Answer
a. $\frac{27\pi}{2}$
b. $\frac{27\pi}{2}$
c. $\frac{72\pi}{5}$
d. $\frac{108\pi}{5}$
Work Step by Step
a. Draw a diagram as shown in the figure. With the enclosed region revolving around $x=2$, using cylindrical shells, we have
$V=\int_{-1}^2 2\pi (2-x)(x+2-x^2)dx=2\pi \int_{-1}^2 (x^3-3x^2+4)dx=2\pi (\frac{1}{4}x^4-x^3+4x)|_{-1}^2=2\pi [(\frac{1}{4}(2)^4-(2)^3+4(2))-(\frac{1}{4}(-1)^4-(-1)^3+4(-1))]=\frac{27\pi}{2}$
b. With the enclosed region revolving around $x=-1$, using cylindrical shells, we have
$V=\int_{-1}^2 2\pi (1+x)(x+2-x^2)dx=2\pi \int_{-1}^2 (2+3x-x^3)dx=2\pi (2x+\frac{3}{2}x^2-\frac{1}{4}x^4)|_{-1}^2=2\pi [(2(2)+\frac{3}{2}(2)^2-\frac{1}{4}(2)^4)-(2(-1)+\frac{3}{2}(-1)^2-\frac{1}{4}(-1)^4)]=\frac{27\pi}{2}$
c. With the enclosed region revolving around the x-axis, using cylindrical shells, we have
$V=\int_{0}^1 2\pi (y)(\sqrt y-(-\sqrt y))dy+\int_{1}^4 2\pi (y)(\sqrt y-(y-2))dy=4\pi \int_{0}^1 (y^{3/2})dy+2\pi \int_{1}^4 (y^{3/2}-y^2+2y)dy=4\pi (\frac{2}{5}y^{5/2})|_{0}^1 +2\pi (\frac{2}{5}y^{5/2}-\frac{1}{3}y^3+y^2)|_{1}^4=4\pi (\frac{2}{5}(1)^{5/2})+2\pi [(\frac{2}{5}(4)^{5/2}-\frac{1}{3}(4)^3+(4)^2)-(\frac{2}{5}(1)^{5/2}-\frac{1}{3}(1)^3+(1)^2)]=\frac{72\pi}{5}$
d. With the enclosed region revolving around $y=4$, using cylindrical shells, we have
$V=\int_{0}^1 2\pi (4-y)(\sqrt y-(-\sqrt y))dy+\int_{1}^4 2\pi (4-y)(\sqrt y-(y-2))dy
=4\pi \int_{0}^1 (4y^{1/2}-y^{3/2})dy+2\pi \int_{1}^4 (8-6y+4y^{1/2}-y^{3/2}+y^2)dy
=4\pi (\frac{8}{3}y^{3/2}-\frac{2}{5}y^{5/2})|_{0}^1 +2\pi (8y-3y^2+\frac{8}{3}y^{3/2}-\frac{2}{5}y^{5/2}+\frac{1}{3}y^3)|_{1}^4
=4\pi (\frac{8}{3}(1)^{3/2}-\frac{2}{5}(1)^{5/2})+2\pi [(8(4)-3(4)^2+\frac{8}{3}(4)^{3/2}-\frac{2}{5}(4)^{5/2}+\frac{1}{3}(4)^3)-(8(1)-3(1)^2+\frac{8}{3}(1)^{3/2}-\frac{2}{5}(1)^{5/2}+\frac{1}{3}(1)^3)]
=\frac{108\pi}{5}$
