Today in my Calculus class my teacher made an example using change of variables using this problem:
Find the volume of the solid inside the cylinder $y^{2} + z^{2} = 2y$ bounded by $x=0$ and the surface of the cone $x^{2}=y^{2} + z^{2}$. With $x \geq 0$ .
He wrote directly the change of variables like this: from (x, y, z) to $(\theta, r, x)$ which is, I think, the same as cylindrical coordinates but with rotated axis (I don't know how to call it)? because instead of $z$ he used $x$ for the height because it was convenient for the problem (if you make a graph, the cylinder lies in the $x$ axis).
The thing is, I didn't understand the bounds for $(\theta, r, x)$. He wrote the answer and it should be this:
$$-\frac{\pi}2 \leq \theta\leq \frac{\pi}2,$$ $$0 \leq r\leq 2cos(\theta),$$ $$0 \leq x \leq r,$$
After the integration, the result should be 32/9.
I've been trying to think all day but I still have troubles. Could you explain to me the change of variables, please? The $\theta$ angle bounds are the most confusing. Please.
I'm sorry if I made a mistake with the equations format.
Note that$$y^2+z^2\leqslant2y\iff(y-1)^2+z^2\leqslant1.$$But then $|y-1|\leqslant1$, which means that $0\leqslant y\leqslant2$. But $y=r\cos\theta$ and so, since $y\geqslant0$, $\theta\in\left[-\frac\pi2,\frac\pi2\right]$. Besides,$$y^2+z^2\leqslant2y\iff r^2\leqslant2r\cos\theta\iff r\leqslant2\cos\theta.$$Furthermore,$$x^2\leqslant y^2+z^2\iff x^2\leqslant r^2\iff0\leqslant x\leqslant r,$$since we are assuming that $x\geqslant0$. So, the computation will be\begin{align}\int_{-\pi/2}^{\pi/2}\int_0^{2\cos(\theta)}\int_0^rr\,\mathrm dx\,\mathrm dr\,\mathrm d\theta&=\int_{-\pi/2}^{\pi/2}\int_0^{2\cos(\theta)}r^2\,\mathrm dr\,\mathrm d\theta\\&=\int_{-\pi/2}^{\pi/2}\frac83\cos^3(\theta)\,\mathrm d\theta\\&=\frac{32}9.\end{align}