The volume of the region enclosed by $x=y^{3}$, $x=0$, $y=2$, revolved around the X-axis is?
Could I solve it with this integral?
$$\int_1^8 \pi(x^{1/3})dx$$
And another question, how would that solid be represented?
2026-04-09 03:51:24.1775706684
The volume of the region enclosed by $x=y^3$, $x=0$, $y=2$
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You want the area between $y=x^{\frac13}, x=0, y=2$ revolved around the x-axis, thus, you get $$\pi \int_{0}^{8}[(2)^2-(x^\frac13)^2]\, dx = \pi(32-\frac35(32))=\frac{64\pi}{5}$$