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I was given this question in class and I was just wondering if I am on the right track… Evaluate: $$I=\iint\left(1-\frac{x^2}{a^2} -\frac{y^2}{b^2} \right)^{3/2} dxdy $$ over the region enclosed by ellipse $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} =1 .$$

First I changed $x=au$ and $y=bv$. Therefore boundary -> $u^2+v^2 \leq1$.

Also I changes $I$ to

$$ I=\iint(1-u^2-v^2)^{3/2} ab \;dudv $$ From here I switched to polar coordinates and solved the integral. $$ ab\int_0^{2\pi}d\phi \int_0^1 r(1-r^2)^{3/2} dr $$

After solving this I ended up with $\frac{2ab}{5}\pi$… Does this look correct? I was just confused about changing variables then applying polar coordinates and was wondering if I made a mistake there. Any help is greatly appreciated.

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    $\begingroup$ It is correct. The problem was done precisely as it should be done. $\endgroup$ – André Nicolas Mar 25 '13 at 18:11
  • $\begingroup$ Kudos to Andre that was able to understand that mess. @user68203, go to the FAQ section to find directions on how to write LaTeX in this site to properly write mathematics in this site. $\endgroup$ – DonAntonio Mar 25 '13 at 18:26
  • $\begingroup$ You may answer your own question. In addition to DonAntonio's comment you find some basic information about writing math at this site here, here, here and here. $\endgroup$ – Américo Tavares Mar 25 '13 at 18:58
  • $\begingroup$ I've changed to $\phi$ (\phi) in the last integral. However as you likely know in polar coordinates the standard variable is $\theta$ (\theta). $\endgroup$ – Américo Tavares Mar 25 '13 at 19:15
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    $\begingroup$ Thank you guys. And yes sorry for the mess. I will check out those sites for next time. $\endgroup$ – user68203 Mar 26 '13 at 0:42

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