# Shell Method Volume

1. Feb 1, 2013

### whatlifeforme

1. The problem statement, all variables and given/known data
Use the shell method to find the volume of the solid generated by revolving the regions bounded by the curves and the lines about the x-axis.

2. Relevant equations
Y=|x|/3 , Y=1

3. The attempt at a solution
1. 2∏∫(0 to 2) (y) (Y/4 - (-Y/4)) dy
2. y^3/6] (0 to 2)

Answer: 8∏/3 (which is wrong). the correct answer is 4∏ but i'm not sure how they go there.

Last edited: Feb 1, 2013
2. Feb 1, 2013

### Staff: Mentor

The above is incorrect. The upper limit of integration is not 2, and your formula for the area of the typical area element is also wrong. How did you get y/4?

As I mentioned in another thread, you need to draw a reasonably accurate sketch of the region that is being rotated, and another sketch of the solid that is produced.
Using the correct setup, I get a volume of 4π as well.

3. Feb 1, 2013

### whatlifeforme

1. 2∏∫(0 to 1) (y) (Y/3 - (-Y/3)) dy

would that be correct. i solved the equation for x=. also, is the upper limit of integration 1?

4. Feb 1, 2013

### whatlifeforme

i also tried integrating just the right side and multiplying by two.

2* 2∏∫(0 to 1) (y) (Y/3) dy

which i get the answer of 4pi/9.

1.if i draw a section parallel to the axis of revolutoin. the distance (radius) is y. thus the first part of the equation is right.

2.the height of the section would be the distance from (-y/3) on the left to (y/3). however, if i take just the right section of the absolute value function (y/3), and integrate, and multiply by two (2). then i should have the right value.

5. Feb 1, 2013

### Staff: Mentor

Better, but still not right. If y = (1/3)|x|, then |x| = 3y, not y/3.
No. It's 2x, which is not equal to 2y/3.
You will if you use the correct value for the width of your shell.