Simple Volume Problem: Solving for x and Rotating a Region around the Y-Axis

[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.
<br /> y=x^3
y=2x-x^2

I solved for x and got
<br /> 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 couldn't 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.

 

[Galileo]

I do have one question here, when given such equations as y=2x-x^2 how do you go about solving for x?

It's a quadratic in x:

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

 

[dextercioby]

At a first glance,your integration seems okay.

Daniel.

 

[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!