Calc5.2

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Find the volume of the solid generated by revolving the region bounded by y=x^{2} and y=4x-x^{2} around the x-axis.

To do this problem, we first need to figure out what the region looks like. The intersection of the two graphs comes when

x^{2}=4x-x^{2}\Rightarrow 2x^{2}-4x=0\Rightarrow 2x(x-2)=0\Rightarrow x=0,2

The function y=4x-x^{2} is the outermost function since we are revolving about the x-axis. Thus, we can find the volume by finding the volume of the solid generated by revolving 4x-x^{2} about the x-axis and subtracting the volume of the solid generated by revolving y=x^{2} about the x-axis.

V=\pi \int _{{0}}^{{2}}(4x-x^{2})^{2}\,dx-\pi \int _{{0}}^{{2}}(x^{2})^{2}\,dx=\pi \int _{{0}}^{{2}}(-8x^{3}+16x^{2})\,dx

=\pi \left[-2x^{4}+{\frac  {16}{3}}x^{3}\right]_{{0}}^{{2}}=\pi \left[-32+{\frac  {128}{3}}\right]={\frac  {32\pi }{3}}


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