Evaluate 2D integral (by change of variable)

Question

The question asks to evaluate integral $$\iint_D \Big[3-\frac12( \frac{x^2}{a^2}+\frac{y^2}{b^2})\Big] \, dx \, dy \ $$ where D is the region $$\frac{x^2}{a^2}+\frac{y^2}{b^2} \le 4 $$

I believe this requires me to convert it to polar coordinates, using the Jacobian etc. but it gets messy really quickly and does not lead me to anything. Any clues/ideas?

PS. The final integral I've got is

$$ \int_0^\pi \frac{a^2b^2}{(bcos(\theta))^2+(asin(\theta))^2} \, d \theta \ $$

which might even be correct, or might have some algebra messed up. but How to approach this?

Answer 1

Split the integral into to parts: from 0 to $\pi/2$ and to $\pi/2$ to $\pi$, the second is equal to the first (you can prove it by the following substitution: $t = \pi - \theta$). To calculate the first integral you can divide all by $(cos(\theta))^2$ and then make the substitution $x = tan(x)$. From there use simple fractions and you're done.

You could calculate the integral easily by $x/a = rcos(\theta)$ and $y/b = rsin(\theta)$. The Jacobian leads to $dxdy = abr$ $drd\theta$. $$ab\iint_{V} (3-1/2*r^{2})r \,dr\,d\theta$$ V is $[0,2\pi]$ x $[0,2]$. This integral is very easy to calculate.

Answer 2

You are integrating over an ellipse, which is hard, relative to integrating over a circle. So first I would change your ellipse into a circle, using $u = \frac{x}{2a}$ and $v = \frac{y}{2b}$. Then your domain of integration is $\mathcal{D}_{1}:= \left\{(u,v): u^2+v^2\le 1\right\}$, and your integral $I$ transforms to \begin{align} I= 4ab \int_{\mathcal{D_{1}}} 3 - 2(u^2 + v^2) \, \mathrm{d}u \mathrm{d}v \end{align} Now polar coordinates are easy:$u = r\cos(\theta)$, $v = r\sin(\theta)$ and \begin{align} I &= 4ab \int_{0}^{2\pi} \int_{0}^{1} (3-2r^2) r \, \mathrm{d}r \mathrm{d}\theta \\ &= 8\pi ab \int_{0}^{1} 3r - 2r^3 \, \mathrm{d}r \\ &= 8\pi ab (3/2 - 1/2) \\ &= 8\pi ab \end{align}