So I have spent about a hour on this problem and figured it was time to ask for some advice. The problem is to find the volume using cylindrical shells by rotating the region bounded
by
$$8y = x^3,\qquad y = 8,\qquad x = 0$$
about the x-axis.
I changed the one on the far left to $x = 2*y^{1/3}$.
I said the region of the integral went from $0$ to $8$ and that the integrand is
$$2 * \pi * y * (2* y^{1/3})\,dy$$
I figured the height would be $x = 2*y^{1/3}$ and that the radius is just $y$. Any help would be greatly appreciate since I can't find any help for this online. I searched around and found some similar problems but it still doesn't make sense to me. Thanks!
Best Answer
I agree with you. The volume is \begin{align*} \int_0^8 2\pi y x\,dy &= \int_0^8 2\pi y \left(2\sqrt[3]{y}\right)\,dy \\ &= 4\pi \int_0^8y^{4/3} \,dy \\ &= 4\pi \left[\frac{3}{7}y^{7/3}\right]^8_0 = \frac{12\pi}{7} \cdot 2^7 = \frac{1536\pi}{7} \end{align*}