# Solving $\iint\frac{x^2}{(8x^2+6y^2)^{\frac 3 2}}$ on the domain $8x^2+6y^2\leq 1$

I need to solve $\displaystyle\iint\frac{x^2}{(8x^2+6y^2)^{\frac 3 2}}$ on the domain $8x^2+6y^2\leq 1$.

I recognise this is an improper integral, so we need a monotonic series of domains $D_n\rightarrow D$ on which we'll calculate the integral. Can I have $D_n=8x^2+y^2+1/n$?

The integral is difficult to solve. I guess there's an easier way to approach it.

• take $x = (\sqrt 3 / 2) u,$ $y=v.$ Then polar coordinates. This does not change behavior about convergence and gives a more familiar setting. – Will Jagy Dec 30 '15 at 18:53
• @WillJagy Good tip, thank you! – Whyka Dec 30 '15 at 19:01

Consider the transformation $$(x,y) = \left(\frac{r}{\sqrt{8}} \cos \theta, \frac{r}{\sqrt{6}} \sin \theta\right),$$ which results in a much simpler integrand. You will notice that the Jacobian $4 \sqrt{3} r$ will cancel out the singularity at $0$.
• Thanks! I think the Jacobian should be $\frac r {4 \sqrt{3}}$ though – Whyka Dec 30 '15 at 19:21