Show that the differential forms in the integrals are exact. Then evaluate the integrals. ∫(0,0,

Show that the differential forms in the integrals are exact....

Key Concepts

Potential Functions

A potential function is a function whose differential equals the given exact differential form. Once a potential function is identified, it simplifies the evaluation of the integral, because the line integral of an exact form between two points depends only on the values of this potential function at those points. This eliminates the need to compute the integral along a specific path.

Fundamental Theorem for Line Integrals

The Fundamental Theorem for Line Integrals states that if a vector field (or differential form) is conservative (or exact), then the line integral along any path between two points depends solely on the values of its potential function at the endpoints. This theorem is a powerful tool that enables one to directly compute line integrals by simply subtracting the value of the potential function at the starting point from its value at the ending point.

Exact Differential Forms

An exact differential form is one that can be expressed as the differential of some scalar function; that is, if ? is a differential form, then there exists a function f such that ? = df. Identifying whether a form is exact often involves checking if the mixed partial derivatives are equal, which is a simple consequence of the theory of differential forms and the Poincaré lemma in the context of simply connected domains.

Transcript

00:01 So you want to do the following integral, the line integral from the point 0 to the point 331 of this 2x d x minus y squared dy minus 4 over 1 plus c squared d c so this can be seen as the line integral from the point let's call this point of a to this point b from a to b of the vector field f the d s where f with f i denote this field as on the first component to x on the second one minus y squared and the third one minus 4 over 1 plus c squared so this would be the doubt product with the ds which gives that so in order to be able to integrate along a path so this to get the 30 this integral is equal to the integral along some path down where this is any path starting starts at a starts at a and ends at b to get this path independence we need a have to be conservative conservative but so that is something that we need to check first so this at this stage that is a question is f conservative.

02:10 So if that is the case, well then we should have, let's call this function m and then this function p.

02:22 For f to be conservative, that is equivalent to the following, the equations to be true.

02:28 So the derivative of n with respect to, with respect to y, should be equal to the derivative of n with respect to x and then also the derivative of n with respect to c should be equal to the derivative of p with respect to y and this next equation that the derivative of m with respect to c should be equal to the derivative of p with respect to x so is in a connected domain this set of equations is the same as asking as is f equal to is f coming from a potential so with this i'm saying this is the partial of some function with respect to x this is this is the partial of the same function with respect to y.

03:55 And this is the partial of some function with respect to c.

04:00 So if that is the case, then we can, if it's conservative, but this is easy to check because partial of n with respect to y, that this part is 0, partial of n with respect to x is 0, partial of n with respect to c is 0, partial of p with respect to y is 0 and then partial of m with respect to x to c is 0 and partial of with respect to x is 0 so those equations are trivially satisfied so satisfied so m is conservative f conservative so uh let's let's do the following path for convenience uh so you have here the point you are doing great integrate this looks sort of like should look like the following so you have this point zero zero zero and we want to rate or up to this point three and one so this point lies over here this is the point three three one so i'm i'm saying that this is the y axis then that is the c and the x so this point three one one here so this point three one is here so we're going to integrate first along that path that goes along x let's call this one gamma 1 then a path that follows varize along y this is gamma 2 and then a path that the path independence this integral this integral over here this integral is equal to the integral along gamma 1 of f d s plus the integral along gamma 2 of f yes plus the integral along gamma 3 of f that yes so along gamma 1 we are only bar x so the integral is going to be just the x component from 0 up to 3 because x along gamma 1 is varying from 0 to 3 and of this of 2x just 2x so we have this this is this integral is that and then the integral along gamma 2 would be more uh over that point um gamma 2 goes from 0 up to 3 the y here is 0 and here y is x0 and here y is the value 3 so it will be that and then the function over there is minus y squared so we're integrating along there minus y squared the y so this corresponds to this gamma 2 the integral along gamma 2 and the integral o on gamma 3 is going to be the integral of this function so it's going to be plus the integral from 0 up to 1 because c only goes up to 1 of minus 4 1 plus is 1 plus c squared this so this integral is equal to these three integrals so this first the integral that would be the total of 2x is x squared evaluating between 0 and 3 for this one the interval of minus y squared is going to be minus y -cub 3rds and that evaluated between 0 and 3 and then for this one well we have it was c squared it's 4 c squared so the integral of this integral is equal to art tangent of c.

09:07 So then that would be equal to, we have the minus 4, so minus 4 times the arc tangent of c and that evaluated between 0 and 1...

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