# Triple integral/volume of a region enclosed by two planes an elliptic hyperboloid

1. Apr 26, 2012

### sydneyw

For positive a and h, let A designate the region of R3 enclosed by the elliptic hyperboloid, x2 +y2 -z2 =a2 and the two planes, z= -h/2 and z=h/2.

Determine the volume of A

So I figure this will be a triple integral in cylindrical coordinates. the first integrand being from -h/2 to h/2, the second from 0 to 2∏ and the third is a transformation to r.

The equation to integrate would be my elliptic hyperboloid equation, correct? So that would look something like r2cos2(θ) +r2sin2(θ)-z2 rdrdθdz....right?

I'm confused on how to go about this problem so I told you what I was able to figure out on my own. Could someone PLEASE explain how I do this?

2. Apr 27, 2012

### sydneyw

i got the triple integral with limits stated above and of r^3 -rz^2 drd(theta)dz..I made my last integrand be from 0 to a..is this correct? I then integrated it all out and ended up with [pi(a^4)h]/2. Is this right?? Could I have possibly solved this???

3. Apr 27, 2012

### sharks

Your surface equation is actually that of a one-sheeted paraboloid. I've attached the graph.

In the plane z=0, r varies from r=a
In the plane z=h/2, r varies to r=$\sqrt{a^2+\frac{h^2}{4}}$

To get the entire volume, just multiply the triple integral by 2.

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4. Apr 27, 2012

### sydneyw

But there is no plane z=0. It's-h/2 and h/2.

5. Apr 27, 2012

### sharks

You have to evaluate the total volume in two parts. First, from z=0 to z=h/2. Then, from z=-h/2 to z=0.

Note that the figure is symmetrical at z=0, which is at its smallest cross-sectional area, where r=a.

Hence, there is no need to calculate two triple integrals. Just find one and then multiply by 2.

6. Apr 27, 2012

### sydneyw

okay so 1. do i need to evaluate in cylindrical coords even?
2. am I using my elliptic hyperboloid equation for the integration?
3. it would be a double integral of that elliptic hyperboloid equation or no?

I'm sorry. I'm very confused now.

7. Apr 27, 2012

### LCKurtz

This is a solid of revolution. No law says you must use a triple integral. Why not just set it up with circular slabs and integrate in the z direction?