# Shell Method Integration (need help)

1. Sep 18, 2011

1. The problem statement, all variables and given/known data
Use the shell method to calculate the volume of rotation about the x-axis for the region underneath the graph.

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

2. Relevant equations

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

3. 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 $\sqrt[3]{x}$-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=$(y+2)^{3}$. I then started plugging it into the integral. I knew that since 8$\leq$x$\leq$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 $\int^{1}_{0}$2pi(y)(27-$(y+2)^{3}$).

From that integral my answer came out to be $\frac{54pi}{5}$ which I know is wrong because when I use the disk method the answer comes out to $\frac{38pi}{5}$ 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.

2. Sep 18, 2011

### LCKurtz

Your integral is set up correctly and will give 38pi/5 if you don't make any arithmetic errors.

3. Sep 18, 2011

I was wondering why every time I do the integral I got a different number.

2pi$\int$y(27-$(y+2)^{3}$)dx=2pi$\int$(27y-y$(y+2)^{3}$)dx=2pi$\int$(27y-y($y^{3}$+$6y^{2}$+12y+8))dx=2pi$\int$(27y-($y^{4}$+$6y^{3}$+$12y^{2}$+8y))dx =2pi$\int$(27y-$y^{4}$-$6y^{3}$-$12y^{2}$-8y)dx=2pi($\frac{27y^{2}}{2}-\frac{y^{5}}{5}-\frac{6y^{4}}{4}-\frac{12y^{3}}{3}-\frac{8y^{2}}{2})^{1}_{0}$=2pi[($\frac{27}{2}$-$\frac{1}{5}$-$\frac{3}{2}$-4-4)-(0)]=2pi(12-$\frac{1}{5}$-8)=2pi(4-$\frac{1}{5}$) =2pi($\frac{20}{5}$-$\frac{1}{5}$)=2pi($\frac{19}{5}$)=$\frac{38pi}{5}$

YESSSSSSS. Thanks so much for your help LCKurtz. I was doing it over and over changing the integral but by knowing that it was an arithmetic error, I realized when I did (y+2)(y+2)(y+2) I multiplied incorrectly leading to the whole thing being wrong. Thanks.

4. Sep 18, 2011

### LCKurtz

That's a dy integral, by the way. Instead of multiplying the cubic out like that, which, as you know, is very error prone, a much easier way to do the integral is to substitute u = y+2, du = dy.

5. Sep 18, 2011