Converting an integral from Cartesian to Polar coordinates.

I'm trying to evaluate the double integral $$\int_0^{10}\int_0^{(5/2)\sqrt{4-x^2}}4-x^2-\frac{4}{25}y^2 \; dy \; dx$$

but using Cartesian coordinates requires the idea of trigonometric substitution and the limits aren't very nice either. So, my question is, how can I change this integral into another one in polar coordinates that would make the evaluation easier? I don't know if this helps, but the original problem was to find the solid bound in the first octant by $25z=100-25x^2-4y^2$. Thank you.

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Something is wrong with the integration domain: $\sqrt{4-x^2}$ exists only when $|x|\leqslant2$ hence $10$ cannot be the upper limit of the integral over $x$. –  Did Mar 31 '12 at 23:48

Let me assume you are interested in $$I=\int_0^{2}\int_0^{(5/2)\sqrt{4-x^2}}\left(4-x^2-\frac{4}{25}y^2\right) \mathrm dy \;\mathrm dx.$$ A key feature of the integration domain is the upper limit $y=5\sqrt{1-(x/2)^2}$. It hints at the fact that a curve of interest is (a part of) the ellipse $(x/2)^2+(y/5)^2=1$, which is parametrized by $x=2\cos(u)$ and $y=5\sin(u)$.

One sees that $0\leqslant y\leqslant5\sqrt{1-(x/2)^2}$ and $0\leqslant x\leqslant2$ can be parametrized by $x=2\cos(u)$ and $y=5r\sin(u)$ with $0\leqslant u\leqslant\pi/2$ and $0\leqslant r\leqslant1$.

The Jacobian of the transformation $(r,u)\mapsto(x,y)=(2\cos(u),5r\sin(u))$ is $10\sin^2(u)$, hence $$I=\int_0^{1}\int_0^{\pi/2}4(1-r^2)\sin^2(u)\;10\sin^2(u) \mathrm du \;\mathrm dr=5\cdot J\cdot K,$$ with $$J=\int_0^1(1-r^2)\mathrm dr=2/3,\qquad K=\int_0^{\pi/2}8\sin^4(u)\mathrm du=3\pi/2.$$ Finally, $I=5\pi$.

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You would be better off using $x=2r\cos\theta$, $y=5r\sin\theta$ and (using the Jacobian) $dx\,dy=10r\,dr\,d\theta$. Then the region becomes $0\le r\le 5$ and $0\le\theta\le\pi/2$.

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How did you know to make those transformations? –  Hautdesert Mar 31 '12 at 23:21

When changing into polar coordinates, $dy dx = r \; dr \; d\theta$ and change the $x$ and $y$ terms using $x = r \cos \theta$, $y = r \sin \theta$, and $x^2 + y^2 = r^2$.

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But what about the region? How does the conversion simplify the limits for the integration? –  Hautdesert Mar 31 '12 at 20:27