# Triple integral xyz coordinates to spherical

1. Apr 27, 2010

### exidez

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

the integral is bounded below by the cone $$z^2=x^x+y^2$$ and above by the sphere $$x^2+y^2+z^2=18$$

$$\int^{3}_{0}\int^{\sqrt{9-y^2}}_{0}\int^{\sqrt{18-x^2-y^2}}_{\sqrt{x^2+y^2}}x^2+y^2+y^2 dzdxdy$$

2. Relevant equations

conversion formulas

3. The attempt at a solution

I am struggling on my limits. Obviously these limits are not true
$$\int^{3}_{0}\int^{\sqrt{9-y^2}}_{0}\int^{\sqrt{18-x^2-y^2}}_{\sqrt{x^2+y^2}}p^2 p^2sin\phi dp d\theta d\phi$$

I know the upper limit for p is going to be $$\sqrt{18}$$ as that is the upper bound of the sphere but the lower bound of p is the cone $$z^2=x^2+y^2$$

is this $$x^2+y^2-z^2 = (psin\phi cos\theta )^2 + (psin\phi sin\theta )^2 - (pcos\phi)^2$$ for the lower limit of p when factoring out to get p = something??? It seems a little long...

//Edit

ok, i had another look at this

Is my domain as followed:

D = {$$(\theta , \phi \ , \rho) | (0 \leq \theta \leq 2\pi , 0 \leq \phi \leq \frac{\pi}{4} , 0 \leq \rho \leq \sqrt{18}$$}

Is this correct for a cone within a sphere? Im not too certain about $$\rho$$

Last edited: Apr 28, 2010