Is there any way to calculate this integral?

Rafa Ariza
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Moved from a technical forum, so homework template missing
upload_2017-5-24_12-59-48.png

I have done it by the parametric form of σ, but if I change σ to implicit form that is G(x,y,z)=x^2+y*2+z^2-R^2=0 I don't know how continue.
The theory is:
upload_2017-5-24_13-3-58.png

where Rxy is the projection of σ in plane xy so it's the circumference x^2+y^2=R^2
 
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Hello Rafa, :welcome:

Maybe the problem statement is correct, but you don't tell what you are actually doing (parametric form of ##\sigma## ?)
Maybe the theory is correct, but you don't explain all the symbols, so to me it's of no practical use.
In the past I learned to change to spherical coordinates for something like this and if I do that here I have no problem coming up with the answwer. How about you ?

[edit] Oh, and: for homework you should post in the homework forum and use the template there.
 
BvU said:
Hello Rafa, :welcome:

Maybe the problem statement is correct, but you don't tell what you are actually doing (parametric form of ##\sigma## ?)
Maybe the theory is correct, but you don't explain all the symbols, so to me it's of no practical use.
In the past I learned to change to spherical coordinates for something like this and if I do that here I have no problem coming up with the answwer. How about you ?

[edit] Oh, and: for homework you should post in the homework forum and use the template there.
Thanks,
I want to calculate this integer but σ:x^2+y^2+z^2-R^2=0
 
Rafa Ariza said:
View attachment 204158
I have done it by the parametric form of σ, but if I change σ to implicit form that is G(x,y,z)=x^2+y*2+z^2-R^2=0 I don't know how continue.
The theory is:
View attachment 204159
where Rxy is the projection of σ in plane xy so it's the circumference x^2+y^2=R^2
You can do the integral in spherical polar coordinates. http://hyperphysics.phy-astr.gsu.edu/hbase/sphc.html
Write x and y and also the surface element in the spherical coordinates.
 
PF requires you to do something too. You said you had done it
Rafa Ariza said:
I have done it by the parametric form of σ,
Show your work

[edit] on a friendlier note (for a first-time poster:smile:): is there something in the link ehild gave that you don't understand ? Do you know what you need from there ?
 
BvU said:
PF requires you to do something too. You said you had done it
Show your work

[edit] on a friendlier note (for a first-time poster:smile:): is there something in the link ehild gave that you don't understand ? Do you know what you need from there ?
eaaaa.jpg

here is my problem but in the other form is resolved yet
 
And what does the theory say about z on ##R_{xy}## ?
 
BvU said:
And what does the theory say about z on ##R_{xy}## ?
Rxy is the projection of S in the plane xy.
 
  • #10
So ##
R_{xy}## is a quarter circle (not just the circumference).
But ##z=0## in the plane xy, so what do you do with that ##R\over z## ? Where does this theoretical formula come from ? Is it applicable ?

Would you be interested to follow the path ehild and I learned long ago and proposed here in #2 and #3 ?
 
  • #11
BvU said:
So ##
R_{xy}## is a quarter circle (not just the circumference).
But ##z=0## in the plane xy, so what do you do with that ##R\over z## ? Where does this theoretical formula come from ? Is it applicable ?

Would you be interested to follow the path ehild and I learned long ago and proposed here in #2 and #3 ?
yes i am interested. here is the resolution by put the sphere in parametric form r(u,v)
exercise.jpg
 
  • #12
my problem is to resolve it with surface in this form: G(x,y,z)=x^2+y^2+z^2-R^2=0
the theory is
upload_2017-5-24_14-59-5.png

or equivalent
upload_2017-5-24_14-59-35.png

or
upload_2017-5-24_14-59-48.png

the real problem is in the f(x,y,z(x,y)) or the other equivalent
 
  • #13
Rafa Ariza said:
yes i am interested. here is the resolution by put the sphere in parametric form r(u,v)
View attachment 204162
Excellent work.

So -- if the theory is correct and applicable -- we (or rather, your helpers) are back to understanding what is needed for the alternative route.
What is meant with ##\left | dG\over dz\right |## in the formula ?
Does it perhaps mean you need to evaluate this factor in the numerator at ##z= \sqrt{R^2 - (x^2+y^2)\,}\ ## ?

[edit] oopsed & fixed.
 
  • #14
BvU said:
Excellent work.

So -- if the theory is correct and applicable -- we (or rather, your helpers) are back to understanding what is needed for the alternative route.
What is meant with ##\left | dG\over dz\right |## in the formula ?
Does it perhaps mean you need to evaluate this factor in the numerator at ##z= \sqrt{R^2 - (x^2+y^2)\,}\ ## ?

[edit] oopsed & fixed.
I will try it, thanks!
 
  • #15
BvU said:
Excellent work.

So -- if the theory is correct and applicable -- we (or rather, your helpers) are back to understanding what is needed for the alternative route.
What is meant with ##\left | dG\over dz\right |## in the formula ?
Does it perhaps mean you need to evaluate this factor in the numerator at ##z= \sqrt{R^2 - (x^2+y^2)\,}\ ## ?

[edit] oopsed & fixed.
but ##x²+y²=R²## so ##z= \sqrt{R^2 - (x^2+y^2)\,}=0##
 
  • #16
this way of solve it is so difficult
 
  • #17
Rafa Ariza said:
but ##x²+y²=R²## so ##z= \sqrt{R^2 - (x^2+y^2)\,}=0##
No. That is on the rim of the circle, not in the interior. The projection of S on the xy plane is the whole quarter circle, not just the edge.
 
  • #18
BvU said:
No. That is on the rim of the circle, not in the interior. The projection of S on the xy plane is the whole quarter circle, not just the edge.
could you write it ?
 
  • #19
So your bounds are e.g. 0-1 for x, 0-##\sqrt{1-x^2\,}## for y
 
  • #20
BvU said:
So your bounds are e.g. 0-1 for x, 0-##\sqrt{1-x^2\,}## for y
o-R for x right?
 
  • #21
Yes. You can work R out of the integral and forget about the R4 -- it's just a scale factor.

re x, y bounds:
But here too there are more suitable coordinates to be chosen
 
  • #22
BvU said:
Yes. You can work R out of the integral and forget about the R4 -- it's just a scale factor.

re x, y bounds:
But here too there are more suitable coordinates to be chosen
so difficult.. i don't know
 
  • #23
Come on... it looks simple enough now:$$\iint xy\ {R\over \sqrt{R^2-(x^2+y^2)\,}} \ dx dy$$
 
  • #24
BvU said:
Come on... it looks simple enough now:$$\iint xy\ {R\over \sqrt{R^2-(x^2+y^2)\,}} \ dx dy$$
$$R\iint {xy\over \sqrt{R^2-(x^2+y^2)\,}} \ dx dy=\int {r³\over \sqrt {R²-r²\,}} \ dr...$$
 
  • #25
BvU said:
You can work R out of the integral and forget about the R4 -- it's just a scale factor.
My mistake. #23 is correct, though.
BvU said:
more suitable coordinates
Polar coordinates: ##\displaystyle{\int_0^R \int_0^{\pi\over 2}... dr\;d\phi}##
 
  • #26
BvU said:
My mistake. #23 is correct, though.
Polar coordinates: ##\displaystyle{\int_0^R \int_0^{\pi\over 2}... dr\;d\phi}##
i edit it
 
  • #27
##\displaystyle{\int_0^R {r³\over \sqrt{R²-r²}\,}\int_0^{\pi\over 2}{\sin \phi\cos \phi}\ dr\;d\phi}## ?
 
  • #28
Why the question marks ? The ##d\phi## part we've seen already. To be honest, I have to look up the ##dr## part :rolleyes:. How about you ?
 
  • #29

BvU said:
Why the question marks ? The ##d\phi## part we've seen already. To be honest, I have to look up the ##dr## part :rolleyes:. How about you ?
YESSSS DONE! THANKS MEN AWESOME
 
  • #30
You are welcome. Not bad for a first thread !
 
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  • #31
BvU said:
You are welcome. Not bad for a first thread !
it was awesome, really love maths
 
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