Answer
$\dfrac{13}{24}$
Work Step by Step
We set up the integral as follows:
Volume, $V=\int_{0}^{ \pi/2 }\int_{0} ^{1}\int_{0} ^{2-r \sin \theta } r^2 \cos \theta dx dr d\theta $
$=\int_{0}^{ \pi/2}\int_{0} ^{1} 2r^2 \cos \theta -r^3 \cos \theta \sin \theta dr d\theta $
$=\int_{0}^{ \pi/2}\int_{0} ^{1} (2/3) \cos \theta -(1/4) \cos \theta \sin \theta d\theta $
Suppose that
$ a= \sin \theta \implies du =\cos \theta d\theta$
So, $Volume =\dfrac{2}{3} \sin \theta|_0^{\pi/2} -\int_0^1 \dfrac{a da}{4}=\dfrac{2}{3}-\dfrac{1}{8}=\dfrac{13}{24}$
