- #1
Destroxia
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Homework Statement
Find the area in the first quadrant that is inside the circle ##r=100sin(\theta)## and outside the leminscate ##r^2=200cos(2\theta)##.
I have graphed the region as I interpreted it below. The area I am trying to find is the non-shaded, white region.
Homework Equations
##\int_{\alpha}^{\beta} \int_{r_1(\theta)}^{r_2(\theta)} f(r, \theta)rdrd\theta##
The Attempt at a Solution
1. Determining the limits of the integrals:
Since the area we are solving for is in the first quadrant, and spans the whole first quadrant...
## \alpha = 0## and ##\beta = \frac {\pi} {2}##
As the leminscate is the first polar region we interact...
##r_1(\theta) = \sqrt{200cos(2\theta)}## and ##r_2(\theta) = 100sin(\theta)##
2. Setting up & Evaluating the Integral:
##\int_{0}^{\frac {\pi} {2}} \int_{\sqrt{200cos(2\theta)}}^{100sin(\theta)} r dr d\theta =##
##= \int_{0}^{\frac {\pi} {2}} [ \frac 1 2 r^2 |_{\sqrt{200cos(2\theta)}}^{100sin(\theta)}]d\theta##
##= \int_{0}^{\frac {\pi} {2}} [5000sin(\theta) - 100cos(2\theta)] d\theta##
##= 5000 \int_{0}^{\frac {\pi} {2}} sin(\theta)d\theta - 100 \int_{0}^{\frac {\pi} {2}} cos(2\theta)d\theta ##
##= 5000 [-cos(\theta) |_{0}^{\frac {\pi} {2}}] - 100 [\frac {sin(2\theta)} {2} |_{0}^{\frac {\pi} {2}}] ##
##= 5000 [0+1] - 100 [0-0]##
##\int_{0}^{\frac {\pi} {2}} \int_{\sqrt{200cos(2\theta)}}^{100sin(\theta)} r dr d\theta = 5000##
Apparently, this is an incorrect answer... I think I am having more of a conceptual problem with the regions than anything else.