Find the Volume of a Solid Revolution around the y axis

Solution 1:

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.

Solution 2:

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}$