Cant understand the mathematical transition

In summary, the integral used in the video and the one you used are different types of integrals and cannot be interchanged.
  • #1
nhrock3
415
0
http://www.youtube.com/watch?v=JUFZbGv6wdM&feature=relmfu

in here in the end
i got an arch that startes at A and end in B
he wrote this integral
[tex]
\intop_{a}^{d}\frac{{dl}}{r^{2}}
[/tex]

but point a and be are not values that i can put in
if my function depends on "r" then i need to change dl to be dr expression somehow
?

i think that the integral should go from 0 to pi
and dl should be d theta expression (instead of dr)
and dl =r*d(theta)
R is constant
i get
[tex]
\intop_{0}^{2\pi}\frac{{Rd\theta}}{R^{2}}
[/tex]
i get 2pi/R

this resolt is different from the resolt he gets

why?
 
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  • #2
It is likely that the instructor was using a different form of the integral than what you used. The integral in the video is a line integral, which involves integrating along a path from point A to point B. The integral you used is a standard integral, which is used for functions of a single variable. These two types of integrals are not interchangeable. The instructor's integral could be written as: \int_{A}^{B}\frac{\mathrm{d}l}{r^2},where $\mathrm{d}l$ is the differential arc length along the curve between points A and B. This form of the integral is appropriate for calculating the arc length of a general curve, and it is not equivalent to the standard integral you wrote.
 

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