Answer
a. $\frac{96\pi}{5}$
b. $\frac{264\pi}{5}$
c. $\frac{336\pi}{5}$
d. $\frac{768\pi}{7}$
e. $\frac{576\pi}{7}$
f. $\frac{936\pi}{7}$
Work Step by Step
a. Draw a diagram as shown in the figure. With the enclosed region revolving around the y-axis, using cylindrical shells, we have
$V=\int_0^2 2\pi x(8-x^3)dx=2\pi\int_0^2 (8x-x^4)dx=2\pi(4x^2-\frac{1}{5}x^5)|_0^2=2\pi(4(2)^2-\frac{1}{5}(2)^5) =\frac{96\pi}{5}$
b. With the enclosed region revolving around $x=3$, using cylindrical shells, we have
$V=\int_0^2 2\pi (3-x)(8-x^3)dx=2\pi\int_0^2 (24-8x-3x^3+x^4)dx=2\pi(24x-4x^2-\frac{3}{4}x^4+\frac{1}{5}x^5)|_0^2=2\pi(24(2)-4(2)^2-\frac{3}{4}(2)^4+\frac{1}{5}(2)^5)=\frac{264\pi}{5}$
c. With the enclosed region revolving around $x=-3$, using cylindrical shells, we have
$V=\int_0^2 2\pi (x+2)(8-x^3)dx=2\pi\int_0^2 (16+8x-2x^3-x^4)dx=2\pi(16x+4x^2-\frac{1}{2}x^4-\frac{1}{5}x^5)|_0^2=2\pi(16(2)+4(2)^2-\frac{1}{2}(2)^4-\frac{1}{5}(2)^5)=\frac{336\pi}{5}$
d. With the enclosed region revolving around the x-axis, using cylindrical shells, we have
$V=\int_0^8 2\pi (y)(x)dy=\int_0^8 2\pi (y)(y^{1/3})dy=2\pi\int_0^8 (y^{4/3})dy=2\pi (\frac{3}{7}y^{7/3})|_0^8=2\pi (\frac{3}{7}(8)^{7/3})=\frac{768\pi}{7}$
e. With the enclosed region revolving around $y=8$, using cylindrical shells, we have
$V=\int_0^8 2\pi (8-y)(x)dy=\int_0^8 2\pi (8-y)(y^{1/3})dy=2\pi\int_0^8 (8y^{1/3}-y^{4/3})dy=2\pi (6y^{4/3}-\frac{3}{7}y^{7/3})|_0^8=2\pi (6(8)^{4/3}-\frac{3}{7}(8)^{7/3})=\frac{576\pi}{7}$
f. With the enclosed region revolving around $y=-1$, using cylindrical shells, we have
$V=\int_0^8 2\pi (y+1)(x)dy=\int_0^8 2\pi (y+1)(y^{1/3})dy=2\pi\int_0^8 (y^{1/3}+y^{4/3})dy=2\pi (\frac{3}{4}y^{4/3}+\frac{3}{7}y^{7/3})|_0^8=2\pi (\frac{3}{4}(8)^{4/3}+\frac{3}{7}(8)^{7/3})=\frac{936\pi}{7}$
