When to use the Jacobian in spherical coordinates?

Click For Summary
SUMMARY

The discussion centers on the application of the Jacobian in spherical coordinates, specifically addressing the surface element in the context of parametrization. The surface area element is derived from the cross product of the partial derivatives of the parametrization function, denoted as ##\boldsymbol{\sigma}##. The Jacobian factor of ##r^2 \sin v## is inherently included in the calculation of the area element, which is expressed as ##dS = |d\boldsymbol{l}_u \times d\boldsymbol{l}_v|##. This confirms that the Jacobian is effectively accounted for in the surface area computation.

PREREQUISITES
  • Understanding of spherical coordinates and their parametrization
  • Knowledge of vector calculus, particularly cross products
  • Familiarity with surface integrals and area elements
  • Basic proficiency in mathematical notation and symbols used in calculus
NEXT STEPS
  • Study the derivation of the Jacobian in spherical coordinates
  • Learn about parametrization techniques in multivariable calculus
  • Explore vector calculus applications in surface integrals
  • Investigate the role of the cross product in determining area elements
USEFUL FOR

Students and professionals in mathematics, particularly those focusing on multivariable calculus, vector calculus, and geometric applications in physics or engineering.

Amaelle
Messages
309
Reaction score
54
Homework Statement
look at the image
Relevant Equations
jacobian is r^2 sinv
Greetings!
1639498591229.png


here is the solution which I undertand very well:
1639498692794.png

1639498769703.png


my question is:
if we go the spherical coordinates shouldn't we use the jacobian r^2*sinv?

thank you!
 

Attachments

  • 1639410956821.png
    1639410956821.png
    10.7 KB · Views: 207
  • 1639411043905.png
    1639411043905.png
    8.3 KB · Views: 183
Physics news on Phys.org
It's sort of already baked in. The surface element comes from a little parallelogram with sides ##d\boldsymbol{l}_u = \frac{\partial \boldsymbol{\sigma}}{\partial u} du## and ##d\boldsymbol{l}_v = \frac{\partial \boldsymbol{\sigma}}{\partial v} dv## and therefore area
\begin{align*}
dS = |d\boldsymbol{l}_u \times d\boldsymbol{l}_v| = \left| \frac{\partial \boldsymbol{\sigma}}{\partial u} \times \frac{\partial \boldsymbol{\sigma}}{\partial v} \right| du dv
\end{align*}The part ##\frac{\partial \boldsymbol{\sigma}}{\partial u} \times \frac{\partial \boldsymbol{\sigma}}{\partial v}## is what your solutions denoted by the vector ##\mathbf{N}##.
 
Last edited:
  • Love
Likes   Reactions: Amaelle
thank you so much!
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 7 ·
Replies
7
Views
6K
Replies
2
Views
1K
Replies
3
Views
2K
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K