Answer to Question #282221 in Calculus for Kkkk

Question

Find the volume of the solid generated by revolving the region bounded by the curves about the y axis y=x^3 , x=0 , y=1

Answer 1

$$y=x^3=>x=\sqrt[3]{y}$$

The solid lies between $y =0$ and $y= 1,$ its volume is

$$V=\displaystyle\int_{0}^1\pi(\sqrt[3]{y})^2dy=\pi[\dfrac{3}{5}y^{5/3}]\begin{matrix} 1 \\ 0 \end{matrix}=0.6\pi({units}^3)$$