Volume of Solid of Revolution Shell Method About Y-Axis? I have been reading up on disk/washer, shell methods to find volume of solids of revolution, but I am having trouble with the following question:
We are being asked to find the volume of the following solid of revolution:
$R$ bounded by the graph of $y=x/2, y=3x, y=4$, revolved about the $y$-axis.
My thoughts so far: Use shell: $V=2\pi\int_a^b xf(x)dx$
I am unsure of how to visualize/set up the integral. Do I integrate with respect to $x$? Any help would be appreciated.
Respectfully,
Jack
 A: HINT
Express $x_{1,2}$ in terms of y.
$$V=\pi\int_0^4 (x_1^2-x_2^2) dy$$
A: Visualization is an important feature of getting the integral set up correctly for volumes of revolution. The shell method is more complicated for this problem because the shell widths vary as differences between two sets of different functions. However, I'll proceed anyway with that method.
A shell method rotated about the $y$ axis will have shell thicknesses of $dx$ so we need to express the integral in terms of $x$. Each shell will be of height $4 - \frac{x}{2}$ from $x=\frac{4}{3}$ to $x=8$ (for the first limit of integration), and height $3x-\frac{x}{2}$ from $x=0$ to $x=\frac{4}{3}$ (for the second limit of integration).
The basic formula is:
$$\int_a^b 2\pi(\text{shell radius})(\text{shell height}) dx$$
We have two separate integrals to add to get the total volume.
$$\int_{\frac{4}{3}}^8 2\pi(x)(4 - \frac{x}{2}) dx + \int_{0}^{\frac{4}{3}} 2\pi(x)(3x-\frac{x}{2}) dx$$
Which simplifies to:
$$2\pi \int_{\frac{4}{3}}^8 4x - \frac{x^2}{2} dx + 2\pi \int_{0}^{\frac{4}{3}} \frac{5x^2}{2}dx$$
$$2\pi\cdot ([2x^2 - \frac{x^3}{6}]_{\frac{4}{3}}^8 + [\frac{5x^3}{6}]_0^{\frac{4}{3}})$$
$$2\pi\cdot (128 - \frac{256}{3} - \frac{32}{9} + \frac{64}{162} + \frac{320}{162})$$
$$= 260.6358$$


