# Revolving region bounded by $x = 16y^2 − 4y^3,\ x = 0$ about the $x$-axis, volume using cylindrical shells

Use the method of cylindrical shells to find the volume $V$ of the solid obtained by rotating the region bounded by the given curves about the $x$-axis. $x = 16y^2 − 4y^3,\ x = 0$

• so first i need to find the intersection $16y^2 -4y^3 \implies 4y^2(4 - y) \implies y = 4$

• then i need to find the radius so since im rotating about the $x$-axis my radius is in terms of $y$ hence my radius is ($y$) (maybe im wrong here, if i am can someone explain how to get the radius correctly?)

• finally, i need the height which is $16y^2 - 4y^2$

my final equation should be something like $\int_0^4 2\pi y(16y^2 - 4y^2)\, dy$

and my final answer would be $\frac{-5632\pi}{5}$ but its wrong and im not sure where i went wrong?

Your integral is set up correctly. However, you may have an error when you evaluated the integral. Check it out: $$2\pi\int_{0}^{4}y\left(16y^{2} - 4y^{3}\right)dy$$ $$2\pi\int_{0}^{4}\left(16y^{3} - 4y^{4}\right)dy$$ $$2\pi\left[4y^{4} - \frac{4}{5}y^{5}\right]_{0}^{4}$$ $$2\pi\left[\left(4 \cdot 4^{4} - \frac{4}{5} \cdot 4^{5}\right) - \left(0\right)\right]$$ $$2\pi \cdot \frac{1024}{5}$$ $$\boxed{\frac{2048\pi}{5}}.$$