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

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SUMMARY

The area of the region bounded by the polar equation r = 3 + 2sin(θ) can be calculated using the formula A = 1/2 ∫ r² dθ. The integral to evaluate is ∫ (3 + 2sin(θ))² dθ, which expands to ∫ (9 + 12sin(θ) + 4sin²(θ)) dθ. The solution involves splitting the integral into manageable parts and substituting sin²(θ) with (1 - cos(2θ))/2 to simplify the calculation. It is essential to determine the correct limits of integration to accurately represent one complete traversal around the boundary.

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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?
 

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