# Cylindrical triple integral

1. Jan 15, 2010

### mathman44

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

Use cylindrical coordinates to evaluate the triple integral (over E) $$sqrt(x^2+y^2) dV$$ , where E is the solid bounded by the circular paraboloid z=9−16(x^2+y^2) and the xy plane.

3. The attempt at a solution

This is really bugging me... Is this the correct setup for the integral?

0 < z < 9 - 16r^2
0 < r < 3/4
0 < theta < 2pi

$$\int_0^{3/4}\int_0^{9-16r^2}\int_0^{2pi}r^2(d\theta,dz,dr)$$

Last edited: Jan 15, 2010
2. Jan 15, 2010

### Fightfish

1. Order of integration - please write your integrals in the correct order, or else you are going to get nonsense for your answer when you integrate outwards.
2. The function you are integrating is r^2, not r.

3. Jan 15, 2010

### mathman44

Yep, that's all fixed now. Learning latex for the first time. Does this look good? It's not giving me the correct answer according to WeBWorK.

Last edited: Jan 15, 2010
4. Jan 15, 2010

### Fightfish

Hm...the order of integration is wrong. You must integrate wrt z first before integrating wrt r, because the bound of z is dependent on the value of r. If you integrate in the order above, you will end up with r in your final expression.

$$\int_0^{2\pi}\int_0^{3/4}\int_0^{9-16r^2}\,\,r\,\,r \,dz\, dr\, d\theta$$

Last edited: Jan 15, 2010
5. Jan 15, 2010

### mathman44

Once again my bad, the E is actually sqrt(x^2 + y^2) = r... :S

I did it your way before (with r^2 as the integrating function) and I still got the wrong answer.

6. Jan 15, 2010

### Fightfish

Hmm....what does the WeBwork thingy say the correct answer is? The integral looks correct to me

Last edited: Jan 15, 2010
7. Jan 15, 2010

### mathman44

It works! Ty.

Last edited: Jan 15, 2010