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

Conservative Fields

A conservative field is one where the line integral between two points is independent of the chosen path. This property is equivalent to the existence of a potential function, and in simply-connected regions, a necessary and sufficient condition for a field to be conservative is that its curl is zero. Recognizing a field as conservative confirms that the associated differential forms are exact.

Fundamental Theorem for Line Integrals

This theorem states that if a vector field is conservative (i.e., if a potential function exists such that the field equals its gradient), then the line integral of the field along any smooth path from one point to another is simply the difference between the potential function evaluated at the end point and at the starting point. This greatly simplifies the computation of such integrals.

Potential Function

The potential function is a scalar function whose gradient yields the original differential form. When a differential form is determined to be exact, finding this potential function allows one to compute the integral simply by taking the difference of the function’s values at the endpoints, thus bypassing the need to perform a potentially complicated integration along a specific path.

Exact Differential Forms

An exact differential form is one that can be expressed as the differential of a scalar function (the potential function). This means that if a differential form ? is exact, there exists a function ? such that ? = d?. Establishing exactness is useful because it immediately implies that the line integral over any path between two points depends only on the endpoints.

Transcript

00:02 The question asked to show that the differential form in the integral are exact and to evaluate the integral.

00:09 So basically this question is asking us to find whether this function inside here is conservative and to evaluate the integral using this knowledge.

00:19 So this function, as we recall, is basically the function of the vector field.

00:25 So we can rewrite this as 2xy, xx, xx, minus z squared, and negative to 1 .1.

00:32 Y z in this vector field format and to find if it's conservative we can check using this test with partial derivatives so we can check by seeing the partial derivative of r respect to y is equivalent to the partial derivative of q in respect to z the partial derivative of r respect to x is equivalent to the partial derivative p in respect to z finally the partial derivative of q in respect to x is equivalent to the partial derivative of p in respect to y.

01:05 So, oh, and p being 2xy, q being x4 minus z squared, and r being negative to y z.

01:15 So the partial derivative of r or partial derivative of negative to y z in respect to y is just negative to z.

01:22 The partial derivative of q or our x squared minus z squared in respect to z is also negative to z.

01:29 So these two are equivalent.

01:31 The partial derivative of r or negative to y z in respect to x is 0.

01:37 The partial derivative of p or r2xy in respect to z is also zero, so these two are equivalent.

01:47 Now the partial derivative of q or x squared minus z square in respect to x is 2x.

01:54 And finally, the partial derivative of p or 2xy in respect to y is 2x, which they are both equivalent.

02:01 Since all three pairs are equivalent, this is conservative.

02:08 And because these differential forms in this integral are conservative, a .k .a.

02:14 Exact, we can use the fundamental theorem of line integrals to evaluate the integral.

02:21 So the fundamental theorem of line integrals state that from points a to b of the gradient of the potential function little f, d .r is equivalent to the potential function of point b minus the potential function of point a.

02:36 So in order to use this fundamental theorem, we have to find what that potential function is and plug in the points and then subtract them.

02:47 And to find what that potential function is, we can recall that the gradient of little f is equivalent to the vector field function, big f.

02:58 So essentially the partial derivative of little f in respect to x is equal to 2xy.

03:06 The partial derivative of little f in respect to y is x squared.

03:14 And the partial derivative of little f for respect to z is equivalent to negative 2y z.

03:23 Negative 2yc.

03:26 So now what we're going to do is take the integral of any of these partials.

03:31 And generally we start with the partial in respect to x.

03:36 So we're going to integrate this in respect to x.

03:38 So this will be the integral of 2xy in respect to x, which is 2x squared over 2 times y in the weekend.

03:50 So this plus some function of y and z.

03:55 Because if we take the partial in respect to x, this function goes away...

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