Because we are looking at revolution about an horizontal axis, using cross-section (See videos 1 through 6 here if you need a refresher) is more natural, so let us start with that.
Here is the region we are rotating about the xx-axis:
![enter image source here]()
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The first thing to understand is where the curves intersect.
y=\frac{1}{x}y=1x and y=2y=2 intersect for x=2x=2 and y=x^2y=x2 and y=\frac{1}{x}y=1x intersect for x=1x=1.
So the expression for the area A(x)A(x) of the cross section by x=cx=c is going to be different for 0\leq x\leq frac{1}{2}0≤x≤12 and for \frac{1}{2}\leq x\leq 112≤x≤1.
Namely, for 0\leq x\leq frac{1}{2}0≤x≤12, the cross-section is a washer with outer-radius 2, and inner-radius x^2x2 so that the area is A(x)=\pi (2^2-(x^2)^2)=\pi(4-x^4)A(x)=π(22−(x2)2)=π(4−x4).
On the other hand if \frac{1}{2}\leq x\leq 112≤x≤1, then the cross section is a washer with outer radius \frac{1}{x}1x and inner radius x^2x2, so that the area is
A(x)=\pi((\frac{1}{x})^2-(x^2)^2)=\pi(\frac{1}{x^2}-x^4)A(x)=π((1x)2−(x2)2)=π(1x2−x4).
As a result, the volume of the resulting solid of revolution is
V=\int_0^1A(x) dx=\int_0^{\frac{1}{2}}A(x)dx+\int_\frac{1}{2}^1A(x)dxV=∫10A(x)dx=∫120A(x)dx+∫112A(x)dx and
\int_0^\frac{1}{2}A(x)dx=\int_0^\frac{1}{2}\pi(4-x^4)dx=\pi[4x-\frac{x^5}{5}]_0^{\frac{1}{2}}=\pi(2-\frac{1}{160})∫120A(x)dx=∫120π(4−x4)dx=π[4x−x55]120=π(2−1160).
On other hand, \int_\frac{1}{2}^1A(x)dx=\int_\frac{1}{2}^1 \pi(\frac{1}{x^2}-x^4)=\pi [-\frac{1}{x}-\frac{x^5}{5}]_\frac{1}{2}^1=\pi(-1-\frac{1}{5}+2+\frac{1}{160})∫112A(x)dx=∫112π(1x2−x4)=π[−1x−x55]112=π(−1−15+2+1160).
Hence, altogether, we obtain
V=\pi(2-\frac{1}{160}+1-\frac{1}{5}+\frac{1}{160})=\pi(3-\frac{1}{5})=\frac{14\pi}{15}V=π(2−1160+1−15+1160)=π(3−15)=14π15.
Considering the length of the answer, I am leaving the cylindrical shell part to another time.
