Calculating Volume of Rotation with Shell Method for Region Underneath Graph

[DeadOriginal]

Homework Statement

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\leqx\leq27

Homework Equations

I was taught that I would set up the integral by using the area which would look something like \int2pi(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 \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\leqx\leq27 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.

 

[LCKurtz]

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

 

[DeadOriginal]

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

2pi\inty(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.

 

[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.

 

[DeadOriginal]

Woops. I totally messed up with the dx's haha. Hmm. I will try with the substitution. Thanks so much for your help.