Use the shell method to find the volume of the solid by rotating the region bounded by the given curves about the y-axis: $y=x^3, y=8$ and $x=0$.
I need to know if I am rotating around the y-axis using the shell method my definite integral points will be 0 and 8 right? What if I used the disk method?
Please someone hels me as I am a little confused about it.
Best Answer
The given region is bounded by the curve $y=x^3$, the horizontal line $y=8$ and the vertical line $x=0$. Using the method of disks, since the solid is obtained by rotating the region about the $y$-axis, the inner radius is 0 and the cross-sectional area is $$ A(y) = \pi\left(\sqrt[3]{y}\right)^2 = \pi y^{2/3}. $$ Hence, the volume is \begin{align*} V & = \int_0^8 A(y)\, dy \\ & = \pi \int_0^8 y^{2/3}\, dy \\ & = \left(\frac{3\pi}{5}\right)8^{5/3} \\ & = \frac{96\pi}{5}. \end{align*}