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I am trying to compute the integral

$$\iint_{R} sin(9x^2 + 4y^2) dA$$ where R is bounded by the region $9x^2 + 4y^2 = 36$, by the change of variables method.

I am having trouble determining the proper transformation $G: \mathbb{R^2} \to R$ so that I can perform my change of variables. I have tried expressing the ellipsoid by transforming it into a circle (with radius 1) but that did not get me anywhere and hence I am really stuck.

I'd appreciate any hints or ideas on how to approach these problems. More so, any general advice that you may have for finding a good transformation/diffeomorphism that works for the change of variables.

Thanks!

TimelordViktorious
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2 Answers2

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If it were $$\iint_R \sin{(u^2+v^2)}dA$$ with $R$ being the region $u^2+v^2\leqslant 1$, then you'd use polar coordinates: $u= r\cos{\theta}$, $v=r\sin{\theta}$, and $dA = du\, dv = rdr\, d\theta$.

Your region is $9x^2 + 4y^2 \leqslant 36$, or $(\frac{x}{2})^2 + (\frac{y}{3})^2 \leqslant 1$, so consider $x=2u = 2r\cos{\theta}$ and $y=3v=3r\sin{\theta}$ with $dA = dx\, dy = 6rdr\, d\theta$.

Catalin Zara
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  • I see what I did wrong now! I dropped the "r" term in the polar coordinate, hence I was getting sin(36) inside the integral which I felt was wrong. Thanks for your help! – TimelordViktorious Mar 10 '17 at 23:02
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You were on the right track. The domain is indeed elliptic: $$ R=\{(x,y)\;|\; 9x^2+4y^2 \le 36 \} $$ With $x=2r \cos \theta$ and $y=3r\sin \theta$, $R$ can be rewritten as follows: $$ R=\{(r,\theta)\;|\; 0 \le r \le 1, 0 \le \theta \le 2\pi \} $$ Therefore the integral equals $$ \int_0^{2\pi}\int_0^{1}\sin( 36 r^2)6r \; dr d\theta = \frac{1}{3}\pi \sin^218 $$

Kuifje
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