Find the Volume of a Solid Revolution around the y axis Having trouble with this question from my OpenStax Calculus Volume 1 Homework, It is question 89 of Chapter 6 about Solid Revolution. I put my math below:
y=4-x, y=x, x=0 Find the volume when the region is rotated around the y-axis.
$$y= 4-x \Rightarrow x = 4 - y \implies R(y) = 4 - y$$
This is rotated around the y-axis should thus follow the general formula (c and d are position on the y axis):
$$V=\pi\int_{c}^{d} R^2(y) \, dy $$
Using this I inserted the relevant values and variables:
$$ 
V= \pi\int_{0}^{4} (4-y-y)^2  \,dy  
= \pi\int_{0}^{4} (4-2y)^2 \,dy 
= \pi\int_{0}^{4} (16 - 16y +4y^2) \,dy 
= \pi\Big[16y - \frac{16y^2}{2} + \frac{4y^3}{3}\Big] \Bigg|_{0}^{4} 
= \pi \ \Big[16(4) - \frac{16(4^2)}{2}+ \frac{4(4^3)}{3} \Big] -\pi\big[0 \big] 
= \pi \Big[64 - 128 - \frac{256}{3} \Big] 
= \pi \Big[\frac{64}{3} \Big]
= \frac{64\pi}{3} 
$$
The Official Answer given in the textbook is $\frac{16\pi}{3} $ but I am not sure how to get there. Some help would be greatly appreciated.

The region is as stated in the diagram and is rotated around y-axis.

If you are using disk method, it should be two integrals:
$ \displaystyle \int_0^2 \pi y^2 ~ dy ~ + \int_2^4 \pi (4-y)^2 ~ dy = \frac{8 \pi}{3} + \frac{8 \pi}{3} = \frac{16 \pi}{3}$
You can alternatively use the shell method which is a single integral in this case. The volume of revolution using shell method is,
$ \displaystyle \int_a^b 2 \pi h(x) \cdot x ~ dx ~, ~$ where $h(x)$ is the height of shell as a function of $x$, and $a$ and $b$ are bounds of $x$.
$h(x) = 4-x-x = 4 - 2 x$
So the integral is $ \displaystyle \int_0^2 2 \pi x \cdot (4-2x) ~ dx = \frac{16 \pi}{3}$
