Answer
$\frac{3 \pi}{2}=A$
Work Step by Step
$r=-\cos \theta+1, 0 \leqslant \theta \leqslant 2 \pi$
so
$\int_{0}^{2 \pi} \int_{0}^{1-\cos \theta} r d r d \theta=A$
Evaluate
$A=\frac{1}{2} \int_{0}^{2 \pi}(-\cos \theta+1)^{2} d \theta$
$A=\frac{1}{2} \int_{0}^{2 \pi}\left(1-2 \cos \theta+\cos ^{2} \theta+\frac{\cos 2 \theta}{2}\right) d \theta$
$A=\frac{1}{2} \int_{0}^{2 \pi}\left(1-2 \cos \theta+\frac{1}{2}+\frac{\cos 2 \theta}{2}\right) d \theta$
$A=\frac{1}{2} \int_{0}^{2 \pi}\left(\frac{3}{2}-2 \cos \theta+\frac{\cos 2 \theta}{2}\right) d \theta$
Integrate
$\frac{1}{2}\left[-2 \sin \theta+\frac{\sin 2 \theta}{4}+\frac{3}{2} \theta\right]_{0}^{2 \pi}=A$
$\left[\frac{6 \pi}{2}-2 \sin 2 \pi+\frac{\sin 4 \pi}{4}\right]\frac{1}{2}-\frac{1}{2}[0]=A$
$\left(\frac{6 \pi}{2}\right)\frac{1}{2}=A$
$\frac{3 \pi}{2}=A$
