# Paraboloid problem help

1. Jul 5, 2010

### EV33

1. The problem statement, all variables and given/known data
The solid enclosed by the paraboloid x=y2+z2 and the plane x=16.

2. Relevant equations
Triple integral in ractangular coordinates

3. The attempt at a solution
I figured out this is a paraboloid that circles the x axis, that starts at the origin and it gets wider and wider as it goes in the x direction until it is stopped by the plane x=16 where there is a circle around the x axis with a radius of

$$\int_{x=y2+z2}^{16}\int_{z=-\sqrt{16-y2}}^{\sqrt{16-y2}}\int_{z=0}^4 dxdzdy$$

I was just curious if my set up looks right.

Thank you.

Last edited: Jul 5, 2010
2. Jul 5, 2010

### EV33

Re: Paraboloid

ok that is not what I meant to write lol, I have to work on this really quick

3. Jul 5, 2010

### EV33

Re: Paraboloid

ok I got what I want up there except for some reason when I try and square stuff, the coding keeps appearing for the stuff in the integrals but not in the first sentence I wrote.

4. Jul 5, 2010

### vela

Staff Emeritus
Re: Paraboloid

In TeX, you use ^ to get a superscript and _ to get a subscript, so x2 would be written x^2 and xmin would be written x_{min}, for example.

5. Jul 5, 2010

### The Chaz

Re: Paraboloid

The integral isn't correct.
Your left-most integral bounds shouldn't have variables in it.
You have used "z=" in two of the integrals, and "y=" in none.

6. Jul 5, 2010

### vela

Staff Emeritus
Re: Paraboloid

You have the integrals backwards, and I assume where you wrote z=0, you meant y=0. You have dy on the end, so its limits should be on the frontmost integral. Similarly, the limits for z should be on the middle integral, and the limits for x on the innermost integral. Other than the wrong order, it looks good.

7. Jul 5, 2010

### EV33

Re: Paraboloid

Ya I meant to write them as dxdzdy, but I was so caught up in trying to use Latex for the first time I didn't even notice I had the order mixed up. Thank you for the help.