Compute Integral Using Jacobian Det in Polar Coordinates

In summary, the conversation discusses determining the Jacobian determinant for "polar" coordinates and using it to compute an integral. The person asks for clarification on why the Jacobian determinant is different when mapping from (x,y) to (theta,r) versus (r,theta) and provides their attempt at a solution. The expert suggests calculating the determinant of the derivative matrix and explains that the result should be rsin2(x) + rcos2(x).
  • #1
Mamooie312
10
0

Homework Statement


Determine the Jacobian determinant for "polar" coordinates and use that to compute the intergral . . . Blah blah blah that's not the point.


Homework Equations



(x,y) maps by T to (r, theta) or (theta, r) detT = jacobian

The Attempt at a Solution



Anyways, first I treated it as a map from (x,y) to (theta, r) and I got the answer in the book but negative and then I did it the other way and I got the answer in the book. Why is this and if I differ how I map the function then I supposed to integrate the opposite way (like from pi to 0 instead of from 0 to pi)
 
Physics news on Phys.org
  • #2
I have no idea what you are saying or what question you are asking. Please show exactly what you did in calculating the Jacobian.
 
  • #3
T: (x,y) to (theta, r) x = rcostheta y = rsintheta derivative matrix of T: t11 = -rsintheta t12 = costheta t21 = rcostheta t22 = sintheta detT = -r(sintheta)(squared) - r(costheta)(squared) = -r but the Jacobian should just be r and if you map T: (x,y) to (r, theta) instead of what I did then it works out to be r, why is this.
 
  • #4
Mamooie312 said:
T: (x,y) to (theta, r) x = rcostheta y = rsintheta derivative matrix of T: t11 = -rsintheta t12 = costheta t21 = rcostheta t22 = sintheta detT = -r(sintheta)(squared) - r(costheta)(squared) = -r but the Jacobian should just be r and if you map T: (x,y) to (r, theta) instead of what I did then it works out to be r, why is this.

take your matrix and do the determinant.

rsin2(x) + rcos2(x) is what you'll get
 

Related to Compute Integral Using Jacobian Det in Polar Coordinates

1. What is the Jacobian determinant in polar coordinates?

The Jacobian determinant in polar coordinates is a mathematical expression used to transform integrals from Cartesian coordinates to polar coordinates. It takes into account the change in variables and allows for the computation of integrals in polar coordinates.

2. How is the Jacobian determinant used to compute integrals in polar coordinates?

The Jacobian determinant is used to transform the integrand and the limits of integration from Cartesian to polar coordinates. This allows for the computation of the integral in terms of polar variables, which can be solved using standard integration techniques.

3. Is the Jacobian determinant always necessary when computing integrals in polar coordinates?

No, the Jacobian determinant is not always necessary when computing integrals in polar coordinates. It is only needed when there is a change of variables involved, such as when transforming from Cartesian to polar coordinates.

4. How is the Jacobian determinant calculated in polar coordinates?

The Jacobian determinant in polar coordinates is calculated by taking the determinant of the Jacobian matrix, which is a matrix of partial derivatives of the transformation equations. In polar coordinates, the Jacobian matrix has the form [r*cos(theta), r*sin(theta); -r*sin(theta), r*cos(theta)], and the determinant is r.

5. Are there any limitations or special considerations when using the Jacobian determinant in polar coordinates?

One limitation of using the Jacobian determinant in polar coordinates is that it only applies to integrals with circular or symmetric boundaries. Additionally, when computing integrals over non-circular regions, the Jacobian determinant may need to be broken into multiple parts to account for the change in variables.

Similar threads

Replies
4
Views
934
  • Calculus and Beyond Homework Help
Replies
24
Views
2K
  • Calculus and Beyond Homework Help
Replies
7
Views
4K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
14
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
396
  • Calculus and Beyond Homework Help
Replies
14
Views
2K
  • Calculus and Beyond Homework Help
Replies
22
Views
2K
Back
Top