## University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{\pi }{2}(9 \pi -16)$
We need to integrate the integral as shown below: $Volume = \pi \int_{0}^{\pi/3} (2-\sin x)^2 dx \\=\pi \int_{0}^{\pi}(4-4 \sin x +\dfrac{1-\cos 2x}{2} ) dx \\=\pi [4x+4 \cos x +\dfrac{x}{2}-\dfrac{\sin 2x}{4}]_{0}^{\pi} \\=\dfrac{\pi }{2}(9 \pi -16)$