# Triple Integral problem (maybe using spherical coordiantes)

## Homework Statement

Integrate the function f(xyz)=−4x+6y over the solid given by the "slice" of an ice-cream cone in the first octant bounded by the planes x=0 and y=x*sqrt(22/3) and contained in a sphere centered at the origin with radius 13 and a cone opening upwards from the origin with top radius 5.

## Homework Equations

x^2+y^2+z^2=p(row)
x=p(sinf(fie)*cost(theta)
y=p*sinfsint

## The Attempt at a Solution

So I changed function 6y-4x into the terms of spherical coordinates. While i attempted to find the limits of the integral, I'm not sure if my drawing is correct:

But I assumed it meant that the region had to be inside the cone such that it was in the first octet (since it's bounded by x=0 which is the yz plane --NOT SHOWN IN PICTURE), but I'm not sure how the ice cream part has to do with this...
Could someone just explain to me what the trace part means and also explain if my drawing is right, I have no idea what inside the circle and the cone means..

I would recommend cylindrical coordinates because there is more cylindrical symmetry than spherical symmetry. If you alter the coordinates so that the axis of the cone is along the z-axis then you have cone itself as given by $z= \sqrt{22/3}r$ and the sphere by r^2+ z^2= 169. To integrate over that, $\theta$ goes from 0 to $2\pi$, r goes from 0 to 5, and z goes from $\sqrt{22/3}r$ to $\sqrt{169- r^2}$. Of course, now the integrand is $f(x, y, z)= -4x+ 6z= -4rcos(\theta)+ 6z$ since we have swapped y and z.