# Write the Integral Equation

1. Oct 27, 2014

### mshiddensecret

1. The problem statement, all variables and given/known data
The volume of the solid obtained by rotating the region bounded by
x=6y^2 [PLAIN]http://msr02.math.mcgill.ca/webwork2_files/jsMath/fonts/cmmi10/alpha/144/char3B.png [Broken] [Broken] y=1 [PLAIN]http://msr02.math.mcgill.ca/webwork2_files/jsMath/fonts/cmmi10/alpha/144/char3B.png [Broken] [Broken] x=0[PLAIN]http://msr02.math.mcgill.ca/webwork2_files/jsMath/fonts/cmmi10/alpha/144/char3B.png [Broken] [Broken] it is rotate about the y- axis.

2. Relevant equations

3. The attempt at a solution

Using the disk method, I figured out by integrating (pi)(6y^2)^2 dy from 0 - 1 and got the answer: 22.619

However, I cannot write an integral equation using the cylindrical shell method. My attempt is:

Integrating (2pi)(sqrt(x/6))dx from x = 0 to 6.

Last edited by a moderator: May 7, 2017
2. Oct 27, 2014

### Staff: Mentor

Check the length of your shells. How long is a shell for some specific x-value? In particular, is it increasing as your approach would suggest?

3. Oct 27, 2014

### mshiddensecret

What do you mean? Isn't it always x?

4. Oct 27, 2014

### HallsofIvy

No! It is not. The whole point of the "shell" method is that the shells are parallel to the axis of revolution. Here, that is parallel to the y-axis.

And if you meant "Isn't it always y?" Again, no it it isn't. x= 6y^2 is the lower boundary, y= 1 is the upper boundary. The length of a shell is the vertical distance between them.

5. Oct 27, 2014

### mshiddensecret

I got it with this:

(2pi)(1-sqrt(x/6))(x)dx

thanks.