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Find the volume of the solid that results when the region enclosed by the curves y=1+x^3 x=1 and y=9 is revolved about the y-axis

Verified Answer :

The volume is given by the integral

\(\displaystyle2\pi\int_1^2x(9-(1+x^3))\,\mathrm dx=2\pi\int_1^2(8x-x^4)\,\mathrm dx\)

with the shell method, and

\(\displaystyle\pi\int_2^9((\sqrt[3]{y-1})^2-1^2)\,\mathrm dy=\pi\int_2^9((y-1)^{2/3}-1)\,\mathrm dy\)

with the washer method. Both integrals give a volume of \(\dfrac{58\pi}5\).