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Triple integral in spherical form

  • Thread starter naspek
  • Start date
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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

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

LCKurtz

Science Advisor
Homework Helper
Insights Author
Gold Member
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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 [itex]x^{2}+y^{2}=9/2[/itex], then this means that the radius is [itex]3/\sqrt{2}[/itex] and i think that you limits are: [itex]3/\sqrt{2}\leqslant r\leqslant 3[/itex] 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

Homework Helper
1,669
11
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 [itex]3/\sqrt{2}[/itex].
 

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