How do you find the area of the region shared by the circles r=2cos(theta) and r=2sin(theta)?

Oct 24, 2015

$\frac{\pi}{2} - 1$

Explanation: Intersection:
$2 \sin \theta = 2 \cos \theta \implies \sin \frac{\theta}{\cos} \theta = 1 \implies \tan \theta = 1 \implies \theta = \frac{\pi}{4}$

$A = \int {\int}_{A} r \mathrm{dr} d \theta$

$A = {\int}_{0}^{\frac{\pi}{4}} d \theta {\int}_{0}^{2 \sin \theta} r \mathrm{dr} + {\int}_{\frac{\pi}{4}}^{\frac{\pi}{2}} d \theta {\int}_{0}^{2 \cos \theta} r \mathrm{dr}$

${A}_{1} = \frac{1}{2} {\int}_{0}^{\frac{\pi}{4}} d \theta \cdot 4 {\sin}^{2} \theta = {\int}_{0}^{\frac{\pi}{4}} \left(1 - \cos 2 \theta\right) d \theta$

${A}_{1} = \left(\theta - \frac{1}{2} \sin 2 \theta\right) {|}_{0}^{\frac{\pi}{4}} = \frac{\pi}{4} - \frac{1}{2}$

${A}_{2} = \frac{1}{2} {\int}_{\frac{\pi}{4}}^{\frac{\pi}{2}} d \theta \cdot 4 {\cos}^{2} \theta = {\int}_{\frac{\pi}{4}}^{\frac{\pi}{2}} \left(1 + \cos 2 \theta\right) d \theta$

${A}_{2} = \left(\theta + \frac{1}{2} \sin 2 \theta\right) {|}_{\frac{\pi}{4}}^{\frac{\pi}{2}} = \frac{\pi}{4} - \frac{1}{2}$

$A = \frac{\pi}{4} - \frac{1}{2} + \frac{\pi}{4} - \frac{1}{2} = \frac{\pi}{2} - 1$