Find the volume of the solid generated by revolving the area bounded by y=4x−x2, y=x, about x=3.
Solution.
V=π(∫03(4x−x2)2dx−∫03x2dx)=π(∫03(16x2−8x3+x4−x2)dx)=π(16x33−8x44+x55−x33)∣03=1085π=2135π.V=π(\int_0^3(4x-x^2)^2dx-\int_0^3x^2dx)=\newline π(\int_0^3(16x^2-8x^3+x^4-x^2)dx)=\newline π(\frac{16x^3}{3}-\frac{8x^4}{4}+\frac{x^5}{5}-\frac{x^3}{3})|_0^3=\newline \frac{108}{5}π=21\frac{3}{5}π.V=π(∫03(4x−x2)2dx−∫03x2dx)=π(∫03(16x2−8x3+x4−x2)dx)=π(316x3−48x4+5x5−3x3)∣03=5108π=2153π.
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