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[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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- Thread starter iNCREDiBLE
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[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]

- #2

saltydog

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iNCREDiBLE said:

[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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Where do you have problems, iNCREDiBLE?

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saltydog

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Hurkyl said:

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).

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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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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.

- #9

Hurkyl

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As usual, it

Your region is the system of equations:

r² + z² < 1

0 < z < r

So, the first question becomes "for which values of

You have suggested 1/√2 < r < 1. Why?

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I understand that the region is

r² + z² < 1

0 < z < r

But after that I'm not following you..

r² + z² < 1

0 < z < r

But after that I'm not following you..

- #11

Hurkyl

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Hurkyl said:r, then for each value ofr, you need to know all possible values ofz... right?

1/√2 < r < 1

0 < z < 1/√2 ?

- #13

Hurkyl

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Okay, how did you get that?

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Well, Hurkyl, I *think* I was wrong...

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]

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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So how did you get your bounds this time?

- #16

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[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.

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Hurkyl said:Looks good to me. And more importantly, it looks good to you.

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.

- #19

Hurkyl

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IMHO you should work it out in cylindrical coordinates too. It's good exercise.

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- #21

Hurkyl

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I'll ask my question again:

For what values of

r² + z² < 1

0 < z < r

can you find a solution to this system?

- #22

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Hurkyl said:

I'll ask my question again:

For what values ofrdoes the system

r² + z² < 1

0 < z < r

can you find a solution to this system?

0 <

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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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

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z can vary between 0 and r?

- #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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