1. Mar 8, 2005

### mewmew

Ok,so I have a simple volume problem that i'm having a little trouble with.

I want to rotate the region in the first quadrant bounded by the following formulas around the Y axis.
$$y=x^3$$
$$y=2x-x^2$$

I solved for x and got
$$x=\sqrt[3]{y}$$
$$x=1- \sqrt{1-y}$$ I know I get two roots but I am pretty sure this is the one I want.

I do have one question here, when given such equations as $$y=2x-x^2$$ how do you go about solving for x? I used Mathematica because I couldnt get it, but anyways back to the main problem.

so I then did $$\pi \int^1 _0 \sqrt[3]{y}^2 - (1- \sqrt{1-y})^2 dy$$

I expanded $$(1- \sqrt{1-y})^2$$ and then integrated the expresion and got the following:

$$\pi [\frac{3y^\frac{5}{3}}{5} - 2y - \frac{4}{3} (1-y)^{\frac{3}{2}}+\frac{y^2}{2}]^1_0$$

This isn't right however and I assume I messed up somewhere in my final integration, any help? Thanks a lot. I am also going to assume it has to do with $$(1- \sqrt{1-y})^2$$, I did U substitution on the term $$2 \sqrt{1-y}$$ and am thinking I messed that up, but once again, am not sure. The only other thing I can think about is maybe I missed up my functions somehow. Thanks for any help.

Last edited: Mar 8, 2005
2. Mar 8, 2005

### Galileo

$$x^2-2x+y=0 \Rightarrow x=1\pm \sqrt{1-y}$$

3. Mar 8, 2005

### dextercioby

At a first glance,your integration seems okay.

Daniel.

4. Mar 8, 2005

### mewmew

Oh, sorry, I think I just messed up evaluating it, I forgot that all the terms aren't 0 when it is evaluated at 0...I make the dumbest mistakes!