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.


It is symmetrical about $y=2$.
For $0 \le y \le 2$, we have $R(y)=y$.
\begin{align}
V &= 2\pi \int_0^2 y^2 \, dy\\
&= 2\pi \left.\frac{y^3}{3}\right|_0^2\\
&= \frac{16\pi}{3}
\end{align}
Note that $R(y)$ should be the radisu when $y$ takes a certain value.
