<s>[INST] 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 = $\frac{ab\pi}{4}(a^{2}+b^{2})$
 [/INST] If we are going to do this without using a transformation. Notice our regions boundary is,
$$(\frac{x}{a})^2+(\frac{y}{b})^2=1$$
So that $x \in [-a\sqrt{1-(\frac{y}{b})^2},a\sqrt{1-(\frac{y}{b})^2}]$ and of course $y \in [-b,b]$. So we are interested in computing,
$$\int_{-b}^{b} \int_{-a\sqrt{1-(\frac{y}{b})^2}}^{a\sqrt{1-(\frac{y}{b})^2}} (x^2+2xy+y^2) dx dy$$
We integrate with respect to $x$ first. Notice that $2xy$ is odd with respect to $x$ so we can drop that term. So then what we get is,
$$=\int_{-b}^{b} \left( \frac{2}{3}(a\sqrt{1-(\frac{y}{b})^2})^3+y^2\left(2a\sqrt{1-(\frac{y}{b})^2}\right) \right) dy$$
$$=\int_{-b}^{b} \left(\frac{2}{3}(a\sqrt{1-(\frac{y}{b})^2})a^2\left(1-(\frac{y}{b})^2\right)+y^2\left(2a\sqrt{1-(\frac{y}{b})^2}\right) \right) dy$$
First let $\frac{y}{b}=u$ so $dy=bdu$ and we have that what we want is the sum of,
$$I_1=\frac{2}{3}a^3b \int_{-1}^{1}(1-u^2)\sqrt{1-u^2} du$$
$$I_2=2ab^3\int_{-1}^{1} u^2\sqrt{1-u^2} du$$
The integrals in each of these are both standard, and related integrals. Find one, you find the other by geometry.
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