# Triple Integral of a cone bounded by a plane.

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1. Nov 17, 2014

### ktvphysics

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

find the volume using spherical coordinates of the region bounded above by z=9 and below by z=sqrt(x^2+y^2) in the first octant.
2. Relevant equations

3. The attempt at a solution
I found this volume using cartesian and cylindrical coordinates, so I know the answer I am looking for, but I can't for the life of me get it in spherical coordinates.

The region that I am solving for is simply a cone with a height of 9, correct? At z =9 the height is 9. at z = 9 the radius is also 9. So by the equation for volume of a cone, it would be pi r^2 h/3, which = 9^3/3*pi, then divided by 4 since we just want the first octant, which is a quarter of the cone, right? I am sure I am over complicating this somehow.

I found rho to be 9/cos(phi) which is 9/cos(pi/4) and (phi = pi/4) and theta is pi/2. So the integral is
∫∫∫ ρ2sinφ dρdφdθ. 0≤ρ≤9/cos(π/4), 0≤φ≤π/4, 0≤θ≤π/2

Am I doing something wrong?

2. Nov 17, 2014

### Dick

Yes, you are doing something wrong. Shouldn't the upper limit of $\rho$ depend on $\phi$?

3. Nov 17, 2014

### ktvphysics

Yes. I figured it out. thank you, though