triple integral and cyclindrical coordinates Find the volume of the region cut from the cylinder $x^2+y^2=4$ by the plane $z=0$ and the plane $x+z=3$.
I know that I will use polar coordinates, however I wonder how to graph it. $z=0$ means $xy$-plane. I cannot solve type of these questions. Is there any person who teaches or helps me? Thank you.
 A: Since you're using cylindrical coordinates the bounds for the integral will be $\theta:[0,2\pi],\;\; r:[0,2]$ and $z:[0,3-r\cos{\theta}]$ Therefore constructing the integral with Jacobian$=r$ we have $$\int_0^{2\pi}\int_0^{2}\int_0^{3-r\cos{\theta}}r \; dzdrd\theta$$ I myself am pretty new to volume integrals, but evaluating this expression is how I would do it.
Here is a visual representation of your volume :) Beware that the picture is different at different orientations, here the $z$ axis is vertical, the $x$ axis horizontal and the $y$ axis out of the page towards us.
 
A: Answer:
Using Cylindrical Coordinates, $z = z, x = rcos\theta, y = rsin\theta$.
The limits for z runs from 0 at the bottom of the cylinder to a plane 3-x.  Now substitute the expression of x in cylindrical cooridates you get upper limit of z to be $3-rcos\theta$.
Now in cylindrical coordinates $x^2 + y^2 = r^2$.  In your equation you have mentioned the cylinder to be $x^2+y^2 = 2^2$.  So limits for r runs from 0 in the centre to 2 at the outer extreme.  Now the you want the volume of the cylinder, so $\theta$ should run from 0 and make one complete revolution and that is $2\pi$.  Thus the limits are from 0 to $2\pi$
Evaluate the the following integral:
$$\int_{0}^{2\pi} \int_{0}^{2} \int_{0}^{3-rcost\theta} r dz dr d\theta$$
$$\int_{0}^{2\pi} \int_{0}^{2} (3-rcos\theta) rdrd\theta$$
$$\int_{0}^{2\pi} [(\frac{3r^2}{2} - \frac{r^3cos\theta}{3})|2,0] d\theta$$
$$ \int_{0}^{2\pi}[6-\frac{8}{3}cos\theta]d\theta$$
$$ [6 - \frac{8}{3} sin\theta|2\pi,0]$$
$$ [12\pi - 0]$$
$$ 12\pi$$
