The question is: The region bounded by $y=\frac{1}{x}, y=0, x=1, x=2$ is rotated about the $y$-axis, thus creating a solid. Compute the volume using the Shell and Slicing method.
This is what I have thus far:
Shell method
Radius of the shell = x
Height of the shell = $\frac{1}{x}$
Volume: $\int_1^2 2π × x ×\frac{1}{x} dx$ = $2π\int_1^2 dx$ = $2π$
Slicing method
I tried to set the outer radius to be $1/y$ and the inner radius is $1$, but I don't know what the limits of the integral. Am I doing it wrong? Any hints?
You will need to break it into two regions, since the functions which maintain the outer and inner radius are not the same throughout $0\le y\le 1$.
$$R_1:\ \ 0\le y\le \dfrac{1}{2}, \ \ \ 1\le x\le 2$$ $$R_2:\ \ \dfrac{1}{2}\le y\le1, \ \ \ 1\le x\le 1/y$$
Then
$$V = \underbrace{\pi\int_0^{1/2} (2)^2-(1)^2\ dy}_{R_1} \ +\ \underbrace{\pi\int_{1/2}^1 (1/y)^2 - (1)^2 \ dy}_{R_2}$$