Finding Volume with Rotational Solids: Cylindrical Shell Method

[mshiddensecret]

Homework Statement

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 y=1 [PLAIN]http://msr02.math.mcgill.ca/webwork2_files/jsMath/fonts/cmmi10/alpha/144/char3B.png x=0[PLAIN]http://msr02.math.mcgill.ca/webwork2_files/jsMath/fonts/cmmi10/alpha/144/char3B.png it is rotate about the y- axis.

Homework Equations

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.
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[mfb]

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?

 

[mshiddensecret]

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

 

[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.

 

[mshiddensecret]

I got it with this:

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

thanks.