Answer
a. $\frac{56\pi}{5}$
b. $\frac{872\pi}{45} $
Work Step by Step
a. Draw a diagram as shown in the figure. With the enclosed region revolving around $x=1$, using cylindrical shells, we have
$V=\int_{-1}^1 2\pi (1-x)(4-3x^2-x^4)dx=2\pi \int_{-1}^1 (4-4x-3x^2+3x^3-x^4+x^5)dx
=2\pi (4x-2x^2-x^3+\frac{3}{4}x^4-\frac{1}{5}x^5+\frac{1}{6}x^6)|_{-1}^1
=2\pi [(4-2-1^3+\frac{3}{4}-\frac{1}{5}+\frac{1}{6}))-(4(-1)-2(-1)^2-(-1)^3+\frac{3}{4}(-1)^4-\frac{1}{5}(-1)^5+\frac{1}{6}(-1)^6)]=\frac{56\pi}{5}$
b. Step 1 With the enclosed region revolving around the x-axis, using cylindrical shells, we have
$V=\int_{0}^1 2\pi (y)(2y^{1/4})dy+\int_{1}^4 2\pi (y)(2\sqrt {\frac{4-y}{3}})dy
=4\pi \int_{0}^1 (y^{5/4})dy+\frac{4\pi\sqrt 3}{3} \int_{1}^4 (y(4-y)^{1/2})dy$
Step 2. The first part is
$V_1=4\pi \int_{0}^1 (y^{5/4})dy=4\pi (\frac{4}{9}y^{9/4})|_{0}^1=\frac{16\pi}{9}$
Step 3. The second part is
$V_2=\frac{4\pi\sqrt 3}{3} \int_{1}^4 (y(4-y)^{1/2})dy$.
Letting $u=4-y$, we have $du=-dy$, $y\to1, u\to3$ and $y\to4, u\to0$.
Thus
$V_2=-\frac{4\pi\sqrt 3}{3} \int_{3}^0 ((4-u)(u)^{1/2})du=\frac{4\pi\sqrt 3}{3} \int_{0}^3 ((4u^{1/2}-u^{3/2}) )du=\frac{4\pi\sqrt 3}{3} (\frac{8}{3}u^{3/2}-\frac{2}{5}u^{5/2}) |_{0}^3=\frac{4\pi\sqrt 3}{3} (\frac{8}{3}(3)^{3/2}-\frac{2}{5}(3)^{5/2})=\frac{88\pi}{5} $
Step 4. Combining the above results, we have
$V=V_1+V_2=\frac{872\pi}{45} $
