Double interal in polar coordinates

**Combinatus** · March 24th 2011, 01:17 AM

I'm trying to evaluate $\iint_D {(4-x^2-4y^2)}\, dA$ over the region enclosed by the ellipse $D: \frac{x^2}{4} + y^2 = 1$ .

Transforming the region to a circle of radius 1 (with the transformation $u=\frac{x}{2}, v=y$ ) and then to polar coordinates gives the result $4 \pi$ , which corresponds to the key of the problem.

However, I figured I would attempt to do the integration again by going directly to polar coordinates, giving, after a bit more work

$\iint_D {(4-x^2-4y^2)}\, dA = \int_0^{2 \pi} d\theta \int_0^{\sqrt{4\cos^2{\theta}+\sin^2{\theta}}} {(4-r^2 \cos^2{\theta} - 4 r^2 \sin^2{\theta})r}\, dr = ... = \frac{115 \pi}{32}$ .

I checked the latter result by integrating the integral after the first equality sign in the row above in Mathematica, giving $\frac{115 \pi}{32}$ . Thus, it seems likely that there's some conceptual misunderstanding on my part. What exactly did I do wrong?

**FernandoRevilla** · March 24th 2011, 01:30 AM

The elipse $x^2/4+y^2=1$ in polar coordinates is $\rho^2\cos^2 \theta+4\rho^2\sin^2 \theta=4$ , so:

$0\leq \rho \leq \sqrt{\dfrac{4}{\cos ^2\theta+4\sin^2\theta}}$

**Combinatus** · March 24th 2011, 01:40 AM

Ah, indeed. I was too hasty, thinking of r being integrated from 0 to the length of the radius in Euclidean coordinates.

Thank you!

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