Triple integral in spherical form

[naspek]

consider this following triple integral
1/(x^2+y^2+z^2)dxdydz

bounded above by sphere z=(9-x^2-y^2)^1/2 and below by the cone z=(x^2+y^2)^1/2

what i have done:

z=Pcospi

P^2=x^2+y^2+z^2

9=x^2+y^2+z^2

P=0 to 3

pi=0 to pi/4

theta=0 to 2pi

is this the correct range?

 

[hunt_mat]

I would first equate the z's to find that $x^{2}+y^{2}=9/2$, then this means that the radius is $3/\sqrt{2}$ and i think that you limits are: $3/\sqrt{2}\leqslant r\leqslant 3$ and all your limits are okay.

 

[LCKurtz]

consider this following triple integral
1/(x^2+y^2+z^2)dxdydz

bounded above by sphere z=(9-x^2-y^2)^1/2 and below by the cone z=(x^2+y^2)^1/2

what i have done:

z=Pcospi

P^2=x^2+y^2+z^2

9=x^2+y^2+z^2

P=0 to 3

pi=0 to pi/4

theta=0 to 2pi

is this the correct range?

I would first equate the z's to find that $x^{2}+y^{2}=9/2$, then this means that the radius is $3/\sqrt{2}$ and i think that you limits are: $3/\sqrt{2}\leqslant r\leqslant 3$ and all your limits are okay.

I think his P is supposed to be the spherical coordinate ρ, so his limits are correct. naspek, don't forget the spherical dV element.

 

[hunt_mat]

I think you're right coming to think of it, the point being that the intersection (which is a point on the sphere) is a circle of radius $3/\sqrt{2}$.