Convering double intgral to polar coordinates

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Homework Statement


Hey am studing for my up coming exam and i am having trouble with transforming double intrgral to polar coordinates i have no idea where to start or anything so can someone explain it to me


Homework Equations


this is example

[tex]\oint^{\infty}_{0}[/tex][tex]\oint^{\infty}_{0}[/tex] exp^-(x^2+y^2) dx dy

The Attempt at a Solution


like i said i have no idea about this so please help

also can someone tell me what I=[tex]\oint^{\infty}_{0}[/tex] exp^(-x^2) dx relates to probability using normal distribution

Thanks :)
 
Last edited:

Answers and Replies

  • #2
LCKurtz
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I don't think you mean circuit integrals, rather just ordinary integrals.

Let x = r cos(θ), y = r sin(θ) and dydx = r dr dθ

and use r,θ limits for the first quadrant.
 
  • #3
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so i get this right
[tex]\oint^{\infty}_{0}[/tex][tex]\oint^{\infty}_{0}[/tex] exp^-(rcos(o)+rsin(o)) dr do
is that in polar coordinates now right? and then i just do the ingeral? and is that what i would always do when tranforming to polar?

when i did that i get left with
[tex]\oint^{\infty}_{0}[/tex] (1/(sin(0)+cos(0)) do
i cant do beause the integral doesn't converge
 
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  • #4
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First of all, use "\int" instead of "\oint" when you're writing an integral, unless its a line integral about a closed curve.

As for your problem, here are the steps you should take when transforming from rectangular to polar coordinates:
  1. Express the domain of integration in polar coordinates. For your integral, [itex]x[/itex] and [itex]y[/itex] both run from [itex]0[/itex] to [itex]\infty[/itex], so the domain of integration is the first quadrant (the top right quadrant). To express this in polar coordinates, notice that no matter how far you are from the origin (i.e. no matter how large you make [itex]r[/itex]), you can always vary [itex]\theta[/itex] between 0 and [itex]\pi/2[/itex] and still stay in the first quadrant. So, [itex]r[/itex] must run from [itex]0[/itex] to [itex]\infty[/itex] and [itex]\theta[/itex] must run from [itex]0[/itex] to [itex]\pi/2[/itex].
  2. Do what LCKurtz suggested above.
 
  • #5
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so i get this right
[tex]\oint^{\infty}_{0}[/tex][tex]\oint^{\infty}_{0}[/tex] exp^-(rcos(o)+rsin(o)) dr do
is that in polar coordinates now right?
No!

First of all, read what I wrote above as to the limits of integration. Second of all, you forgot to square [itex]r\cos\theta[/itex] and [itex]r\sin\theta[/itex].

Also, to give you an example of LaTeX, enclose
Code:
\int_0^\infty \int_0^\infty e^{-(x^2+y^2)} dxdy
[code]
in "TEX" tags to typeset your original integral.
 
  • #6
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No!

First of all, read what I wrote above as to the limits of integration. Second of all, you forgot to square [itex]r\cos\theta[/itex] and [itex]r\sin\theta[/itex].

Also, to give you an example of LaTeX, enclose
Code:
\int_0^\infty \int_0^\infty e^{-(x^2+y^2)} dxdy
[code]
in "TEX" tags to typeset your original integral.[/QUOTE]

sorry for the troble but is this what it looks like then?

[tex]\int^{pi/2}_{0}[/tex][tex]\int^{\infty}_{0}[/tex] e^-(rcos^2[tex]\theta[/tex] + rsin^2[tex]\theta[/tex]) dr d[tex]\theta[/tex]

= [tex]\int^{pi/2}_{0}[/tex] 1 d[tex]\theta[/tex]

= pi/2
 
  • #7
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sorry for the troble but is this what it looks like then?

[tex]\int^{pi/2}_{0}[/tex][tex]\int^{\infty}_{0}[/tex] e^-(rcos^2[tex]\theta[/tex] + rsin^2[tex]\theta[/tex]) dr d[tex]\theta[/tex]

= [tex]\int^{pi/2}_{0}[/tex] 1 d[tex]\theta[/tex]

= pi/2
Nope. Let me ask you this: If I give you a point (x,y), can you tell me how far away from the origin it is?

Also, remember that [itex] dx\,dy = r\,dr\,d\theta[/itex], not [itex] dr\,d\theta[/itex].
 
  • #8
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Nope. Let me ask you this: If I give you a point (x,y), can you tell me how far away from the origin it is?

Also, remember that [itex] dx\,dy = r\,dr\,d\theta[/itex], not [itex] dr\,d\theta[/itex].
No not rly u can say its [tex]\sqrt{(x^2+y^2}[/tex]
 
  • #9
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No not rly u can say its [tex]\sqrt{(x^2+y^2}[/tex]
First of all, this is not text messaging, so please use full words. Anyway, this is correct, the distance is [tex]\sqrt{x^2+y^2}[/tex]. Now, what you want to keep in mind is that the 'r' in polar coordinates is defined to be the distance from the origin--in other words, [tex]r=\sqrt{x^2+y^2}[/tex].
 
  • #10
11
0
First of all, this is not text messaging, so please use full words. Anyway, this is correct, the distance is [tex]\sqrt{x^2+y^2}[/tex]. Now, what you want to keep in mind is that the 'r' in polar coordinates is defined to be the distance from the origin--in other words, [tex]r=\sqrt{x^2+y^2}[/tex].
Sorry for annoying you like this so the does the integral look this

[tex]\int^{pi/2}_{0}[/tex][tex]\int^{\infty}_{0}[/tex] e^-(rcos^2[tex]\theta[/tex]+rsin^2[tex]\theta[/tex]) r dr d[tex]\theta[/tex]

what am i doing wrong?
 

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