07 Area Enclosed by r = 2a cos θ and r = 2a sin θ

$\displaystyle A = 4a^2 \int_0^{\pi/2} \cos^2 \theta \, d\theta$

$\displaystyle A = 4a^2 \int_0^{\pi/2} \frac{1}{2}(1 + \cos 2\theta) \, d\theta$

$\displaystyle A = 2a^2 \int_0^{\pi/2} (1 + \cos 2\theta) \, d\theta$

$A = 2a^2 \left[ \theta + \dfrac{\sin 2\theta}{2} \right]_0^{\pi/2}$

$A = 2a^2 \left[ \left( \dfrac{\pi}{2} + \dfrac{\sin \pi}{2} \right) - \left( 0 + \dfrac{\sin 0}{2} \right) \right]$

$A = 2a^2 \left[ \left( \dfrac{\pi}{2} + 0 \right) - \left( 0 + 0 \right) \right]$

$A = \pi a^2$       answer

$\displaystyle A = 4a^2 \int_0^{\pi/2} \sin^2 \theta \, d\theta$

$\displaystyle A = 4a^2 \int_0^{\pi/2} \frac{1}{2}(1 - \cos 2\theta) \, d\theta$

$\displaystyle A = 2a^2 \int_0^{\pi/2} (1 - \cos 2\theta) \, d\theta$

$A = 2a^2 \left[ \theta - \dfrac{\sin 2\theta}{2} \right]_0^{\pi/2}$

$A = 2a^2 \left[ \left( \dfrac{\pi}{2} - \dfrac{\sin \pi}{2} \right) - \left( 0 - \dfrac{\sin 0}{2} \right) \right]$

$A = 2a^2 \left[ \left( \dfrac{\pi}{2} - 0 \right) - \left( 0 - 0 \right) \right]$

$A = \pi a^2$       answer