What are the steps for finding the area of a region bounded by a polar equation?

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Homework Statement



find the area of the region bounded by r = 3 + 2sin(theta)

heres a pic of the intended equations i have so far: http://img263.imageshack.us/img263/8804/untitledpk1.jpg

Homework Equations



A = 1/2 B integral A r^2 d(theta)

The Attempt at a Solution



2 * 1/2 integral (3 + 2sin(theta))^2 d(theta

this is as far as i have gotten with this, i am not good with integrals, can someone help me with the rest of this please. thanks.
 
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ok Well basically you need help with this integral:

\int^b_a (3 + 2\sin \theta)^2 d\theta.

Expand the binomial. (I'm going to ignore the bounds for now, account for them later for me please :P)

\int 9 + 12\sin\theta + 4\sin^2 \theta d\theta.

Split the integral up. the first 2 parts are very easy, you should get this.

9\theta - 12\cos\theta + 4\int \sin^2\theta d\theta.

We know that \sin^2 \theta = \frac { 1-\cos {2\theta}}{2}. Just sub that in, take out the factor of 1/2, split it up again, easy work. GOGOGO!
 
Becareful when you are doing the part Gib Z did NOT do- determine the limits of integration. What values of \theta will take you exactly once around the boundary?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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