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.