# Solid Volume rotation around y-axis

• Oct 11th 2011, 05:09 PM
paroxsitic
Solid Volume rotation around y-axis
I need to find the volume of the solid formed when rotating f(x) = 4x-x^2 and y=4 and x=0 about the y-axis.

Using the disc/washer method.

I already did it using the shell method and got 128/3 pi.

I am getting confused on how to do this, I believe it will need to be setup in 2 parts? Refer to my image sketch: http://k.minus.com/je3SvPydQUFsp.png Where I think the 2 parts are split by x=2. From x=0 to x=2 I see a disc method and then from x =2 to x=4 I see the washer method. However because the disc/washer method is perpendicular to the axis of rotation (y) that means that the thicknes of the disc/washers will be dy, which means the limits of integration would be in terms of y. Therefore I don't see how I can set it up with 2 equations with different integrals.
• Oct 12th 2011, 06:54 AM
tom@ballooncalculus
Re: Solid Volume rotation around y-axis
You're reading it as the volume of revolution of the shaded area, yes? In that case there should be mention at least of the line x = 4.

Edit: sorry, I do agree with your 128/3 pi.

$\displaystyle x = 2 - \sqrt{4 - y}$

and then the washer distance will be the same again. So you don't need to split it. Double $\displaystyle 2 - \sqrt{4 - y}$ is what you need to square before integrating with respect to y. Edit: no, pardon me. Doubling won't do. Didn't think it through. Sorry. More below.
• Oct 12th 2011, 07:02 AM
paroxsitic
Re: Solid Volume rotation around y-axis
Quote:

Originally Posted by tom@ballooncalculus
You're reading it as the volume of revolution of the shaded area, yes? In that case there should be mention at least of the line x = 4.

I get 32/3 pi from washers/discs as well as shells.

$\displaystyle x = 2 - \sqrt{4 - y}$

and then the washer distance will be the same again. So you don't need to split it. Double $\displaystyle 2 - \sqrt{4 - y}$ is what you need to square before integrating with respect to y.

I figured it that y=4 was the upper and 4x-x^2 was the lower so I did the area under y=4 and cutout the bottom function. I added x=4 myself because thats where the lower function ends but now I am debating to myself if the shaded area should really just be 4x-x^2 given they never tell me x=4.
• Oct 12th 2011, 07:08 AM
tom@ballooncalculus
Re: Solid Volume rotation around y-axis
I take it you mean under 4x - x^2. But then why mention y = 4? What's the precise wording?

Perhaps they mean the disc without the washer.
• Oct 12th 2011, 07:15 AM
paroxsitic
Re: Solid Volume rotation around y-axis
"Use the shell method and then the disc/washer method to find the volume of the solid formed by revolving the region bounded by the graph of: f(x)=4x-x^2, y=4 and x=0 about the y-axis."
• Oct 12th 2011, 07:22 AM
tom@ballooncalculus
Re: Solid Volume rotation around y-axis
Sounds like the disc without the washer. Perhaps just as well. I was wrong about doubling it. Pardon me.
• Oct 12th 2011, 07:41 AM
paroxsitic
Re: Solid Volume rotation around y-axis
Quote:

Originally Posted by tom@ballooncalculus
Sounds like the disc without the washer. Perhaps just as well. I was wrong about doubling it. Pardon me.

So the y=4 is redundant information or is my shaded region from my picture correct?

I used the following for my shell method:

thickness = dx
r = x
h = 4x-x^2

Integral from 0 to 4 [ (2*pi*x) (4x-x^2) ] dx
2pi * Integral from 0 to 4 [ (4x^2-x^3) ] dx
2pi * [ (4/3)x^3 - (1/4)x^4 ] -> 2pi*[(4/3)*4^3-(1/4)*4^4] = 2Pi*[64/3] = 128/3 pi

I guess that's incorrect?
• Oct 12th 2011, 09:09 AM
tom@ballooncalculus
Re: Solid Volume rotation around y-axis
But what you've got there is the area under the graph, right? If so, fine.

If you want the shaded region, then with shells it's

$\displaystyle 2 \pi \int _0^4 x [ 4 - (4x - x^2)]\ dx = 2 \pi \int _0^4 4x - 4x^2 + x^3\ dx$

for both parts, and obviously then just

$\displaystyle 2 \pi \int _0^2 4x - 4x^2 + x^3\ dx$

for the inside part. With discs, the inside part is

$\displaystyle \pi \int _0^4 (2 - \sqrt{4 - y})^2\ dx = \pi \int _0^4 8 - y - 4\sqrt{4 - y}\ dx$

and the other, which is a washer, is

$\displaystyle \pi \int _0^4 4^2 - (2 + \sqrt{4 - y})^2\ dx = \pi \int _0^4 8 + y - 4\sqrt{4 - y}\ dx$