The volume of the solid obtained by rotating the region enclosed by $y=\frac{1}{x^2} , y=0, x=3, x=8$ about the line $y=-1$.
How do I find the volume? I need help.
I tried $\pi\int_8^3(\frac{1}{x^2}-(-1))^2dx$ but i did not find the correct answer
2026-03-25 11:13:21.1774437201
find the volume using the method of disks or washers via an integral
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The disk integral is
$$\pi\int^8_3[ (\frac{1}{x^2}-(-1))^2-(0-(-1))^2]dx = \pi\left( -\frac1{3x^3}-\frac2x\right)\bigg|_3^8$$