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$$\pi\int\limits_0^2(4y^2-y^4)dy=\pi\left(\frac{32}{3}-\frac{32}{5}\right)=\frac{64\pi}{15}$$

Added: Look at both curves as functions of the form $\,x(y)\,$. Draw a diagram if necessary and find out where they meet.

The function $\,x=2y\,$ is above the function $\,x=y^2\,$ on $\,y\in [0,2]\,$ , i.e. $\,2y\geq y^2\,\,,\forall\,y\in [0,2]\,$ , so the volume of revolution is Pi times the integral on the given interval of the difference of squares of the given functions.

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Thanks, Andre. I really need to go to sleep now. –  DonAntonio Jan 17 '13 at 4:10
    
The interval of integration is [0, 2], not [0, 1]; the result seems correct. –  pharmine Jan 17 '13 at 6:52
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