# Homework Help: Triple Integral Help

1. Nov 13, 2007

### beckyroar

Triple Integral Help :(

Can anyone help me with this triple integral problem? I'm sorry I don't know how to post the script properly; I'm a complete newb.

It's a surface integral problem- that part is not important- I have to calculate a triple integral where S is the surface of the volume bounded by z=1-x^2 and the planes z=0, y=0, and y+z=2

I am having difficulty setting this up- if anyone can post any relevant information, or direct me to any sites that are phenomenal in helping with triple integrals, in would be greatly appreciated.

2. Nov 13, 2007

### EnumaElish

What are the variables of integration for the 3 integrals?

3. Nov 13, 2007

### beckyroar

That's where I'm having the problem, finding the bounds of the three integrals. The computation itself I feel as though I can work out, if I have the right bounds of integration

4. Nov 13, 2007

### beckyroar

If anyone can calculate the triple integral of 3y for the bounds that I mentioned above, let me know if you get 144/35....if you want to do that for fun? ahaha :(
The bounds that I used were x ranges from -1 to 1, Z ranges from 0 to 1-x^2, and y ranges from 0 to 2-Z. Does anyone else htink that this looks correct?

5. Nov 14, 2007

### HallsofIvy

I hadn't had time to look at this before. z= 1- x^2 is a "parabolic cylinder" with axis running along the y-axis. It's highest point (z max) comes when x= 0 and is at z= 2. y+ z= 2 is a plane crossing the xz-plane in the line z= 2 (above the parabola) and the xy-plane at y= 2. The plane crosses the parabolic cylinder when z= 2- y = 1- x^2 or y= x^2+ 1. The fact that that does not cross the y= 0 threw me for a moment! What that means is that you have to do the integral in two different parts. As long as we are below the parabola y= x^2+ 1, our upper boundary is z= 1- y^2, the parabolic cylinder. After that the upper boundary is z= 2- y, the plane.
You will need to do this as two integrals. The first integral will have x from -1 to 1, for each x, y from 0 to x^2+ 1, for each (x,y), z from 0 to 1- y^2. The second integral will have x from -1 to 1, for each x, y from x^2+ 1 to 2, for each (x,y), z from 0 to 2- y.