How do you find the area of the region cut from the second quadrant by the cardioid r=1-cosθ?

Jul 6, 2016

$= \frac{3}{4} \pi + 2$

Explanation:

that's

$A = {\int}_{\frac{\pi}{2}}^{\pi} {r}^{2} \setminus d \theta$

$= {\int}_{\frac{\pi}{2}}^{\pi} {\left(1 - \cos \theta\right)}^{2} \setminus d \theta$

$= {\int}_{\frac{\pi}{2}}^{\pi} d \theta q \quad 1 - 2 \cos \theta + {\cos}^{2} \theta$

$= {\int}_{\frac{\pi}{2}}^{\pi} d \theta q \quad 1 - 2 \cos \theta + \frac{\cos 2 \theta + 1}{2}$

$= {\left[\frac{3}{2} \theta - 2 \sin \theta + \frac{1}{4} \sin 2 \theta\right]}_{\frac{\pi}{2}}^{\pi}$

$= \left[\frac{3}{2} \pi\right] - \left[\frac{3}{4} \pi - 2\right]$

$= \frac{3}{4} \pi + 2$