Double integral over area bounded by ellipse Evaluate the double integral of $\int \int(x+y)^{2} dxdy$ over the area bounded by the ellipse $(\frac{x}{a})^2+(\frac{y}{b})^2=1$
Please check answer = $\frac{ab\pi}{4}(a^{2}+b^{2})$

\begin{align}
I= \iint_A (x+y)^{2} dxdy = \iint_A (x^2 +y^2) dxdy + 2\iint_A xy\ dxdy
\end{align}
The latter integral vanishes due to odd symmetry in $x$ and $y$. We didn't need odd symmetry in both for the integral to vanish; just one would have sufficed.
\begin{align}
\iint_A xy\ dxdy = \int_{-b}^b\int_{-a\sqrt{1-(y/b)^2)}}^{a\sqrt{1-(y/b)^2)}} xy\ dxdy =0
\end{align}
The former integral is best evaluated in polar, but we  proceed in Cartesian. It turns into a trigonometric integral shortly. First, note the even symmetry in both $x$ and $y$, leading to simplified bounds and a factor of $2^2$ in front.
\begin{align}
I= \iint_A (x^2 +y^2) dxdy &=  \int_{-b}^b\int_{-a\sqrt{1-(y/b)^2)}}^{a\sqrt{1-(y/b)^2)}} (x^2 +y^2) dxdy
\\
 &=  4\int_0^b\int_0^{a\sqrt{1-(y/b)^2)}} (x^2 +y^2) dxdy
\\
 &=  4\int_0^b \left(
\frac13a^3\left(1-\frac{y^2}{b^2}\right)^{3/2} +ay^2\sqrt{1-\frac{y^2}{b^2}} 
\right)dy
\end{align}
After reducing to a single integral, we need to make a trig substitution. The trigonometric integrals are then evaluated through repeated use of the half angle formula. 
\begin{align}
I &= 4\int_0^{\pi/2} \left(
\frac{a^3b}{3}\cos^4t +ab^3\sin^2 t\cos^2 t 
\right)  dt
\qquad\text{(letting } \sin t=y/b)
\\
&= 4\int_0^{\pi/2} \left(
\frac{a^3b}{3}\cos^4t +\frac{ab^3}{4}\sin^2(2t) 
\right)  dt
\qquad\text{(double angle formula)}
\\
 &=  4\left(
\frac{a^3b}{3}\frac{3\pi}{16} +\frac{ab^3}{4}\frac{\pi}{4}\right) 
\qquad\text{(repeated use of half angle formulae)}
\\
 &=  \frac{ab\pi}{4}\left(a^2 +b^2\right)
\end{align}
