Calculating Polar Areas
Key Questions
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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 self-intersections 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/
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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}/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.
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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 double-integral#A=int_{theta_1}^{theta_2}int_0^{r(theta)}rdrd theta# .
I hope that this was helpful.