Answer
a. $\frac{16\pi}{15} $
b. $\frac{56\pi}{15} $
c. $\frac{64\pi}{15} $
Work Step by Step
a. Draw a diagram as shown in the figure. With the enclosed region revolving around $y=1$, use disk approximation and symmetry to get
$V=2\int_0^1 \pi (1-x^2)^2 dx=2\pi (1-2x^2+x^4) dx=2\pi (x-\frac{2}{3}x^3+\frac{1}{5}x^5)|_0^1=2\pi (1-\frac{2}{3}(1)^3+\frac{1}{5}(1)^5)=\frac{16\pi}{15} $
b. With the enclosed region revolving around $y=2$, use washer approximation and symmetry to get
$V=2\int_0^1\pi((2-x^2)^2-(2-1)^2)dx=2\pi\int_0^1(3-4x^2+x^4)dx=2\pi (3x-\frac{4}{3}x^3+\frac{1}{5}x^5)|_0^1=2\pi (3-\frac{4}{3}(1)^3+\frac{1}{5}(1)^5)=\frac{56\pi}{15} $
c. With the enclosed region revolving around $y=-1$, use washer approximation and symmetry to get
$V=2\int_0^1\pi((1+1)^2-(1+x^2)^2)dx=2\pi\int_0^1(3-2x^2-x^4)dx=2\pi (3x-\frac{2}{3}x^3-\frac{1}{5}x^5)|_0^1=2\pi (3-\frac{2}{3}(1)^3-\frac{1}{5}(1)^5)=\frac{64\pi}{15} $
