Calc5.3

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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 x-axis.

First, we need to find the intersection points of the two graphs. These are easy to find with these particular points. The intersections come at x=0,1. Since x^{2} is greater than x^{3} between 0 and 1, and because we are revolving this region about the x-axis, our outer function is y=x^{2}. So, we will find the volume of the solid of revolution using y=x^{2} and subtract from this the volume of the solid of revolution using y=x^{3}. That is,

V=\pi \int _{{0}}^{{1}}\left[(x^{2})^{2}-(x^{3})^{2}\right]\,dx=\pi \int _{{0}}^{{1}}(x^{4}-x^{6})\,dx=\pi \left[{\frac  {1}{5}}x^{5}-{\frac  {1}{7}}x^{7}\right]_{{0}}^{{1}}={\frac  {2\pi }{35}}


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