Volumes with Cylindrical Shell Method

[Ivan92]

Homework Statement

Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by the given curves about the x-axis.

y= 4x², y=-2x+6

Homework Equations

y= 4x², y=-2x+6

These 2 equations meet at x= -3/2 and x=1

integral from a to b of (2∏x*f(x)) dx

The answer is 250pi/3.

The Attempt at a Solution

Been spending over an hour with this and can't figure out a way to figure this out by shells. I can do this by discs and got the answer right. However, I want to figure this out by shells. First I drew the graph of these 2 equations to find where they meet at. I then did this:

\int^{-3/2}_{1} 2\pix((-2x-6)-(4x²)) dx

I put this in Wolfram and I did not get the intended answer. I work this out by hand and I am getting something very different from Wolfram. Guidance please! Thanks in advance.

 

[SammyS]

Do do this problem using cylindrical shells, you need to express x as a function of y. The integral will be with respect to y. In other words, your differential has dy in it.

If the region to be rotated is indeed above y = 4x² and below y = -2x+6, then you will need to break your integral into to pieces; one piece from y = 0 to 4, the other from y=4 to 9.

 

[Ivan92]

Ha I figured using y from 4 to 9 but I do not see why 0 to 4. You would be integrating the function itself. Then I don't see what you would do with the 2 integrals.

 

[SammyS]

Draw the graph.

x = ±(√y)/2

So for y = 0 to 4, use (√y)/2 - (-√y)/2

What did you use for y = 4 to 9 ?

 

[Ivan92]

I didn't use them, I just found the zeros and got 4 through 9 before, but I kind of didn't know what to do with it. Wouldn't the above just be √y by itself? Here is what I would do now

∫ from 0 to 4 (2πy√y)dy - ∫ from 4 to 9 (2pi*y*((√y)/2+.5y-3)dy)

Would this be right? Thanks for your assistance Sammy S.

 

[SammyS]

Yes, that's right.

 

[Ivan92]

Sweet! Thanks for your guidance Sammy!