Answer
a. $16\pi$
b. $32\pi$
c. $28\pi$
d. $24\pi$
e. $60\pi$
f. $48\pi$
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
$V=\int_0^2 2\pi x(3x-0)dx=6\pi\int_0^2 (x^2)dx=2\pi(x^3)|_0^2 =16\pi$
b. With the enclosed region revolving around $x=4$, using cylindrical shells, we have
$V=\int_0^2 2\pi (4-x)(3x-0)dx=6\pi\int_0^2 (4x-x^2)dx=6\pi(2x^2-\frac{1}{3}x^3)|_0^2=6\pi(2(2)^2-\frac{1}{3}(2)^3) =32\pi$
c. With the enclosed region revolving around $x=-1$, using cylindrical shells, we have
$V=\int_0^2 2\pi (x+1)(3x-0)dx=6\pi\int_0^2 (x^2+x)dx=\pi(2x^3+3x^2)|_0^2=\pi(2(2)^3+3(2)^2)=28\pi$
d. With the enclosed region revolving around the x-axis, using cylindrical shells, we have
$V=\int_0^6 2\pi (y)(2-x)dy=\int_0^6 2\pi (y)(2-\frac{1}{3}y)dy=2\pi\int_0^6 (2y-\frac{1}{3}y^2)dy=2\pi (y^2-\frac{1}{9}y^3)|_0^6=2\pi ((6)^2-\frac{1}{9}(6)^3)=24\pi$
e. With the enclosed region revolving around $y=7$, using cylindrical shells, we have
$V=\int_0^6 2\pi (7-y)(2-x)dy=\int_0^6 2\pi (7-y)(2-\frac{1}{3}y)dy=2\pi\int_0^6 (14-\frac{13}{3}y+\frac{1}{3}y^2)dy=2\pi (14y-\frac{13}{6}y^2+\frac{1}{9}y^3)|_0^6=2\pi (14(6)-\frac{13}{6}(6)^2+\frac{1}{9}(6)^3)=60\pi$
f. With the enclosed region revolving around $y=-1$, using cylindrical shells, we have
$V=\int_0^6 2\pi (y+2)(2-x)dy=\int_0^6 2\pi (y+2)(2-\frac{1}{3}y)dy=2\pi\int_0^6 (4+\frac{4}{3}y-\frac{1}{3}y^2)dy=2\pi (4y+\frac{2}{3}y^2-\frac{1}{9}y^3)|_0^6=2\pi (4(6)+\frac{2}{3}(6)^2-\frac{1}{9}(6)^3)=48\pi$
