How do you find the area of the region bounded by the polar curves $r=3+2cos(theta)$ and $r=3+2sin(theta)$ ?

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Answer 1 · 2014-11-08T11:06:33

Let us look at the region bounded by the polar curves, which looks like:

enter image source here

Red: $y=3+2cos theta$
Blue: $y=3+2sin theta$
Green: $y=x$

Using the symmetry, we will try to find the area of the region bounded by the red curve and the green line then double it.

$A=2int_{pi/4}^{{5pi}/4}\int_0^{3+2cos theta}rdrd theta$

$=2int_{pi/4}^{{5pi}/4}[r^2/2]_0^{3+2cos theta} d theta$

$=int_{pi/4}^{{5pi}/4}(9+12cos theta+4cos^2theta)d theta$

by $cos^2theta=1/2(1+cos2theta)$ ,

$=int_{pi/4}^{{5pi}/4}(11+12cos theta+2cos2theta)d theta$

$=[11theta+12sin theta+sin2theta]_{pi/4}^{{5pi}/4}$

$={55pi}/4-6sqrt{2}+1-({11pi}/4+6sqrt{2}+1)$

$=11pi-12sqrt{2]$

Hence, the area of the region is $11pi-12sqrt{2}$ .

I hope that this was helpful.

How do you find the area of the region bounded by the polar curves r=3+2cos(theta)r=3+2cos(theta) and r=3+2sin(theta)r=3+2sin(theta) ?