$R$ is a region that is defined by the curve $y=1/(4+x^2)$ and the lines $x=0$, $y=0$ and $x=2$. I need to find the volume of solid of revolution about the $y$-axis.
We know that $x^2 = (1-4y)/y$ and the boundaries are $y=0$ and $y=4$.
Then we know that the $$V = \pi \int_0^4(1-4y)/y \text{ }dy.$$
The answer should be $\pi \ln{2}$ which is not equal to my volume. What went wrong?
Notes:
First of all, at $x=2, ~y = (1/8)~~$ and at $~~x = 0, ~y = (1/4).$
Next, for $y$ between $0$ and $(1/8)$, the radius will be 2.
For $y$ between $(1/8)$ and $(1/4)$, the radius will be
$$ \sqrt{\left[\frac{1}{y} - 4\right]}.$$
Next, as you integrate between $y = 0$ and $y = (1/4),$ the cross sectional area is
$$\pi \times (\text{radius})^2.$$
Therefore the volume is
$$\pi \times \int_0^{(1/8)} (2)^2 dy $$
$$ + ~~\pi \times \int_{(1/8)}^{(1/4)} \left[\frac{1}{y} - 4\right] dy.$$
The $\pi \ln 2$ answer given is correct.
Can you take it from here?