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 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/

  • 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.

  • 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.

Questions