How do you integrate vectors, given vector components?

[jaguar7]

Homework Statement

We're trying to integrate a vector, E, with components, Ex = 6x^2y, Ey = 2x^3 + 2y and Ez = 0, from the origin to x = -0.4, y = 6.1, z = 0.0.

Homework Equations

integral E dot dr = V

The Attempt at a Solution

i don't know how

 

[lanedance]

$$\int \vec{E} \bullet \vec{dr}$$

start by parameterising (in terms of say t) the line you want to integrate along and find the form of dr(t), and infinitesimal line elemnt. then when you take the dot product it should reduce to a normal integral over t.

 

[HallsofIvy]

Homework Statement

We're trying to integrate a vector, E, with components, Ex = 6x^2y, Ey = 2x^3 + 2y and Ez = 0, from the origin to x = -0.4, y = 6.1, z = 0.0.

My first thought was "Along what path"? Generally, the integral of a vector between two points depends upon the particular path you take between the two points. You have to choose a parameterization for the particular path you want to use as landance said.

However, for some special vector functions, that is not true. And this is an example.
What happens here is that
[itex]\vec{E}\cdot d\vec{r}= \left((6x^2y)\vec{i}+ (2x^3+ 2y)\vec{j}+ 0\vec{k}\right)[/tex]$$\left(dx\vec{i}+ dy\vec{j}+ 0\vec{k}\right)= 6x^2ydx+ (2x^3+ y)dy$$<br /> <br /> is an "exact" differential (some texts use "phyics" terminology and say it is "conservative" as a conservative force field is such a vector function).<br /> <br /> That is because there exists a function, F(x,y), such that <br /> $$dF= \frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial y}dy= 6x^2ydx+ (2x^3+ 2y)dy$$<br /> <br /> Whenever there exist such a function, then differentiating the part multiplying dx with respect to y, and the part multiplying dy by x we get<br /> $$\frac{\partial }{\partial y}\left(\frac{\partial F}{\partial x}\right)= \frac{\partial^2 F}\partial y\partial x}$$<br /> and<br /> $$\frac{\partial}{\partial x}\left(\frac{\partial F}{\partial y}\right)= \frac{\partial^2 F}{\partial x\partial y}$$<br /> and for well behaved functions, those "mixed" second derivatives must be equal. Here<br /> $$\frac{\partial 6x^2y}{\partial y}= 6x^2= \frac{\partial (2x^3+ 2y)}{\partial x}$$. One method of finding the integral, since it is "independent of the path" is to choose whatever path you want, say the straight line between (0, 0, 0) and (-0.4, 6.1, 0.0) or the "broken line" from (0, 0, 0) to (-0.4, 0, 0) and then to (0.4, 6.1, 0).<br /> <br /> But when we have such an "exact" differential, we can integrate in exactly the way we do in Calculus I- find an "anti-derivative" (the "F" above) and evalulate between the two endpoints.<br /> <br /> That is, we want to find F(x,y) such that<br /> $$\frac{\partial F}{\partial x}= 6x^2y$$<br /> and<br /> $$\frac{\partial F}{\partial y}= 2x^3+ 2y$$<br /> <br /> Integrating <br /> $$\frac{\partial F}{\partial x}= 6x^2y$$<br /> gives<br /> $$F(x,y)= 2x^3y+ C$$<br /> <b>except</b> that, since the derivative is with respect to x only, treating y as a constant, that "constant of integration" might be a function of y. We really have <br /> $$F(x,y)= 2x^3y+ g(y)$$<br /> where g can be any function of y. <br /> From that<br /> $$\frac{\partial F}{\partial y}= 2x^3+ g'(y)$$<br /> Note that, since g is a function of y only, g'(y) is the ordinary derivative.<br /> But we must have<br /> $$\frac{\partial F}{\partial y}= 2x^3+ g'(y)= 2x^3+ 2y$$<br /> so that the "$2x^3$" terms cancel (that happens precisely because the mixed second derivatives are equal) and we are left with <br /> $$g'(y)= 2y$$<br /> so that <br /> $$g(y)= y^2+ C$$<br /> where "C" now really is a constant.[/itex]

 

[lanedance]

nice one missed it was a conservative function

 

[Char. Limit]

Watch out... the problem actually states 2x³+2y, not +y. Also, I think your power for y in the final solution is too high.

Also, we're not supposed to give out full solutions like that, are we?

 

[HallsofIvy]

Watch out... the problem actually states 2x³+2y, not +y. Also, I think your power for y in the final solution is too high.

Right. Fortunately, it is still "exact:" so I ducked that bullet. I have gone back and edited to correct those.

Also, we're not supposed to give out full solutions like that, are we?

Well, I didn't! I left the evaluation to be done.
(And, when I edited, I removed the last part of what I did.)

 

[Char. Limit]

Excellent, just fact-checking a few things.