Trouble understanding meaning of triple integral in spherical coordinates

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  • #1
zjmarlow
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Homework Statement


Evaluate [tex] \iiint\limits_B e^{x^2 + y^2 + z ^2}dV [/tex] where [tex] B [/tex] is the unit ball.


Homework Equations


See above.


The Attempt at a Solution


Does this evaluate the volume of [itex] f(x, y, z) [/itex] within the unit ball (i.e. anything falling outside the unit ball is discarded)? Also, since [itex] f [/itex] is a function of three independent variables, I'm confused - wouldn't that be four dimensional?
 
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  • #2
SammyS
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Homework Statement


Evaluate [tex] \iiint\limits_B e^{x^2 + y^2 + z ^2}dV [/tex] where [tex] B [/tex] is the unit ball.

Homework Equations


See above.

The Attempt at a Solution


Does this evaluate the volume of [tex] f(x, y, z) [/tex] within the unit ball (i.e. anything falling outside the unit ball is discarded)? Also, since [tex] f [/tex] is a function of three independent variables, I'm confused - wouldn't that be four dimensional?
Yes, you are integrating the function, [itex]\displaystyle e^{x^2 + y^2 + z ^2}[/itex] over the volume of a unit ball.

Remember that [itex]\displaystyle x^2 + y^2 + z ^2=\rho^2\ .[/itex]
 
  • #3
zjmarlow
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Thank you for your reply. The solution is worked out: [itex] \frac{4\pi}{3}(e - 1) [/itex]. My problem now is visualizing [itex] e^{x^2 + y^2 + z^2} [/itex]. Could you explain what this would look like?
 
  • #4
zjmarlow
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Or is [itex] \frac{4\pi}{3}(e - 1) [/itex] like a scalar field that says how much each position in the unit ball is "worth"?
 
  • #5
SammyS
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Or is [itex] \frac{4\pi}{3}(e - 1) [/itex] like a scalar field that says how much each position in the unit ball is "worth"?

How did you arrive at that result?

I looks as if you have you have made some gross simplification.
 
  • #6
zjmarlow
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Sorry, that should be [itex] f(x, y, z) [/itex] (in general) or [itex] e^{x^2 + y^2 + z^2} [/itex] (in this particular case).
 
  • #7
SammyS
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Show what integral you have in spherical coordinates.
 
  • #8
zjmarlow
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Ok, I copied the example problem wrong. It should be
[tex] \iiint\limits_B e^{(x^2 + y^2 + z^2)^{(3/2)}}dV [/tex]
which becomes
[tex] \int_0^{2\pi} \int_0^\pi \int_0^1 e^{(\rho^2)^{(3/2)}} \cdot \rho^2 sin \phi d\rho d\phi d\theta [/tex]
[tex] = \int_0^{2\pi} \int_0^\pi \frac{e - 1}{3} sin \phi d\phi d\theta [/tex]
[tex] = \int_0^{2\pi} \frac{2}{3}(e - 1) d\theta [/tex]
[tex] = \frac{4\pi}{3}(e - 1) [/tex]

I guess I see how the integral without the function calculates volume (the book has a nice picture of a curved box) but what's the meaning of the function you integrate? What does something like [itex] e^{(x^2 + y^2 + z^2)^{(3/2)}} [/itex] or [itex] e^{\rho^3} [/itex] even look like?
 
  • #9
SammyS
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Ok, I copied the example problem wrong. It should be
[tex] \iiint\limits_B e^{(x^2 + y^2 + z^2)^{(3/2)}}dV [/tex]
which becomes
...

I guess I see how the integral without the function calculates volume (the book has a nice picture of a curved box) but what's the meaning of the function you integrate? What does something like [itex] e^{(x^2 + y^2 + z^2)^{(3/2)}} [/itex] or [itex] e^{\rho^3} [/itex] even look like?
Well, that makes more sense.

The integral [itex]\displaystyle \int e^{\rho^3}\rho^2 \,d\rho[/itex] is pretty straight forward.

The integral [itex]\displaystyle \int e^{\rho^2}\rho^2 \,d\rho[/itex] involves the error function.
 
  • #10
HallsofIvy
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Your original question was,
Does this evaluate the volume of f(x,y,z) within the unit ball (i.e. anything falling outside the unit ball is discarded)? Also, since f is a function of three independent variables, I'm confused - wouldn't that be four dimensional?

I don't know what you mean by "volume" of a function! An integral, in general, doesn't have any specific geometric or physical interpretation- it depends upon the application. One possible application is to take f(x,y,z) to be the density of an object. Then the integral would be the total mass of the object. And, no it would not be four dimensional. Since x, y, and z are distances, measured, say, in meters, then the integral would have units of "cubic meters" 'times' whatever units f has.

If, as before, f(x,y,z) is density, measured in kilograms per cubic meter, then the integral will have units of [itex](kg/m^3)(m^3)= kg[/itex]
 

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