Find volume using triple integration with cylindrical coordinates Find volume of $E$ using triple integration and cylindrical coordinates, when $E$ is bounded by $$x^2+y^2=x,\quad y=0,\quad y=x,\quad z=0,\quad z=\sqrt{x^2+y^2}$$
I know that in cylindrical coordinates $$x=r\cos\varphi,\quad y=r\sin\varphi,\quad z=h,\quad \text{ where } r\geq0 \text{ and } 0\leq\varphi\leq2\pi$$
but I'm very confused how to set up this integral. Cylindrical coordinates are quite new for me and it's hard to understand, how to make this conversion. So I would be grateful if anyone can help me with this.
 A: $x^2+y^2=x$ in polar form is just $r=\cos\varphi$. The lines $y=0$ and $y=x$ intersect this curve at $\varphi=0$ and $\varphi=\frac\pi4$ respectively. So our outer integrals are
$$\int_{\varphi=0}^{\pi/4}\int_{r=0}^{\cos\varphi}r\,dr\,d\varphi$$
The bounds of $z$ are just $z=0$ to $z=r$ in cylindrical coordinates, so our final answer is
$$\int_{\varphi=0}^{\pi/4}\int_{r=0}^{\cos\varphi}\int_{z=0}^rr\,dz\,dr\,d\varphi$$
A: I imagine the projection of your figure on $Oxy$ as upper segment cropped from circle by line $y=x$. So, firstly let's consider volume in Cartesian coordinates
$$\int\limits_{0}^{\frac{1}{2}}\int\limits_{x}^{\sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2}}\int\limits_{0}^{\sqrt{x^2+y^2}}dzdydx$$
So to find borders for cylindrical coordinates we need to solve inequalities
$$\begin{array}{}
0 \leqslant r \cos \phi \leqslant \frac{1}{2} & \\
r \cos \phi \leqslant r \sin \phi \leqslant \sqrt{\frac{1}{4} -\left(r \cos \phi-\frac{1}{2}\right)^2 } & \\
0 \leqslant z \leqslant r
 \end{array}$$
From 1-st and 2nd line inequalities we have $0 \leqslant \cos \phi \leqslant \sin \phi$, so we obtain $\phi \in \left[\frac{\pi}{4},\frac{\pi}{2}  \right]$ and $0 \leqslant r \leqslant \cos \phi$
$$\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}}\int\limits_{0}^{\cos \phi}\int\limits_{0}^{r}rdzdrd\phi$$
Addition:
In case where we consider "sector" of circle i.e. part of circle between lines $y=x$ and $y=0$, then volume in Cartesian coordinates will be
$$\int\limits_{0}^{\frac{1}{2}}\int\limits_{0}^{x}\int\limits_{0}^{\sqrt{x^2+y^2}}dzdydx+\int\limits_{\frac{1}{2}}^{1}\int\limits_{0}^{\sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2}}\int\limits_{0}^{\sqrt{x^2+y^2}}dzdydx$$
Good news in this case is, that we can calculate this volume again using the founded volume in previous , circle segment, case  subtracting it from the volume over the entire semicircle, which is
$$\int\limits_{0}^{1}\int\limits_{0}^{\sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2}}\int\limits_{0}^{\sqrt{x^2+y^2}}dzdydx=\int\limits_{0}^{\frac{\pi}{2}}\int\limits_{0}^{\cos \phi}\int\limits_{0}^{r}rdzdrd\phi$$
so we obtain
$$\int\limits_{0}^{\frac{\pi}{2}}\int\limits_{0}^{\cos \phi}\int\limits_{0}^{r}rdzdrd\phi-\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}}\int\limits_{0}^{\cos \phi}\int\limits_{0}^{r}rdzdrd\phi = \int\limits_{0}^{\frac{\pi}{4}}\int\limits_{0}^{\cos \phi}\int\limits_{0}^{r}rdzdrd\phi$$
