Answer
a. $8\pi$
b. $\frac{32\pi}{5}$
c. $\frac{8\pi}{3}$
d. $\frac{224\pi}{15}$
Work Step by Step
a. Draw a diagram as shown in the figure. With the enclosed region revolving around the x-axis, using the washer approximation, we have
$V=\int_0^4\pi(2^2-(\sqrt x)^2)dx=\pi\int_0^4(4-x)dx=\pi(4x-\frac{x^2}{2})|_0^4=8\pi$
b. With the enclosed region revolving around the y-axis, using the cylinder approximation, we have
$V=\int_0^4 2\pi x(2-\sqrt x)dx=2\pi\int_0^4 (2x-x^{3/2})dx=2\pi (x^2-\frac{2}{5}x^{5/2})|_0^4=2\pi (4^2-\frac{2}{5}(4)^{5/2})=\frac{32\pi}{5}$
c. With the enclosed region revolving around $y=2$, using the disk approximation, we have
$V=\int_0^4 \pi (2-\sqrt x)^2 dx= \pi\int_0^4 (4-4\sqrt x+x) dx= \pi (4x-\frac{8}{3}x^{3/2}+\frac{x^2}{2}) |_0^4=\pi (4(4)-\frac{8}{3}(4)^{3/2}+\frac{(4)^2}{2})=\frac{8\pi}{3}$
d. With the enclosed region revolving around $x=4$, using the cylinder approximation, we have
$V=\int_0^4 2\pi (4-x)(2-\sqrt x)dx=2\pi\int_0^4 (8-4\sqrt x-2x+x^{3/2})dx=2\pi (8x-\frac{8}{3}x^{3/2}-x^2+\frac{2}{5}x^{5/2})|_0^4=2\pi (8(4)-\frac{8}{3}(4)^{3/2}-(4)^2+\frac{2}{5}(4)^{5/2})=\frac{224\pi}{15}$
