# Using polar coordinates to solve a double integral

I am working through some double integral questions in practice for my upcoming exam. The question I am stuck on is the following: $\iint_R x^2y^2dxdy$ where $R \equiv x^2/a^2 + y^2/b^2 = 1$. I am told to use $x=ar\cos(\theta)$ and $y = br\sin(\theta)$ and polar coordinates. I am more interested in the method than the answer if anyone could be kind enough to shed some light!

So far I have tried substituting the $x$ and $y$ equations into the equation for region R which simplified to give $r^2=1$. I then substituted the $x$ and $y$ equations into the original integral to give me: $a^2b^2\int^{\pi}_0\cos^2(\theta)\sin^2(\theta)\int^1_0r^4$ - I am slightly apprehensive about the limits, I cant help thinking I need to set the r integral from -1 to 1 but cant see the logic behind this? I am also confused about changing the variable of integration because I can only take partial derivatives of the $x$ and $y$ equations?

Many thanks for any help!

You're on the right track, but by the multivariate change of variables formula you need to account for the Jacobian matrix of this change of variables and tweak your bounds a bit, //i.e.//

\begin{align*} \iint_Rx^2y^2\,dxdy &= \iint_{\{(r,\theta) : r = 1\}} (ar\cos\theta)^2(br\sin\theta)^2\left|\begin{pmatrix}\frac{\partial}{\partial r}(ar\cos\theta) & \frac{\partial}{\partial r}(br\sin\theta) \\ \frac{\partial}{\partial \theta}(ar\cos\theta) & \frac{\partial}{\partial \theta}(br\sin\theta)\end{pmatrix}\right|\,dr d\theta\\ &= \int_0^{2\pi}\int_0^1 a^3b^3r^5\cos^2\theta\sin^2\theta\,drd\theta\\ &= \ldots \end{align*}

Note here that you can make the $\theta$ bound range from $0$ to $2\pi$ (not just $\pi$) to account for the whole disk $R$ rather than think about negative $r$!

• Thank you! I will read into it
– Tech
Apr 12 '16 at 16:03
• After working through it using the link you kindly provided I had an r^5 term not an r^3 - I got r^4 from the expansion of the brackets plus an abr term from the change of variables - Have I missed something?
– Tech
Apr 13 '16 at 14:26
• Nope! Good catch! That should be an $r^5$. Sorry for the typo!
– Dan
Apr 13 '16 at 14:28
• Ah okay! Many thanks again for helping me with this!
– Tech
Apr 13 '16 at 14:29