Area polar coordinates.

**Ulysses** · August 16th 2011, 06:50 AM

Hi. I wanted to know if I did this right:
Using double integrals, set in polar coordinates, the integral that gives the area for the region $D: \{x^2+y^2\leq{9},(x-2)^2+y^2\geq{7}$

From the picture:

The pink region represents the area I want.
The intersection is given for x=3/2, where: $y=\displaystyle\frac{3\sqrt[ ]{3}}{2}$

Then $\tan \theta_1=9\sqrt[ ]{3}$

Tita 1 represents the limit on the angle for the circle that isn't centered at the origin. Is the limit for the brown region.

$\tan \theta_2=3\displaystyle\frac{\sqrt[ ]{3}}{4}$
This is the limit on the angle for the orange region, for the circle centered at the origin.

Finally
$y=\displaystyle\frac{-3\sqrt[ ]{3}}{2}(x+3)$
With $x=rcos\theta,y=r\sin \theta$
Then $r=\frac{-15\sqrt[ ]{3}}{2(\sin\theta+\frac{3\sqrt[ ]{3}}{2}\cos\theta)}$

And the area:

$\displaystyle\int_{0}^{2\pi}\int_{0}^{3}rdrd\theta-2\displaystyle\int_{0}^{tan^{-1}9\sqrt[ ]{3}}\int_{0}^{\sqrt[ ]{7}}rdrd\theta-2\displaystyle\int_{0}^{tan^{-1}\displaystyle\frac{3\sqrt[ ]{3}}{4}}\int_{\frac{-15\sqrt[ ]{3}}{2(\sin\theta+\frac{3\sqrt[ ]{3}}{2}\cos\theta)}}^{3}rdrd\theta$

The first integral represents the entire circle, the second the brown region and the third the orange region, both multiplied on two by simmetry

**SammyS** · August 16th 2011, 08:51 AM

My suggestion is to integrate by having r go from the brown circle to the blue circle as θ goes from θ_1 to π, then double the answer.

**Ulysses** · August 16th 2011, 04:41 PM

How would you do that? I tried but I couldn't get an expression for r.

**SammyS** · August 16th 2011, 05:04 PM

In polar coordinates, the equation of the larger circle is obvious, hopefully.

Find the equation of the smaller circle in polar coords.

$(x-2)^2+y^2=7$

$x^2-4x+4+y^2=7$

$r^2-4r\cos(\theta)-3=0$

Solve for r using the quadratic formula or by completing the square.

We usually like to have r be positive, so use the positive sign with the square root. (Either sign will give the same polar graph.)

**Ulysses** · August 17th 2011, 03:41 AM

Great, didn't realized about that. That simplifies things pretty much. Anyway, the answer I give is right or did I made a mistake? Just wanna know, I know its horrendous, but anyway I'd like to know if instead of that its ok.

This is the solution I get using your suggestion:

$\displaystyle\int_0^{2\pi}\int_{\frac{-2\cos\theta+\sqrt[]{4\cos^2\theta+12}}{2}}^{3} rdrd\theta$
Nice

**SammyS** · August 17th 2011, 05:36 PM

You have some algebra error(s) in your solution to the quadratic equation.

**Ulysses** · August 18th 2011, 02:41 AM

You're right, I've made some mistakes. The lower bound would be $\frac{4\cos\theta+\sqrt[]{16\cos^2\theta+12}}{2}$

Thank you very much SammyS.

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