03 Area Enclosed by Cardioids: r = a(1 + sin θ); r = a(1 - sin θ), r = a(1 + cos θ), r = a(1 - cos θ)

$\displaystyle A = a^2 \int_{0}^{\pi} (1 - \cos \theta)^2 \, d\theta$

$\displaystyle A = a^2 \int_{0}^{\pi} (1 - 2\cos \theta + \cos^2 \theta) \, d\theta$

$A = a^2 {\displaystyle \int_{0}^{\pi}} \left[ 1 - 2\cos \theta + \frac{1}{2}(1 + \cos 2\theta) \right] \, d\theta$

$A = a^2 {\displaystyle \int_{0}^{\pi}} \left( \dfrac{3}{2} - 2\cos \theta + \dfrac{\cos 2\theta}{2} \right) \, d\theta$

$A = a^2 \left[ \dfrac{3\theta}{2} - 2\sin \theta + \dfrac{\sin 2\theta}{4} \right]_{0}^{\pi}$

$A = a^2 \left[ \left( \dfrac{3\pi}{2} - 2\sin \pi + \dfrac{\sin 2\pi}{4} \right) - \left( 0 - 2\sin 0 + \dfrac{\sin 0}{4} \right) \right]$

$A = a^2 \left[ \left( \dfrac{3\pi}{2} - 0 + 0 \right) - \left( 0 - 0 + 0 \right) \right]$

$A = \frac{3}{2}\pi a^2$       answer

$\displaystyle A = a^2 \int_{0}^{\pi} (1 + \cos \theta)^2 \, d\theta$

$\displaystyle A = a^2 \int_{0}^{\pi} (1 + 2\cos \theta + \cos^2 \theta) \, d\theta$

$A = a^2 {\displaystyle \int_{0}^{\pi}} \left[ 1 + 2\cos \theta + \frac{1}{2}(1 + \cos 2\theta) \right] \, d\theta$

$A = a^2 {\displaystyle \int_{0}^{\pi}} \left( \dfrac{3}{2} + 2\cos \theta + \dfrac{\cos 2\theta}{2} \right) \, d\theta$

$A = a^2 \left[ \dfrac{3\theta}{2} + 2\sin \theta + \dfrac{\sin 2\theta}{4} \right]_{0}^{\pi}$

$A = a^2 \left[ \left( \dfrac{3\pi}{2} + 2\sin \pi + \dfrac{\sin 2\pi}{4} \right) - \left( 0 - 2\sin 0 + \dfrac{\sin 0}{4} \right) \right]$

$A = a^2 \left[ \left( \dfrac{3\pi}{2} + 0 + 0 \right) - \left( 0 + 0 + 0 \right) \right]$

$A = \frac{3}{2}\pi a^2$       answer

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

$\displaystyle A = a^2 \int_{-\pi/2}^{\pi/2} (1 - 2\sin \theta + \sin^2 \theta) \, d\theta$

$A = a^2 {\displaystyle \int_{-\pi/2}^{\pi/2}} \left[ 1 - 2\sin \theta + \frac{1}{2}(1 - \cos 2\theta) \right] \, d\theta$

$A = a^2 {\displaystyle \int_{-\pi/2}^{\pi/2}} \left( \dfrac{3}{2} - 2\sin \theta - \dfrac{\cos 2\theta}{2} \right) \, d\theta$

$A = a^2 \left[ \dfrac{3\theta}{2} + 2\cos \theta - \dfrac{\sin 2\theta}{4} \right]_{-\pi/2}^{\pi/2}$

$A = a^2 \left[ \left( \dfrac{3\pi}{4} + 2\cos (\pi/2) - \dfrac{\sin \pi}{4} \right) - \left( -\dfrac{3\pi}{4} + 2\cos (-\pi/2) + \dfrac{\sin (-\pi)}{4} \right) \right]$

$A = a^2 \left[ \left( \dfrac{3\pi}{4} + 0 - 0 \right) - \left( -\dfrac{3\pi}{4} + 0 - 0 \right) \right]$

$A = \frac{3}{2}\pi a^2$       answer

$\displaystyle A = a^2 \int_{-\pi/2}^{\pi/2} (1 + \sin \theta)^2 \, d\theta$

$\displaystyle A = a^2 \int_{-\pi/2}^{\pi/2} (1 + 2\sin \theta + \sin^2 \theta) \, d\theta$

$A = a^2 {\displaystyle \int_{-\pi/2}^{\pi/2}} \left[ 1 + 2\sin \theta + \frac{1}{2}(1 - \cos 2\theta) \right] \, d\theta$

$A = a^2 {\displaystyle \int_{-\pi/2}^{\pi/2}} \left( \dfrac{3}{2} + 2\sin \theta - \dfrac{\cos 2\theta}{2} \right) \, d\theta$

$A = a^2 \left[ \dfrac{3\theta}{2} - 2\cos \theta - \dfrac{\sin 2\theta}{4} \right]_{-\pi/2}^{\pi/2}$

$A = a^2 \left[ \left( \dfrac{3\pi}{4} - 2\cos (\pi/2) - \dfrac{\sin \pi}{4} \right) - \left( -\dfrac{3\pi}{4} - 2\cos (-\pi/2) + \dfrac{\sin (-\pi)}{4} \right) \right]$

$A = a^2 \left[ \left( \dfrac{3\pi}{4} - 0 - 0 \right) - \left( -\dfrac{3\pi}{4} - 0 - 0 \right) \right]$

$A = \frac{3}{2}\pi a^2$       answer