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.

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.

Your disc radius, i.e. the length of the strip in your pic, should be

and then the washer distance will be the same again. So you don't need to split it. Double 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.

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.

Your disc radius, i.e. the length of the strip in your pic, should be

and then the washer distance will be the same again. So you don't need to split it. Double

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.

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.

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."

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.

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?

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

for both parts, and obviously then just

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

and the other, which is a washer, is