Question

Find the volume of the solid formed by revolving the region bounded by the graphs of y = x^3, x = 2, and y = 1 about the y-axis.

I got 93pi/5 but it was wrong. Can anyone explain?

Answer 1

If you're using the shell method, the setup would be

`\displaystyle2\pi\int_1^2x(x^3-1)\,\mathrm dx`

Reasoning: consider just one shell (see the image below). The height of the shell is determined by the vertical distance between the curve
`y=x^3` and the line
`y=1`. The radius of each shell is determined by the value of
`x`, which can only range from 1 to 2 within the region of interest, and the height is determined exactly by
`x^3-1` at this
`x`. Multiply by
`2\pi` to get the surface area of this infinitesimally thin cylindrical shell, then integrate along the interval [1, 2] to get the volume.

We have

`\displaystyle2\pi\int_1^2x(x^3-1)\,\mathrm dx=2\pi\int_1^2(x^4-x)\,\mathrm dx=2\pi\left(\frac{x^5}5-\frac{x^2}2\right)\bigg|_1^2=\frac{47\pi}5`