# Find V of cone z=ksqrt(x^2+y^2) bounded by z=h using spherical coordinates

• Mar 26th 2013, 08:41 PM
lytwynk
Find V of cone z=ksqrt(x^2+y^2) bounded by z=h using spherical coordinates
We were given this exercise in class to take home but I am a bit confused with it. If anyone could help I would appreciate it.
Let C be a conical solid bounded above by z=h and below by the cone z=k*sqrt{x^2+y^2}, k>0. We are supposed to find the triple integral for the volume of the cone in spherical coordinates.

V=ρ2sin(θ)dρdϕdθ
I am a bit confused at what the boundaries will be. I know 0<=θ<=2π however I am confused at what the boundaries for ϕ and ρ will be. If someone could shed some light on this I would really appreciate it.
• Mar 27th 2013, 05:11 AM
HallsofIvy
Re: Find V of cone z=ksqrt(x^2+y^2) bounded by z=h using spherical coordinates
Since $\displaystyle z= \rho cos(\phi)$, the plane z= h becomes $\displaystyle \rho cos(\phi)= h$. The cone [tex]z= k\sqrt{x^2+ y^2}]/tex] becomes $\displaystyle \rho cos(\phi)= k\sqrt{\rho^2cos^2(\theta)sin^2(\phiz)+ \rho^3sin^2(\theta)sin^2(\phi)}= k\rho sin(\phi)$ which reduces to $\displaystyle \phi= arctan(1/k)$.

Are you required to do this in spherical coordinates? I think that cylindrical coordinates would be simpler.
• Mar 27th 2013, 07:32 AM
lytwynk
Re: Find V of cone z=ksqrt(x^2+y^2) bounded by z=h using spherical coordinates
Yes I am required to do this in spherical coordinates. Kind of a bummer but whatever.

So ok, yes I can now see what you are saying... So for the boundaries now $0<=\theta<=2\pi$, then 0<=\phi<=arctan(1/k) and finally 0<=\rho<= h/cos\phi. Is this correct?

If so then the Integral would then be:

$$\int_0^{2\pi} d\theta \int_0^{arctan(1/k)} sin\phi d\phi \int_0^{h/cos\phi} \rho^2 d\rho$$

Is this correct?

I don't know if my text format works on this site. So if not forgive me it its too messy.
• Mar 27th 2013, 07:56 AM
lytwynk
Re: Find V of cone z=ksqrt(x^2+y^2) bounded by z=h using spherical coordinates
Nvm. I figured it out. Thanks for your help!