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Triple integral, limits of integration

  • Thread starter Gauss M.D.
  • Start date
153
1
Pretty general question.

Integrate f(x,y,z) dxdydz over the area defined by:

[itex]x^{2} + y^{2} + z^{2} \leq 4[/itex]
[itex]x \leq 0[/itex]
[itex]y \leq 0[/itex]
[itex]z \leq 0[/itex]

It is immidiately apparent that it is 1/8 of a sphere with r=2. So from that geometrical intuition we can do a variable substitution to spherical coordinates and use the following limits of integration.

0 < r < 2
0 < θ < pi/2
0 < σ < pi/2

Or something. What I'm wondering is: how would you go about finding these limits algebraically?? Let θ be the angle to the z axis and σ be the angle between the x and y axis and you would get

0 < θ < pi/2

But how would you figure out the angle between x and y?
 

Zondrina

Homework Helper
2,063
136
Pretty general question.

Integrate f(x,y,z) dxdydz over the area defined by:

[itex]x^{2} + y^{2} + z^{2} \leq 4[/itex]
[itex]x \leq 0[/itex]
[itex]y \leq 0[/itex]
[itex]z \leq 0[/itex]

It is immidiately apparent that it is 1/8 of a sphere with r=2. So from that geometrical intuition we can do a variable substitution to spherical coordinates and use the following limits of integration.

0 < r < 2
0 < θ < pi/2
0 < σ < pi/2

Or something. What I'm wondering is: how would you go about finding these limits algebraically?? Let θ be the angle to the z axis and σ be the angle between the x and y axis and you would get

0 < θ < pi/2

But how would you figure out the angle between x and y?
Spherical coordinates :

##x = rcosθsin \phi##
##y = rsinθsin \phi##
##z = rcos \phi##
##x^2 + y^2 + z^2 = r^2##
##|J| = r^2sin \phi##

For starters, your limits for r are incorrect. Check them again.
 
153
1
What? How are my limits for r incorrect?

x^2 + y^2 + z^2 < 2^2
r^2 < 2^2
r < 2
 

Zondrina

Homework Helper
2,063
136
What? How are my limits for r incorrect?

x^2 + y^2 + z^2 < 2^2
r^2 < 2^2
r < 2
##x^2 + y^2 + z^2 = r^2##

So you get ##r^2 ≤ 4## which implies that ##|r| ≤ 2##.
 

LCKurtz

Science Advisor
Homework Helper
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Gold Member
9,482
721
Pretty general question.

Integrate f(x,y,z) dxdydz over the area defined by:

[itex]x^{2} + y^{2} + z^{2} \leq 4[/itex]
[itex]x \leq 0[/itex]
[itex]y \leq 0[/itex]
[itex]z \leq 0[/itex]

It is immidiately apparent that it is 1/8 of a sphere with r=2. So from that geometrical intuition we can do a variable substitution to spherical coordinates and use the following limits of integration.

0 < r < 2
0 < θ < pi/2
0 < σ < pi/2
No, you can't use those limits without knowing some kind of symmetry conditions on the integrand ##f(x,y,z)##. Changing to spherical coordinates is appropriate, but draw a 3d coordinate system and use it to figure out the appropriate ##(\rho,\phi,\theta)## limits for that back octant.
 

tiny-tim

Science Advisor
Homework Helper
25,790
249
It is immidiately apparent that it is 1/8 of a sphere with r=2. So from that geometrical intuition we can do a variable substitution to spherical coordinates and use the following limits of integration.

0 < r < 2
0 < θ < pi/2
0 < σ < pi/2
yes, as you say it's an octant, they're all the same, so you certainly can choose the "positive" octant and integrate to find the volume of that (if you don't already know the formula for the volume of a sphere! :wink:)
What I'm wondering is: how would you go about finding these limits algebraically?? Let θ be the angle to the z axis and σ be the angle between the x and y axis and you would get

0 < θ < pi/2

But how would you figure out the angle between x and y?
isn't it obviously π/2 ? and haven't you already written 0 < σ < π/2 ? :confused:

(perhaps i'm misunderstanding your question)​

as an exercise, what do you think the limits are for the given ("negative") octant? :smile:
 

LCKurtz

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yes, as you say it's an octant, they're all the same, so you certainly can choose the "positive" octant and integrate to find the volume of that (if you don't already know the formula for the volume of a sphere! :wink:)
But they aren't all the same unless you have special properties in the integrand ##f(x,y,z)##.
 
153
1
##x^2 + y^2 + z^2 = r^2##

So you get ##r^2 ≤ 4## which implies that ##|r| ≤ 2##.
But it's a radius. It doesn't have a negative radius.
 
153
1
yes, as you say it's an octant, they're all the same, so you certainly can choose the "positive" octant and integrate to find the volume of that (if you don't already know the formula for the volume of a sphere! :wink:)


isn't it obviously π/2 ? and haven't you already written 0 < σ < π/2 ? :confused:

(perhaps i'm misunderstanding your question)​

as an exercise, what do you think the limits are for the given ("negative") octant? :smile:
Well yes it is obvious if you appeal to geometric intuition, I just wanted to know if there was a way to do it algebraically.
 

tiny-tim

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Well yes it is obvious if you appeal to geometric intuition, I just wanted to know if there was a way to do it algebraically.
Sorry, but I've no idea what you mean by "algebraically". :confused:

How could the angle between x and y be anything other than π/2 ?
 

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