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

Fundamental Theorem for Line Integrals

The fundamental theorem for line integrals states that if a vector field is conservative (or equivalently, its corresponding differential form is exact), then the line integral of that vector field along any smooth curve depends only on the values of its potential function at the endpoints. This theorem allows one to bypass the calculation of the integral along the specific path by simply calculating the difference in the potential function at the endpoints.

Exact Differential Forms

An exact differential form is one that can be expressed as the differential of a scalar function. In other words, if a differential form ? is equal to d? for some function ?, then ? is considered exact. This is an important concept in multivariable calculus because exact forms have the property that their integrals over any path depend only on the endpoints, not on the particular path taken.

Potential Function

A potential function is the scalar function ? whose gradient gives rise to the exact differential form; that is, d? equals the differential form in question. Once a potential function is identified, the evaluation of a line integral over a path from one point to another becomes straightforward by computing the difference ?(final) - ?(initial). This simplifies many problems in vector calculus and physical applications involving conservative forces.

Transcript

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

00:10 So basically this question is asking us to find whether this function here is conservative and then to evaluate the integral using this knowledge.

00:20 So this function, as we recall, is a function of a vector field.

00:24 So this is basically a vector field f can be written as 2x.

00:32 To y and to z.

00:38 And to find whether it's conservative, we can check using the partial derivative of r in respect to y is equivalent to the partial derivative of q in respect to z.

00:52 The partial derivative r in respect to x is equal to the partial derivative p in respect to z.

00:58 And finally, the partial derivative of q in respect to x is equal to the partial derivative p in respect to y, with the partial derivative of x, the 2x being p, q is 2y, and r is 2.

01:09 Z.

01:11 So the partial derivative of r respect to y or the partial derivative of 2z in respect to y is 0.

01:17 The partial derivative of q in respect to z or partial to y is again 0.

01:23 So these are equivalent.

01:26 The partial derivative of x, the partial derivative of r respect to x or 2z is 0.

01:32 The partial derivative of p respect to z or 2x is also zero.

01:38 So these two are equivalent.

01:40 And finally, the partial derivative of q in respect to x or the partial derivative to y respect to x is zero.

01:48 And the partial derivative of p respect to y or the partial derivative of 2x is again zero.

01:54 So all these are equivalent to each other.

01:58 Therefore, this is conservative.

02:00 So now that we know that this differential form is conservative, aka exact, we can use the funzo mental theorem of line integrals.

02:15 So the funzel mental theorem of line integral state that the integral from a to b of the gradient potential function, little f, dr, is equal to the potential function of point b minus a potential function of point a.

02:35 So what we want to do is then find what the potential function is.

02:40 So we can plug in these points and evaluate the integral.

02:45 So recall that the potential or the gradient of little f is equal to the vector field function f.

02:54 So we can rewrite this as the partial of little f respect to x equals p or r2x.

03:06 The partial of f for respect to y is equal to q or our two y and the partial of little f respect to z is going to be equal to our r or r to z just going to check real quick yes um so now the goal is to again find the potential function of little f so that the gradient of the potential function of little f is equal to the vector field f so we're going to what we're going to do is now we're going to take the integral of these partials, any of these partials, and generally we will start with the partial in respect to x.

03:47 So we're going to integrate this in respect to x, so the integral of 2x in respect to x equals 2x squared over 2, and we can just cancel these out, plus some function of y and z...

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