Finding the volume bounded by a cylinder and a plane I have been given the following equations:
$$x^2 + z^2 = 9$$
$$ x = 0 $$
$$ y = 0 $$
$$ z = 0 $$
$$ x + 2y = 2 $$
and have been asked to find the volume of the bounded region. I understand the principle of the integration and how to apply it in this scenario, but I tend to have trouble determining the bounds of integration for a three dimensional region such as the one described above. I have tried sketching the cylinder and the plane, but to no avail; I'm consistently unable to determine the bounds. In general, how would one go about determining the integral bounds for a region such as the above?
 A: Sketching is indeed a good idea when you deal with non-trivial boundaries. In this case, the main boundaries are the coordinate planes $x=0$, $y=0$, and $z=0$. Then, by inspection, we notice that the given plane $x = 2- 2y$ makes the bound region (in $x$-$y$ plane) in the first octant, which means that our boundaries become
$x\in[0, 2-2y]$, $y\in [0, 1]$ and $z \in [0, \sqrt{9-x^2}]$.
The latter boundary appears due to the fact that the cylinder is oriented along the $z$-axis. 
I guess practice of working with this type of problems bring essential experience and then each new problem will look easier and easier.
Hope this helps.
A: OP: Just to provide some perspective on what others (such as @Pavel) have said:
The desired volume is bounded by the planes $x=0$, $y=0$ and $z=0$.
Since the equation of the cylinder is given by $x^2+z^2=9$, this means that the axis of the cylinder is parallel to the $y$ axis, and it has a radius $r=3$.
Now, $x+2y=2$ is the equation of a plane that lies parallel to the $z$ axis, and hence does not depend on $z$. Also, where this plane intersects the $xy$ plane, the equation of the line (in the $xy$ plane) is $y=\frac{1}{2}\left(2-x\right)$. Also, since this plane lies in the positive quadrant, it can be inferred that $x \ge 0$ and $y \ge 0$ is desired.
Finally, because of the symmetry of the problem, I can assume $z \ge 0$.
Therefore, the desired region is bounded above by the cylinder $z=\sqrt(9-x^2)$ and below by the plane $z=0$.
Also, projecting into the $xy$ plane, the area is bounded by $0 \le y \le \frac{1}{2}\left(2-x\right)$, and $0 \le x \le 2$.
$$\therefore V= \int_0^2{\int_0^{\frac{1}{2}\left(2-x\right)}{\int_0^{\sqrt(9-x^2)} {1}\mathrm{d}z}\mathrm{d}y}\mathrm{d}x$$ 
where $V$ is the desired volume.

As others have pointed out, it is best to be able to sketch these regions (by hand) and use a CAD tool (e.g. Matlab, Mathematica, if available) to confirm correctness. This will build your confidence that subsequent sketches are correct.
A: Apologies in advance, for bad English. You need to start off, by integrating left, integrating right and combining. For example, z component vector, you can integrate first in this case (because it is involved in the least equations). Then, you may go about, triple integrate the other ones (for example, x component first then y component). You will easily be able to determine the integral bounds after this, as you've integrated all three components.
