{3}{2}\left(y-y^3\right)\:dy = \frac{3}{4} - \frac{3}{8} = \frac{3}{8}$$
Don't always immediately go to polar coordinates whenever you see conic sections. The integrand made this nice since we had odd powers of $x$ and $y$ which, when integrated, will cancel out the square roots and give us nice polynomials.
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