Convering double intgral to polar coordinates

In summary, the student is trying to solve an equation in polar coordinates, but is having trouble. He reads and follows the instructionsLCKurtz suggests using the integral formula, and then calculates the derivative using the chain rule. After doing that, he is able to solve for r and θ.
  • #1
11
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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 :)
 
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  • #2
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
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 can't do beause the integral doesn't converge
 
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  • #4
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
oceansoft said:
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
foxjwill said:
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
oceansoft said:
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
foxjwill said:
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
oceansoft said:
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
foxjwill said:
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?
 

1. What is the process for converting a double integral to polar coordinates?

To convert a double integral to polar coordinates, follow these steps:

  • 1. Identify the limits of integration in the Cartesian coordinates.
  • 2. Substitute each coordinate in terms of polar coordinates (r and θ).
  • 3. Change the integrand to a function in terms of r and θ.
  • 4. Find the new limits of integration in terms of r and θ.
  • 5. Rewrite the double integral in polar coordinates.

2. What is the purpose of converting a double integral to polar coordinates?

Converting a double integral to polar coordinates can make the integration process easier and more efficient in certain situations. It is particularly useful when dealing with circular or symmetric regions, as it can simplify the integrand and reduce the number of variables.

3. Can a double integral always be converted to polar coordinates?

No, not all double integrals can be converted to polar coordinates. It depends on the shape and limits of the region being integrated over. In some cases, it may be more complicated or impossible to express the region and integrand in terms of polar coordinates.

4. How does the Jacobian affect the conversion of a double integral to polar coordinates?

The Jacobian, which represents the change in variables, is an important factor in converting a double integral to polar coordinates. It is equal to the determinant of the transformation matrix, which is used to convert the integrand from Cartesian to polar coordinates. The Jacobian must be included when rewriting the integral in polar form.

5. What are some potential challenges of converting a double integral to polar coordinates?

One potential challenge is identifying the correct limits of integration in polar coordinates. It may also be difficult to express the integrand in terms of r and θ, especially if it involves square roots or trigonometric functions. Additionally, the Jacobian calculation can be complicated and may require extensive algebraic manipulation.

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