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## Homework Statement

Use the shell method to calculate the volume of rotation about the x-axis for the region underneath the graph.

y=[itex]\sqrt[3]{x}[/itex]-2, 8[itex]\leq[/itex]x[itex]\leq[/itex]27

## Homework Equations

I was taught that I would set up the integral by using the area which would look something like [itex]\int[/itex]2pi(r)(h).

## The Attempt at a Solution

I saw that the maximum height was 27 and so to get the height of the cyclinder I thought that I would have to subtract the height of the equation [itex]\sqrt[3]{x}[/itex]-2 from 27 to get the height I need to set up the integral. Since I was rotating around the x-axis I first converted the equation so that it would be in terms of y which came out to be x=[itex](y+2)^{3}[/itex]. I then started plugging it into the integral. I knew that since 8[itex]\leq[/itex]x[itex]\leq[/itex]27 I would be integrating from (0,1) because those are the possible y values for this equation within that range. My integral came out to be [itex]\int^{1}_{0}[/itex]2pi(y)(27-[itex](y+2)^{3}[/itex]).

From that integral my answer came out to be [itex]\frac{54pi}{5}[/itex] which I know is wrong because when I use the disk method the answer comes out to [itex]\frac{38pi}{5}[/itex] which also happens to be the answer in the textbook. Could someone please check my integral to see if I have it correct? Thanks a lot.