Answer
$$\pi ( 3 \pi -8)$$
Work Step by Step
We need to integrate the integral to compute the volume.
$$Volume= \int_{0}^{\pi/2} \pi (2-2 \sin x)^2 \space dx \\ = \int_{0}^{\pi/2} \pi (4- 8 \sin x+4(\sin x)^2) dx \\= \int_{0}^{\pi/2} 4\pi (1-2 \sin x+\dfrac{1-\cos 2x}{2}] dx \\=\dfrac{12 \pi 3}{2} +2 \cos x \cdot 4 \pi-4 \pi \times \dfrac{\sin (2x)}{4} \\=\pi ( 3 \pi -8)$$
