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# Scalar field line integral independent of path direction

Showing that the line integral of a scalar field is independent of path direction. Created by Sal Khan.

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• Sal mentions that a standard,run of the mill integral equals minus the same integral with swapped boundaries.He explains it has to be so because the dx's are negative.

But what is the intuition?It is just the area under a curve.How can it be negative when we calculate it the other way? •   Remember how when you first learned about definite integrals, they probably talked about those little approximating rectangles? If there were n rectangles, we said that the width of each rectangle was ∆x=(b-a)/n. But if you switch the a and b, the width is ∆x=(a-b)/n, which is the negative of the original thing. Now if you summed up all the negative rectangles, you would get the negative of the original answer.
• How can these integrals at the end (approx: ) be the same? In the first integral, u=t, in the second, u=a+b-t. So integral(a..b,f(x(t),y(t))*sqrt(x'(t)^2+y'(t)^2))dt is not equal to int(a..b, f(x(u),y(u))*sqrt(x'(u)^2+y'(u)^2))du • The integral(u=a..b,g(u)du) equals integral(t=a..b,g(t)dt). They both calculate the area from g(a) to g(b), it doesn't matter how 'u' and 't' are related. Keep in mind, when you have "u=a" at the bottom of the integral sign and "u=b" at the top of the integral sign, it means u is being evaluated for all values between a and b. It's not a single value, it's all values in that range.

Example: g(x) = x. integral(u=a..b,g(u)du) = [0.5*u^2](u=a,b) = 0.5*b^2 - 0.5*a^2. There is no "u" in the result, it's purely in terms of 'a' and 'b'. Similarly, integral(t=a..b,g(t)dt) = [0.5*t^2](t=a,b) = 0.5*b^2 - 0.5*a^2. Same result, there is no "t" in the result, it's purely in terms of 'a' and 'b'.
• anyone knows where can i find relevant exercises? thanks • Is it possible to have a line integral of a scalar field with respect to a position vector function ?

Why is it that we use a scalar parametrization of a curve for a scalar field line integral while we use a vector parametrization (i.e. position vector function) for a vector field integral ? • I think that if we substitute u = a + b - t, and then u = t, we have a collision of signs used. Clearly a + b - t is not equal to t, so the first u cannot be the second u. We probably should create a new variable, say v = t. When doing so, I no longer see that these two integrals are equal. Of course, I trust the intuition but this whole u-substitution things seems a bit blurry... Can someone clear this out for me? • A "regular" integral becomes negative when the limits of integration (and thus the direction) are switched (). For line integrals, switching the limits of integration entails changing x(t) to x(a+b-t) and doing the same for y(t), as Sal shows in this video; doing so does not change the sign of the integral.

But what if the limits of integration a and b are switched in a line integral? Is that the same as changing x(t) and y(t) to x(a+b-t) and y(a+b-t)? Does the integral still stay positive, unlike a "regular" integral? • if you just switch the extremes without changing parametrization, you'll wind up with something negative. that's the equivalent of the switching in the "regular" integral. in the line integral, that means that you "walk" the line in a direction, but you integrate in the other: it doesn't make as much intuitive sense as switching start and finish and reversing parametrization too.
• what is the difference between a scalar field and a vector field?
(1 vote) • The math makes sense; however, I'm still confused. With regular integrals, direction matters...but I still feel like if you're going in the opposite direction shouldn't the area be negative? Like if you pretend your'e finding the area of the curtain with regular integration?   