Confused about line integral.

In summary: So this means we have to resolve down to a constant vector ( or whatever you call it) that have constant direction. WHICH in our common rectangular, cylindrical and spherical coordinates, we have NO CHOICE but have to transform the vector into x,y,z coordinates no matter what.
  • #1
yungman
5,708
240
I want to integrate around a closed circular path on xy plane around the origin. Say the radius is b. So

##\oint d\vec l## where ##d\vec l=\hat{\phi}b d\phi##

1) If I just use polar( or spherical or even cylindrical) coordinates. R=b and
[tex]\oint d\vec l\;=\;\hat{\phi}\int_0^{2\pi} b d\phi\;=\;2\pi b[/tex]

2) Using rectangular co where ##\hat{\phi}=-\hat x \sin \phi +\hat y \cos \phi##
[tex]\oint d\vec l\;=\;b\int_0^{2\pi} (-\hat x \sin\phi+\hat y \cos \phi) d\phi\;=\;0[/tex]

What did I do wrong?
 
Physics news on Phys.org
  • #2
yungman said:
I want to integrate around a closed circular path on xy plane around the origin. Say the radius is b. So

##\oint d\vec l## where ##d\vec l=\hat{\phi}b d\phi##

1) If I just use polar( or spherical or even cylindrical) coordinates. R=b and
[tex]\oint d\vec l\;=\;\hat{\phi}\int_0^{2\pi} b d\phi\;=\;2\pi b[/tex]
This is a common mistake when you are first learning these things so don't worry. What you did wrong was pull out the ##\hat\phi## from the ##d\phi## integral. ##\hat\phi## is not a constant vector like the standard cartesian basis vectors! You can't simply pull out ##\hat\phi## from the ##d\phi## integral because ##\hat\phi## varies as ##\phi## varies. What you did in (2) was the correct way to evaluate integrals in such coordinates. Cheers!
 
  • #3
WannabeNewton said:
This is a common mistake when you are first learning these things so don't worry. What you did wrong was pull out the ##\hat\phi## from the ##d\phi## integral. ##\hat\phi## is not a constant vector like the standard cartesian basis vectors! You can't simply pull out ##\hat\phi## from the ##d\phi## integral because ##\hat\phi## varies as ##\phi## varies. What you did in (2) was the correct way to evaluate integrals in such coordinates. Cheers!

Thanks for the pointers. So this means we have to resolve down to a constant vector ( or whatever you call it) that have constant direction. WHICH in our common rectangular, cylindrical and spherical coordinates, we have NO CHOICE but have to transform the vector into x,y,z coordinates no matter what.
 
  • #4
Yessiree
 
  • #5
Thanks
 

What is a line integral?

A line integral is a type of integral in calculus that is used to calculate the total value of a function along a specific curve or path.

How is a line integral different from a regular integral?

A regular integral calculates the area under a curve, whereas a line integral calculates the total value of a function along a specific path.

What are the applications of line integrals?

Line integrals have many applications in physics and engineering, such as calculating work done by a force along a specific path, calculating fluid flow, and finding the center of mass of an object.

What is the formula for calculating a line integral?

The formula for calculating a line integral is ∫C f(x,y) ds = ∫ab f(x(t),y(t)) √(x'(t)^2 + y'(t)^2) dt, where f(x,y) is the function being integrated, C is the curve or path, a and b are the starting and ending points on the curve, and x(t) and y(t) are the parametric equations for the curve.

How do I know which direction to integrate in for a line integral?

The direction of integration is determined by the orientation of the curve. To integrate in the opposite direction, simply change the limits of integration from b to a in the formula. This is known as reversing the orientation of the curve.

Similar threads

Replies
3
Views
488
  • Calculus
Replies
7
Views
1K
  • Calculus
Replies
3
Views
2K
  • Calculus
Replies
3
Views
2K
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
21
Views
1K
Replies
3
Views
1K
Replies
2
Views
137
  • Calculus and Beyond Homework Help
Replies
3
Views
474
Replies
8
Views
266
Back
Top