Triple integral in spherical form

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Homework Help Overview

The discussion revolves around evaluating a triple integral in spherical coordinates, specifically the integral of the function 1/(x^2+y^2+z^2) over a region bounded above by a sphere and below by a cone. The problem involves determining the correct limits of integration in spherical coordinates.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the setup of the integral and the appropriate limits for spherical coordinates, questioning the ranges proposed by the original poster. There is an exploration of the relationship between the bounding surfaces and the implications for the limits of integration.

Discussion Status

The discussion is active, with participants providing insights on the limits of integration and confirming the original poster's approach. Some participants suggest that the spherical coordinate notation may have been misinterpreted, while others affirm the correctness of the proposed limits.

Contextual Notes

There is mention of the intersection of the bounding surfaces, which raises questions about the geometric interpretation of the limits in the context of the problem. The original poster's notation and the use of spherical coordinates are also under scrutiny.

naspek
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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?
 
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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.
 
naspek said:
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 said:
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.
 
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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