I'm trying to find the volume of the region enclosed by the curves above rotated along the $y$-axis. I know that my lower and upper bounds will be $a=0$ and $b=1$. I'm not sure if I should set each curve equal to $x$, and even then I'm not sure how to go about setting up the integral.
I asked a similar question to this yesterday but it was along the $x$-axis. Thank you!
Best Answer
hint: Due to symmetry, one can find the volume of the upper part by dishwasher method and double the answer. We have: $V = 2\displaystyle \int_{0}^1 \pi((\sqrt[3]{y})^2 - y^2)dy$. Can you continue?