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Find the volume of the solid generated by revolving the region bounded by y=x^{2} and y=x^{3} around the y-axis.

By inspection, we can see that these two curves intersect at x=0,1. Also, in revolving this region about the y-axis, the function y=x^{3} is the outside function. All we need to do the integral is to change our functions into functions of y, instead of x.

V=\pi \int _{{0}}^{{1}}\left[(y^{{{\frac  {1}{3}}}})^{2}-(y^{{{\frac  {1}{2}}}})^{2}\right]\,dy=\pi \int _{{0}}^{{1}}\left[y^{{{\frac  {2}{3}}}}-y\right]\,dy=\pi \left[{\frac  {3}{5}}y^{{{\frac  {5}{3}}}}-{\frac  {1}{2}}y^{2}\right]_{{0}}^{{1}}={\frac  {\pi }{10}}

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