Area of polar curves

**mmattson07** · December 10th 2009, 09:37 PM

Trying to the find this area
The equation is $r= 3+2sin\theta$

Heres what i did:
$A= \frac{1}{2}\int_0^\pi \! (3+2sin\theta)^2 \, d\theta$
$A= \frac{1}{2}\int_0^\pi \! (9+6sin\theta+4sin^2\theta) \, d\theta$
$A= \frac{1}{2}\int_0^\pi \! (9+6sin\theta+4*\frac{1}{2}(1-cos2\theta) \, d\theta$
$A= \frac{1}{2}\int_0^\pi \! (11+6sin\theta-2cos2\theta) \, d\theta$
$A= \frac{1}{2}[11\theta-6cos\theta-sin2\theta] \bigg|_0^\pi$
$A= \frac{1}{2}[11\pi+12]$
$A= \frac{11\pi}{2}+6$

Where did i go wrong?

**simplependulum** · December 11th 2009, 12:19 AM

I think the upper limit is $\frac{\pi}{2}$

and the lower limit is $- \frac{\pi}{2}$

so $6$ should disappear since $\cos(\theta) \bigg|_{-\pi/2}^{\pi/2} = 0$

Other is OK ( Happy)

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