Cylindrical Triple Integral: Evaluating Over a Bounded Solid

[mathman44]

Homework Statement

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.

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

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

 

[Fightfish]

Your integral looks funny...might want to adjust the formatting somewhat.
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.

 

[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.

 

[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.

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

 

[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.

 

[Fightfish]

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

 

[mathman44]

It works! Ty.