Evaluate the triple integral (with spherical coordinates)

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Homework Help Overview

The discussion revolves around evaluating a triple integral using spherical coordinates. The integral in question involves an exponential function and is set over the first octant of three-dimensional space.

Discussion Character

  • Exploratory, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to understand how to determine the angles theta and phi for spherical coordinates in the context of the given integral.

Discussion Status

Some participants provide insights into the definitions of the spherical coordinate angles, suggesting ranges for phi and theta based on the constraints of the problem. The discussion includes clarifications on the limits of integration for the spherical coordinates.

Contextual Notes

The original poster expresses uncertainty about the angles and their intervals, indicating a need for further clarification on spherical coordinates in relation to the integral setup.

melihaltintas
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Homework Statement


Firstly sorry for my bad english,i have a one question for you(İ try it but i didn't solve it )

Homework Equations


The Attempt at a Solution


i know problem will be solved spherical coordinates but i don't know how i get angles (interval) theta and fi ?
[tex]\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty}e^{-\sqrt{x^2+y^2+z^2}}\,dxdydz[/tex]

Mod note: Fixed LaTeX. Read what tiny-tim said below.[/color]
 
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You get the angles by thinking about what the spherical coordinate variables mean. [itex]\phi[/itex] is the angle from the positive z-axis to the negative x-axis and, to cover all space, would normally go from [itex]0[/itex] to [itex]\pi[/itex]. Since you want z to stay positive, you want [itex]\phi[/itex] to go from [itex]0[/itex] to [itex]\pi/2[/itex]. [itex]\theta[/itex] goes around a complete circle in the xy-plane in going from [itex]0[/itex] to [itex]2\pi[/itex]. Since the first quadrant is 1/4 of that, [itex]\theta[/itex] goes from [itex]0[/itex] to [itex]\pi/2[/itex]. Finally, there is no upper limit on the distance from the origin to any point in the first octant so [itex]\rho[/itex] goes from [itex]0[/itex] to [itex]\infty[/itex].
 
welcome to pf!

hi melihaltintas! welcome to pf! :smile:

don't try to put forum tags inside latex! :wink: …​
melihaltintas said:
[tex]\int_{0}^{\infty}\int_{0}^{ \infty}\int_{0}^{\infty}e^{-\sqrt{x^2+y^2+ z^2}}\,dxdydz[/tex]
 
thanks a lot :)
 

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