Find the volume bounded by a cylinder. Find the volume bounded by the cylinder $x^2 + y^2=1$ and the planes $y=z , x=0 ,z=0$ in the first octant.
How do I go about doing this?
 A: In general the volume of a  region $S$ bounded above by a surface defined by a function of two variables $z=f(x,y)\ge 0$, whose domain is $R$,  and below by the plane $z=0$, is given by:


*

*the double integral $\iint_R f(x,y)\,dA=\iint_R f(x,y)\,dx\,dy$, 

*or by the triple integral   $\iiint_S \,dV=\iiint_S \,dx\,dy\,dz.$ 


The following figure represents the given region of the first octant ($x\ge 0,y\ge 0,z\ge 0$) bounded by the cylinder $x^{2}+y^{2}=1$ (gray) and the plane $y=z$ (green).

The domain $R$ of the function $z=f(y)=y$ is defined in Cartesian coordinates $x,y$ by
\begin{eqnarray*}
R &=&\left\{ (x,y)\in\mathbb{R}^{2}:0\leq x^{2}+y^{2}\leq 1,x\ge 0,y\ge 0\right\} .
\end{eqnarray*}
or in polar coordinates $r,\theta$, with $x=r\cos \theta , y=r\sin \theta $, by
\begin{eqnarray*}
T &=&\left\{ (r,\theta )\in \mathbb{R}^{2}:0\leq r\leq 1,0\leq \theta \leq \pi /2\right\} .
\end{eqnarray*}
As such the requested volume $V$  may be given in Cartesian coordinates by the double integral
\begin{equation*}
V=\iint_{R}z\,dx\,dy=\iint_{R}y\,dx\,dy.
\end{equation*}
Using polar coordinates, since the Jacobian determinant of the transformation from Cartesian to polar coordinates is $J=\frac{\partial (x,y)}{\partial (r,\theta )}=r$, the integral is converted into the following separable one:
\begin{eqnarray*}
V &=&\iint_{T}(r\sin \theta)\, |J|  \,dr\,d\theta =\int_{r=0}^{1}\int_{\theta =0}^{\pi
/2}r^{2}\sin \theta \,dr\,d\theta  \\
&=&\left( \int_{0}^{1}r^{2}dr\right) \left( \int_{0}^{\pi /2}\sin \theta
\,d\theta \right) =\left. \frac{1}{3}r^{3}\right\vert _{0}^{1}\left. \left(
-\cos \theta \right) \right\vert_{0}^{\pi /2} =\frac{1}{3}.
\end{eqnarray*}
A: This might be easier in cylindrical coordinates. You just want the region in which $r$ runs from $0$ to $1$, $\theta$ runs from $0$ to $\frac{\pi}2$, and $z$ runs from $0$ to $y=r\sin\theta$. That gives:
$$\int_0^{\frac\pi2}\int_0^1\int_0^{r\sin\theta}r\,dz\,dr\,d\theta$$,
Which I believe comes out to $\frac13$.
The hard part of these things is visualizing the shape and setting up the integral. Do you see where the limits on each of those variables comes from?
