Shell Method Versus Disc Method

Question

I'm trying to find the volume of a solid formed by revolving the curve $y=x^2+2$ around the y axis between $y=2$ and $y=4$, but I'm getting a different answer using the shell method and the disc method.

Using Disc Method: $V=\pi\int_{2}^{4}(\sqrt{y-2})^2dy = 2\pi$

Using Shell method: $V=2\pi\int_{0}^{\sqrt{2}}\left(x^{2}+2\right)x\ dx = 6\pi$

The solution says the answer is $2\pi$, but WolframAlpha says the volume is $6\pi$. Any explainations for this discrepancy would be great.

Answer 1

The solid whose volume you calculated using the disk method is not the solid whose volume you calculated using the shell method. If you used shells on the first solid, your shells would run from the curve $y=x^2+2$ up to $y=4$, and the height of the shell would be $4-(x^2+2)=2-x^2$; you, however, have taken the shells from the $x$-axis up to the curve. Had you computed that volume using the disk method, you’d have had annuli, not disks, with inner radius $\sqrt{y-2}$ and outer radius $\sqrt2$.

The two volumes combine to form a cylinder of radius $\sqrt2$ and height $4$, with volume $8\pi$, the sum of your two figures.

Answer 2

The disc method is correct. The shell method gives $2\pi\int_0^\sqrt2(2-x^2)x\rm dx=2\pi$.

What you inputted for Wolfram Alpha was different. It was the volume (swept out by the region) below the curve (and inside $x=\sqrt2$).