Is there any way to calculate this integral?

In summary, the theory is that if you change the form of the parameter σ to implicit form, then you need to know how to convert between the two forms. However, the theory does not explain what the symbols mean or how to use them.
  • #1
Rafa Ariza
18
0
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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  • #2
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.
 
  • #3
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
 
  • #4
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.
 
  • #6
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 ?
 
  • #7
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
 
  • #8
And what does the theory say about z on ##R_{xy}## ?
 
  • #9
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 !
 
  • Like
Likes SammyS
  • #31
BvU said:
You are welcome. Not bad for a first thread !
it was awesome, really love maths
 

FAQ: Is there any way to calculate this integral?

Can this integral be solved analytically?

It depends on the specific integral in question. Some integrals can be solved using known mathematical formulas or techniques, while others may require numerical methods.

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Is there a software or tool that can calculate this integral for me?

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