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Limits and Jacobian for Gaussian Integrals

Two Multiple IntegralsJacobian or No Jacobian - Surface IntegralsSetting limits on a Triple IntegralConverting coordinatesHow to convert this cartesian double integral to polar$I = int_0^infty t^2 e^-t^2/2 dt$Integral of Multivariable Gaussian across a Circular Domain.time dependant Gaussian integralDouble Integral from Polar to Cartesian CoordinatesJustification for why Jacobian determinant is new area of area element after transformation

2

$begingroup$

Ahoy everyone!
I am new to Gaussian Integrals and my teachers cannot help me out (because they don't get it). So I turn to the Internet for answers. I have very basic doubts and would really appreciate a clear explanation.
1) How do we explain $dxdy = rcdot dr dtheta$? I am looking for either a simple and intuitive geometric interpretation and/or an algebraic proof which starts from the very basics.
2)While converting cartesian coordinates to polar, I believe the limits for '$r$' are the same as those of $x$ (and $y$) but how do we exactly decide the limits of theta?
I would appreciate any help.
Thanks.

calculus multivariable-calculus gaussian-integral

share|cite|improve this question

edited Mar 14 at 16:41

Max

9071318

asked Mar 14 at 16:06

Hardik RajpalHardik Rajpal

132

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
  $endgroup$
  – dantopa
  Mar 14 at 16:41
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Ahoy everyone!
I am new to Gaussian Integrals and my teachers cannot help me out (because they don't get it). So I turn to the Internet for answers. I have very basic doubts and would really appreciate a clear explanation.
1) How do we explain $dxdy = rcdot dr dtheta$? I am looking for either a simple and intuitive geometric interpretation and/or an algebraic proof which starts from the very basics.
2)While converting cartesian coordinates to polar, I believe the limits for '$r$' are the same as those of $x$ (and $y$) but how do we exactly decide the limits of theta?
I would appreciate any help.
Thanks.

calculus multivariable-calculus gaussian-integral

share|cite|improve this question

edited Mar 14 at 16:41

Max

9071318

asked Mar 14 at 16:06

Hardik RajpalHardik Rajpal

132

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
  $endgroup$
  – dantopa
  Mar 14 at 16:41
  
  

  
  
  
  

add a comment | 

$begingroup$

Ahoy everyone!
I am new to Gaussian Integrals and my teachers cannot help me out (because they don't get it). So I turn to the Internet for answers. I have very basic doubts and would really appreciate a clear explanation.
1) How do we explain $dxdy = rcdot dr dtheta$? I am looking for either a simple and intuitive geometric interpretation and/or an algebraic proof which starts from the very basics.
2)While converting cartesian coordinates to polar, I believe the limits for '$r$' are the same as those of $x$ (and $y$) but how do we exactly decide the limits of theta?
I would appreciate any help.
Thanks.

calculus multivariable-calculus gaussian-integral

share|cite|improve this question

edited Mar 14 at 16:41

Max

9071318

asked Mar 14 at 16:06

Hardik RajpalHardik Rajpal

132

$endgroup$

share|cite|improve this question

edited Mar 14 at 16:41

Max

9071318

asked Mar 14 at 16:06

Hardik RajpalHardik Rajpal

132

share|cite|improve this question

edited Mar 14 at 16:41

Max

9071318

asked Mar 14 at 16:06

Hardik RajpalHardik Rajpal

132

edited Mar 14 at 16:41

Max

9071318

edited Mar 14 at 16:41

Max

9071318

asked Mar 14 at 16:06

Hardik RajpalHardik Rajpal

132

asked Mar 14 at 16:06

Hardik RajpalHardik Rajpal

132

1 Answer
1

active

oldest

votes

0

$begingroup$

The limits of theta ($theta$) are $0$ and $infty$, i.e. one full circle. And you should look at the following: https://www.math24.net/double-integrals-polar-coordinates/

I hope this helps you.

share|cite|improve this answer

edited Mar 14 at 16:36

answered Mar 14 at 16:22

MaxMax

9071318

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The website you shared clears up the doubts I had about the limits but the first question still feels foggy. I am not quite clear about why we put dxdy = Jacobian drdtheta because I don't understand what the Jacobian is exactly. Could you dumb it down for me?
  $endgroup$
  – Hardik Rajpal
  Mar 15 at 2:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Hardik Rajpal Do you know why the jacobian is used in a double integral? Because if you do, the answer from the site seems straightforward. So should your question be rephrased?
  $endgroup$
  – Max
  Mar 15 at 7:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @hardik rajpal PS the jabobian is used because you convert one area to another area. To even out the difference in size you use the determinant of the jacobian. See en.m.wikipedia.org/wiki/Determinant for more information about the determinant, and areas.
  $endgroup$
  – Max
  Mar 15 at 7:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I appreciate the linear algebra involved but it would take me a few months to get the deep understanding I value so much with that road. Is it fine if I just consider dxdy to be an elementary dA (A=area) in cartesian coordinates and rdrdtheta to be the equivalent dA in polar coordinates? It makes a lot more sense with a picture but I am new to the website.
  $endgroup$
  – Hardik Rajpal
  Mar 16 at 4:11
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed, you could say that. The extra $r$ is ‘just’ a scalar that accurs because you convert one area to another. Take the following example: let’s describe the whole plane with $x^2+y^2$ but in polar form you need the extra $r$, otherwise it would be the unit circle (not the whole plane).
  $endgroup$
  – Max
  Mar 16 at 7:56
  
  

  
  
  
  

add a comment | 

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0

$begingroup$

The limits of theta ($theta$) are $0$ and $infty$, i.e. one full circle. And you should look at the following: https://www.math24.net/double-integrals-polar-coordinates/

I hope this helps you.

share|cite|improve this answer

edited Mar 14 at 16:36

answered Mar 14 at 16:22

MaxMax

9071318

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The website you shared clears up the doubts I had about the limits but the first question still feels foggy. I am not quite clear about why we put dxdy = Jacobian drdtheta because I don't understand what the Jacobian is exactly. Could you dumb it down for me?
  $endgroup$
  – Hardik Rajpal
  Mar 15 at 2:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Hardik Rajpal Do you know why the jacobian is used in a double integral? Because if you do, the answer from the site seems straightforward. So should your question be rephrased?
  $endgroup$
  – Max
  Mar 15 at 7:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @hardik rajpal PS the jabobian is used because you convert one area to another area. To even out the difference in size you use the determinant of the jacobian. See en.m.wikipedia.org/wiki/Determinant for more information about the determinant, and areas.
  $endgroup$
  – Max
  Mar 15 at 7:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I appreciate the linear algebra involved but it would take me a few months to get the deep understanding I value so much with that road. Is it fine if I just consider dxdy to be an elementary dA (A=area) in cartesian coordinates and rdrdtheta to be the equivalent dA in polar coordinates? It makes a lot more sense with a picture but I am new to the website.
  $endgroup$
  – Hardik Rajpal
  Mar 16 at 4:11
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed, you could say that. The extra $r$ is ‘just’ a scalar that accurs because you convert one area to another. Take the following example: let’s describe the whole plane with $x^2+y^2$ but in polar form you need the extra $r$, otherwise it would be the unit circle (not the whole plane).
  $endgroup$
  – Max
  Mar 16 at 7:56
  
  

  
  
  
  

add a comment | 

0

$begingroup$

The limits of theta ($theta$) are $0$ and $infty$, i.e. one full circle. And you should look at the following: https://www.math24.net/double-integrals-polar-coordinates/

I hope this helps you.

share|cite|improve this answer

edited Mar 14 at 16:36

answered Mar 14 at 16:22

MaxMax

9071318

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The website you shared clears up the doubts I had about the limits but the first question still feels foggy. I am not quite clear about why we put dxdy = Jacobian drdtheta because I don't understand what the Jacobian is exactly. Could you dumb it down for me?
  $endgroup$
  – Hardik Rajpal
  Mar 15 at 2:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Hardik Rajpal Do you know why the jacobian is used in a double integral? Because if you do, the answer from the site seems straightforward. So should your question be rephrased?
  $endgroup$
  – Max
  Mar 15 at 7:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @hardik rajpal PS the jabobian is used because you convert one area to another area. To even out the difference in size you use the determinant of the jacobian. See en.m.wikipedia.org/wiki/Determinant for more information about the determinant, and areas.
  $endgroup$
  – Max
  Mar 15 at 7:23
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I appreciate the linear algebra involved but it would take me a few months to get the deep understanding I value so much with that road. Is it fine if I just consider dxdy to be an elementary dA (A=area) in cartesian coordinates and rdrdtheta to be the equivalent dA in polar coordinates? It makes a lot more sense with a picture but I am new to the website.
  $endgroup$
  – Hardik Rajpal
  Mar 16 at 4:11
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed, you could say that. The extra $r$ is ‘just’ a scalar that accurs because you convert one area to another. Take the following example: let’s describe the whole plane with $x^2+y^2$ but in polar form you need the extra $r$, otherwise it would be the unit circle (not the whole plane).
  $endgroup$
  – Max
  Mar 16 at 7:56
  
  

  
  
  
  

add a comment | 

$begingroup$

The limits of theta ($theta$) are $0$ and $infty$, i.e. one full circle. And you should look at the following: https://www.math24.net/double-integrals-polar-coordinates/

I hope this helps you.

share|cite|improve this answer

edited Mar 14 at 16:36

answered Mar 14 at 16:22

MaxMax

9071318

$endgroup$

share|cite|improve this answer

edited Mar 14 at 16:36

answered Mar 14 at 16:22

MaxMax

9071318

answered Mar 14 at 16:22

MaxMax

9071318

answered Mar 14 at 16:22

MaxMax

9071318