# Rotating the region bounded by y=x^3, y=0, x=2 around the line y=8.

Makenna Booker 2022-07-21 Answered
Finding volume of solid using shell method.
Rotating the region bounded by $y={x}^{3},y=0,x=2$ around the line $y=8$.
I just want to double check if my initial formula is on the right track.
${V}_{y}={\int }_{0}^{8}\left(8-y\right)\left({y}^{\frac{1}{3}}\right)dy$
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Ali Harper
Explanation:
You are on the right track, but I think you want
$V={\int }_{0}^{8}2\pi r\left(y\right)h\left(y\right)dy={\int }_{0}^{8}2\pi \left(8-y\right)\left(2-{y}^{1/3}\right)dy$

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termegolz6
Step 1
Since you are rotating around the line $y=8$, which is parallel to the x-axis, it makes sense to integrate along the x-axis (i.e., integrate dx) rather than along y. The outer radius at the point x (let's call it R(x)) will be from $y=8$ to $y=0$ and the inner radius at the point x (let's call it r(x)) will be from $y=8$ to $y=0$ and the inner radius at the point x (let's call it r(x)) will be from $y=8$ to $y={x}^{3}$; this all happens over the region where x goes from 0 to 2.
Step 2
${\int }_{0}^{2}\pi \left[R\left(x{\right)}^{2}-r\left(x{\right)}^{2}\right]\phantom{\rule{thinmathspace}{0ex}}dx={\int }_{0}^{2}\pi \left[\left(8-0{\right)}^{2}-\left(8-{x}^{3}{\right)}^{2}\right]\phantom{\rule{thinmathspace}{0ex}}dx=\cdots$

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