How do I find the volume of the following function by using the disc method? I know it's easier with the shell method, but I'm required to do it with disc method and I'm missing something but can't figure out what? $$y = \frac{1}{{4+x^2}}{}$$ bounded by $x=0, x=2, y=0$.
I've used the following formula to calculate the volume of revolution $$\pi\int_{a}^{b}(f(y))^2 dy$$
If I do this, I first figured out $a$ and $b$ by just plug in $x=0$ and $x=2$ in $f(x)$, then calculated the inverse of the function $$x = \sqrt{\frac{1}{y}+4}$$ and then calculated the integral
$$\pi \int_{1/8}^{1/4}\left(\sqrt{\frac{1}{y}+4}\right)^2 dy$$
The solution I got was $\pi(\ln2-1/2)$, while it should just be $\pi\ln2$. What am I missing? What don't I understand?
Hint: $$y = \frac{1}{\sqrt{x^2+4}} \implies x^2 = \frac{1}{y^2}-4.$$