# Volume of a region bounded by a surface and planes

1. Jul 13, 2012

### forestmine

1. The problem statement, all variables and given/known data

Find the volume of the region bounded by the cylinder x^2 + y^2 =4 and the planes z=0, and x+z=3.

2. Relevant equations

V = ∫∫∫dzdxdy

V=∫∫∫rdrdθ

3. The attempt at a solution

Alright, so I feel as though I'm missing a step somewhere along the way, but here's what I've gotten so far.

So I know that I've got a cylinder centered around the z-axis, bounded by z=0 (which would be the xy-plane) and x+z=3.

If I attempt a triple integral in Cartesian coordinates, beginning with the z limits of integration, I enter the region at z=0, and exit the region at z=3-x? I'm having a hard time picturing the x=3-x part since y=0, and so we aren't actually looking at the whole cylinder?

Well from there, I looked at the circle cast by the cylinder on the xy-plane, x^2+y^2 = 4. I solved for y for my y limits of integration, -(4-x^2)^1/2 to + (4-x^2)^1/2.

And then my x limits are simply -2 to 2.

I feel like I'm not taking into account the z=3-x for my y and x limits, though I'm not sure how I would...

Here's my attempt in cylindrical coordinates.

Since x^2+y^2 = r^2=4, I said my r limits are from r=0 to r=2.

My z limits are from z=0 to z=3-rcosθ.

And my limits for θ are from 0 to 2pi.

I feel like I'm missing something...Any help in the right direction would be greatly appreciated!

Thanks!

2. Jul 14, 2012

### LCKurtz

z = 3-x is the top of the surface, so that part is OK.
Yes, although after you do the dz integral you might want to use polar coordinates for the dydx integral for ease of calculation.

It's OK. z as a function of x was taken care of in the inside dz limits
Your cylindrical limits are Ok as long as you are sure to integrate dz before dr because the z limits depend on r.

3. Jul 14, 2012

### forestmine

Oh wow, thank you! I couldn't help but feel like I didn't quite have it.

Thanks for checking it out!