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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!

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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 $r$!

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  • $\begingroup$ Thank you! I will read into it $\endgroup$
    – Tech
    Apr 12 '16 at 16:03
  • $\begingroup$ 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? $\endgroup$
    – Tech
    Apr 13 '16 at 14:26
  • $\begingroup$ Nope! Good catch! That should be an $r^5$. Sorry for the typo! $\endgroup$
    – Dan
    Apr 13 '16 at 14:28
  • $\begingroup$ Ah okay! Many thanks again for helping me with this! $\endgroup$
    – Tech
    Apr 13 '16 at 14:29

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