Calculating Polar Areas
Key Questions

The area inside a polar curve is approximately the sum of lots of skinny wedges that start at the origin and go out to the curve, as long as there are no selfintersections for your polar curve.
Each wedge or slice or sector is like a triangle with height
#r# and base#r# #dθ# , so the area of each element is
#dA = 1/2 b h = 1/2 r (r dθ) = 1/2 r^2 dθ.# So add them up as an integral going around from θ=0 to θ=2π, and using a double angle formula, we get:
#A = 1/2 int_0 ^(2π)(2 + cos(2θ))^2 dθ# #A = 1/2 int_0 ^(2π) [4 + 4 cos(2θ) + cos^2(2θ)] dθ# #A = 1/2 int_0 ^(2π) [4 + 4 cos(2θ) + (1 + cos(4θ))/2] dθ.# Now do the integral(s) by subbing u = 2θ and then u = 4θ, and remember to change limits for the "new u." I'll let you evaluate those to get practice integrating! Remember our motto,
"Struggling a bit makes you stronger." \dansmath/

Let us look at the region bounded by the polar curves, which looks like:
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}/46sqrt{2}+1({11pi}/4+6sqrt{2}+1)# #=11pi12sqrt{2]# Hence, the area of the region is
#11pi12sqrt{2}# .
I hope that this was helpful.

If the region is bounded by a polar curve
#r=r(theta)# from#theta=theta_1# to#theta_2# , then its area#A# can be found by the doubleintegral#A=int_{theta_1}^{theta_2}int_0^{r(theta)}rdrd theta# .
I hope that this was helpful.