When the region enclosed by the graphs of $y = 2x$ and $y = 6x-x^2$ is revolved around the $y-axis$, the volume of the solid generated by is given by?
2026-03-28 01:06:02.1774659962
A volume integral question
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The graphs intersect at $x=0,4$. Hence, by the shell method, $$V=2\pi\int^4_0x\left((6x-x^2)-(2x)\right)dx=2\pi\int^4_0\left(4x^2-x^3\right)dx$$ You can do the integral.