Answer
a) $8 \pi$
b) $8 \pi$
Work Step by Step
a) The volume can be computed as:
$V=\pi \int_0^{4} (\sqrt {4 -y})^2 \ dy$
or, $=\pi \int_0^{4} (4-y) \ dy$
or, $=\pi |(4 y-\dfrac{1}{2}y^2)|_0^4$
or, $=\pi [4 (4-0)-(\dfrac{1}{2}(4)^2-0)]$
or, $=8 \pi$
b) The volume can be computed as:
$V=2\pi \int_0^{2} x(4-x^2) \ dx$
or, $=2\pi \int_0^{2} (4x-x^3) \ dx$
or, $=2\pi (2x^2-\dfrac{1}{4}x^4)|_0^2$
or, $=2 \pi [2(4 -0)-(\dfrac{1}{4}(2)^4-0)]$
or, $=8 \pi$
