Find the volume of the solid obtained by rotating the region bounded by

y=x-x^2
y=0

About the line x = 2

Not sure if it's a solid or a shell, but it's confusing me.

Help?

It's gonna be a solid. And remember that y=0 is a HORIZONTAL line.

You end up with something like this:


...crap, had a graph on here but it won't work..



It'll look something like a D profile ring.

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Right, I know that, but I have to find the area around x=2, not the area between y=x-x^2 and y=0.

So I have to get y=x-x^2 in terms of x, and that is proving difficult.

It would look like the upper half of a toroid (upper half of a donut).

The y=x-x^2 is the upper profile of the shape.

The y=0 is the lower (planar) surface of the shape.

x = 2 defines the center of the hole.

Use the shell method and add up the multiple cylindrical integrated 'shells'. Which will constitute the shape.

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Um,

y(x) = x - x^2 = x(1 - x)

y = 0 when x = 0 or x = 1


did you want x(y)?

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I think I figured it out, that or I completely botched it.

If x=0 and y=0, and y=x-x^2 then y=x and y=x^2, as x^2=x for 0 and 1.

So my y's are y, and root y.

Probably incorrect, but that's about as good of a guess as I could come up with. Area was pi/2.

The problem didn't ask for the area. It asked for the volume.

I am confused about the conditions set in the original problem. You said:

y=x-x^2
y=0

I think you need a range of y's. Otherwise, the volume would be zero because the solid would only have 2 dimensions (x and z). It's got to have a range of y's to give it 3 dimensional volume. Does that make sense? I may have missed the point.

Sorry I meant volume.

Yeah it's pretty confusing, I mean I know how to find volume of solids and shells, but the y=x-x^2 is what threw me off.

I'll see if I can scan what I did to show you if I did it correctly or not.

http://faculty.eicc.edu/bwood/math150supnotes/supplemental26.html