Ok,so I have a simple volume problem that i'm having a little trouble with.(adsbygoogle = window.adsbygoogle || []).push({});

I want to rotate the region in the first quadrant bounded by the following formulas around the Y axis.

[tex]

y=x^3[/tex]

[tex]y=2x-x^2[/tex]

I solved for x and got

[tex]

x=\sqrt[3]{y}[/tex]

[tex]x=1- \sqrt{1-y}[/tex] 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 [tex]y=2x-x^2[/tex] 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 [tex]\pi \int^1 _0 \sqrt[3]{y}^2 - (1- \sqrt{1-y})^2 dy[/tex]

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

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

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 [tex](1- \sqrt{1-y})^2[/tex], I did U substitution on the term [tex]2 \sqrt{1-y}[/tex] 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.

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# Homework Help: Simple volume help please

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