Line Integral question


by mathsciguy
Tags: integral, line
mathsciguy
mathsciguy is offline
#1
Dec5-12, 08:00 AM
P: 132
Suppose I have a vector V and I want to compute for the line integral from point (1,1,0) to point (2,2,0) and I take the path of the least distance (one that traces the identity function).

The line integral is of the form:
[tex] \int _a ^b \vec{V} \cdot d\vec{l} [/tex]

Where:

[tex] x=y, \ d\vec{l} =dx \hat{x} + dx \hat{y} [/tex]

Thus the integral can be computed purely in terms of x (can also be y), which looks something like this:
[tex] \int _a ^b V(x)dx [/tex]

What I don't exactly understand is why is it okay to use the limits like this:
[tex] \int _1 ^2 V(x)dx [/tex]

Why can we use the limits from 1 to 2 if we express the line integral in terms purely of x. I have a very vague idea of why it is, but I'd rather take it from people who actually know this to explain this to me. Thanks.
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Vorde
Vorde is offline
#2
Dec5-12, 06:30 PM
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P: 784
If I understand you correctly, the function you end up integrating is only in terms of ##x##, and therefore you don't need to parametrize it (or you can look at it by saying you are using ##x## as your parameter). Either way, because your function is dependent only on ##x##, all you have to do is integrate along the x-axis, which is from 1 to 2.
mathsciguy
mathsciguy is offline
#3
Dec7-12, 06:03 AM
P: 132
Actually, I've found out that to 'parametrize' the variables into x=t, y=t is a more comforting method to do it. At least intuitively, I see it as tracing the path of integration when we set the x and y variables into that parametric equation.

Edit: Yes, I didn't see it, but I was using x as the parameter. Thanks.


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