Since , the plane z= h becomes . The cone [tex]z= k\sqrt{x^2+ y^2}]/tex] becomes which reduces to .
Are you required to do this in spherical coordinates? I think that cylindrical coordinates would be simpler.
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)dρdϕdθI 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.
Since , the plane z= h becomes . The cone [tex]z= k\sqrt{x^2+ y^2}]/tex] becomes which reduces to .
Are you required to do this in spherical coordinates? I think that cylindrical coordinates would be simpler.
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.