Answer
$\dfrac{4}{3}+\dfrac{5\pi}{8}$
Work Step by Step
Our aim is to integrate the integral as follows:
$ Volume=2\int^{\pi/2}_0 \int^{1+cos\theta}_1 r^2 \times \cos\theta \space dr \space d\theta $
or, $=\dfrac{2}{3} \int^{\pi/2}_0 (3 \cos^2\theta+3 \cos^3\theta +cos^4\theta) \space d\theta $
or, $=\dfrac{2}{3}[\dfrac{15\theta}{8}+\sin (2\theta)+3 \sin (\theta)-\sin^3 (\theta)+\dfrac{\sin (4\theta) }{32}]^{\pi/2}_0$
or, $=\dfrac{4}{3}+\dfrac{5\pi}{8}$
