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Triple Integral-problem.

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  • #1
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How do I solve this triple integral,
[tex]\int\int\int_{\Omega}^{}zdxdydz, \hspace{8} \Omega = \{(x,y,z): x^2 + y^2 + z^2 \leq 1, \hspace{6} 0 \leq z \leq \sqrt{x^2+y^2}\}[/tex]
 

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  • #2
saltydog
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iNCREDiBLE said:
How do I solve this triple integral,
[tex]\int\int\int_{\Omega}^{}zdxdydz, \hspace{8} \Omega = \{(x,y,z): x^2 + y^2 + z^2 \leq 1, \hspace{6} 0 \leq z \leq \sqrt{x^2+y^2}\}[/tex]
Convert to spherical coordinates. Integrate phi from 0 to pi/2, all around from 0 to 2pi, and let rho go from 0 to 1. Oh yea, then multiply by 2. Can you convert this into a real integral?
 
  • #3
Hurkyl
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salty: you notice the region of integration isn't a sphere? :smile: (And if it was, there's an easier way to do it...)

Where do you have problems, iNCREDiBLE?
 
  • #4
saltydog
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Hurkyl said:
salty: you notice the region of integration isn't a sphere? :smile: (And if it was, there's an easier way to do it...)

Where do you have problems, iNCREDiBLE?
No I didnt' Hurkyl. Looked like a sphere to me. Still does. Suppose I should just work it through. Sorry Incredible if I got it wrong (Hurkly is no doubt right in some kind of way I need to figure out).
:blushing:
 
  • #5
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saltydog said:
Convert to spherical coordinates. Integrate phi from 0 to pi/2, all around from 0 to 2pi, and let rho go from 0 to 1. Oh yea, then multiply by 2. Can you convert this into a real integral?
Are you sure about this? :surprised
 
  • #6
Hurkyl
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Note the extra condition 0 < z < &radic;(x² + y²). It's the part of the upper half-ball that's outside of a cone.
 
  • #7
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I can't see what the region of integration is. But my gut says i should integrate over z first. Then convert to polar coordinates where [tex]\frac{1}{\sqrt{2}} \leq r \leq 1?[/tex]
 
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  • #8
saltydog
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Hurkyl said:
Note the extra condition 0 < z < √(x² + y²). It's the part of the upper half-ball that's outside of a cone.
Alright, it's a cone. Thanks Hurkyl. Sorry Incredible. I'll work it through for my own education but will stay out of the thread so as not to make it any worst for Incredible.
:blushing:
 
  • #9
Hurkyl
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You have a good gut! I too would integrate over z first, and change it into cylindrical coordinates. Now, you just have to work out the details. :smile:


As usual, it really helps if you can draw a picture of the region of integration, or at least visualize it really well. But, we can always grind it out algebraically.


Your region is the system of equations:

r² + z² < 1
0 < z < r

So, the first question becomes "for which values of r can we solve this equation for the rest of the variables?"

You have suggested 1/√2 < r < 1. Why?
 
  • #10
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I understand that the region is
r² + z² < 1
0 < z < r

But after that I'm not following you.. :cry:
 
  • #11
Hurkyl
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You need to find the bounds: you need to know all possible values of r, then for each value of r, you need to know all possible values of z... right?
 
  • #12
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Hurkyl said:
You need to find the bounds: you need to know all possible values of r, then for each value of r, you need to know all possible values of z... right?
1/√2 < r < 1
0 < z < 1/√2 ?


o:)
 
  • #13
Hurkyl
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Okay, how did you get that?
 
  • #14
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Well, Hurkyl, I think I was wrong... :wink:
I've been thinking a lot and here's how I wanna do it now.

Spherical coordinates:

[tex]x = rsin\theta cos\phi[/tex]
[tex]y = rsin\theta sin\phi[/tex]
[tex]z = rcos\theta[/tex]

[tex] 0 \leq r \leq 1[/tex]
[tex] \frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}[/tex]
[tex] 0 \leq \phi \leq 2 \pi[/tex]


[tex]\int\int\int_{\Omega}^{}zdxdydz = \int\int\int rcos\theta r^2 sin\theta drd\theta d\phi = \frac{1}{2} \int\int\int r^3 sin(2\theta) drd\theta d\phi = \pi \int\int r^3 sin(2\theta) drd\theta = \frac{\pi}{4} \int sin(2\theta) d\theta = \frac{\pi}{8}[/tex]
 
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  • #15
Hurkyl
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You were! Notice that you could still solve for z for any 0 < r < 1... it's just that 1/&radic;2 is the switching point between the two different expressions for z. (So you'd have to break the integral up into two pieces)


So how did you get your bounds this time?
 
  • #16
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We had
[tex]x^2 + y^2 + z^2 \leq 1[/tex]
[tex] 0 \leq z \leq \sqrt{x^2+y^2}[/tex]

After switching to Spherical Coordinates we have

[tex]r^2 \leq 1[/tex]
[tex]0 \leq rcos\theta \leq \sqrt{(rsin\theta cos\phi)^2+(rsin\theta sin\phi)^2}=|rsin\theta|=rsin\theta[/tex]
<=>
[tex]0 \leq r \leq 1[/tex]
[tex]0 \leq cos\theta \leq sin\theta[/tex]

The last inequality is true for [tex]\frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}[/tex]
 
  • #17
Hurkyl
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Looks good to me. And more importantly, it looks good to you. :smile:
 
  • #18
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Hurkyl said:
Looks good to me. And more importantly, it looks good to you. :smile:
Yeah! I'm preparing for an exam. This is a problem from a previous exam and I really hate the fact that our textbook don't discuss this sort of integrals, just those that are way simpler and not even one of our excercises has been of this sort. :devil:
 
  • #19
Hurkyl
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IMHO you should work it out in cylindrical coordinates too. It's good exercise. :smile:
 
  • #20
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The problem is that I have no clue about how the region of integrations looks like.. So when you say that I would have to break the integral into two pieces I wouldn't know why :redface:
 
  • #21
Hurkyl
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That's why I think it's a good exercise! It fills in a gap in your ability!

I'll ask my question again:

For what values of r does the system

r² + z² < 1
0 < z < r

can you find a solution to this system?
 
  • #22
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Hurkyl said:
That's why I think it's a good exercise! It fills in a gap in your ability!

I'll ask my question again:

For what values of r does the system

r² + z² < 1
0 < z < r

can you find a solution to this system?
0 < r < 1.

Hmm, the solution will be trigonometric.
r = acos(b)
z = asin(b)
a < 1 and 0 < b < Pi/4
 
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  • #23
Hurkyl
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You could look at it that way... but that's really just using spherical coordinates again.

If you saw that from the algebra (rather than from the fact you've already done it with spherical coordinates), then that's good!

(Be careful, though -- I suspect you're about to visit every point twice, and get pi/4... don't forget that r is nonnegative, by definition of polar coordinate!)


However, I was trying to set up an iterated integral where you integrate with respect to z, then with respect to r. So, now that you've determined that 0 < r < 1, the only remaining question is, for a particular value of r, in what range z can vary?
 
  • #24
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z can vary between 0 and r? :rolleyes:
 
  • #25
Hurkyl
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You do have 0 < z < r... but that's only one of your constraints on z!
 

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