Understanding Line Integrals with Respect to Delta X and Delta Y

[Cyrus]

Oh, I see. Then the function has to be one such that it is z= f(x,y). and the curve has to be defined as x=x(t); y=y(t); z=z(t). Now the curve does not lie on a plane, but none the less, you can still associate a point z=f(x,y) above the curve. (As long as the curve does not twist over itself, then things to go poopoo :-) .)

 

[Hurkyl]

So anyways, when you get ths integral set up right, you'll find that you're not integrating with respect to the arclength of the curve in 3-d space -- you'll integrate with respect to arclength in the projection on the xy plane. The key lesson I was trying to impart is that not all of the dimensions of the problem are relevant to the quantities you're trying to compute! That's why doing it WRT arclength isn't the "inherently right" way to do things.

 

[Cyrus]

I think I finally found the anwser I was looking for. I asked about line integrals with respect to x and y and why it was different from line integrals with respect to arc length. This was the anwser I was looking for: When you have a line integral with respect to a VECTOR field along a curve, then it can be reduced to three line integrals with with respect to x, y and z that are for a SCALAR function. This is where line integrals with respect o x,y and z come from. I kept asking if they are the of result doing a line integral with respect to arc length, and it turns out they are. Another mistake I was making was the fact that I kept asking about changing the line integral with respect to ds into a line integral with respect to dx or dy in order to solve it. But that was a big no no, because when I changed it to dx or dy, it was no longer a line integral. The integral was not along a curve, but now from end points a to b. I think that was the big thing I did not notice. When you reduce a line integral along a curve C into an integral with respect to x or y, its no longer a line integral. It becomes a regular definite integral. I was confusing this definite integral with that of the line integral with respect to x or y or z. They are two totally different things. It seems that line integrals with respect to x,y and z are the result of line integrals with respect to arclength of vector functions ONLY, Thats what they are and where the come from. But thank for trying to help. I think Galileo nailed it dead on, but I just dident process it until now. Thanks for all your help though dex and hurk.

 

[Hurkyl]

When you reduce a line integral along a curve C into an integral with respect to x or y, its no longer a line integral. It becomes a regular definite integral.

Only for special curves -- remember the example I gave before, where the area of a region R is computed by:

\int_C x \, dy

where C is the boundary of R. This isn't a regular definite integral at all!


Or, for example, the integral \int_C \, ds for the arclength of the unit circle becomes (after some very careful algebra):

\int_C -1/y \, dx

Which, again, is not just a regular definite integral.

 

[Cyrus]

Yes, hurkyl, but \int_C x \, dy does NOT come from the reduction of a line integral with respect to arclength for a scalar valued function, that's my point. It comes from the reduction of a VECTOR valued function when integrating with respect to arclength. Sorry If I did not type it clearly.

Anyways, another question:

In my book they write the following:

\int_c \nabla f * dr = \int^b_a \del f(r(t)) * r'(t) dt
= \int^b_a (\frac{\delta f dx}{\delta x dt} + \frac{\delta f dy}{\delta y dt} + \frac{\delta f dz}{\delta z dt}) dt

I have a little trouble with their notation. If r(t) = x(t)i + y(t)j +z(t)k,

then that means:

\frac{dr}{dt} = \frac{dx}{dt}i + \frac{dy}{dt}j +\frac{dz}{dt}k

or:

dr = ( \frac{dx}{dt}i + \frac{dy}{dt}j +\frac{dz}{dt}k )*dt.

As for the gradient, shouldent It be written as:

\nabla f(r(t)) = \nabla f(x(t),y(t),z(t)) = ( \frac{\delta f}{\delta x} \frac{dx}{dt} + \frac{\delta f}{\delta y} \frac{dy}{dt} +\frac{\delta f}{\delta z} \frac{dz}{dt} )

Wont dotting (*) by the r'(t)dt term just add a redudant dx/dt dy/dt dz/dt to the gradient, no longer making both sides of the expression equal?!?

It looks as if you do the gradient of \nabla f(r(t)) alone is enough to equal \int_C f * dr

 

[Hurkyl]

As for the gradient, shouldent It be written as:

Nope; the gradient is simply the vector whose entries are the partial derivatives with respect to the corresponding argument. What you've written is the (total) derivative of f with respect to t.

 

[Cyrus]

AH, I see. I thought that when you did the gradient you had to do it with respcet to x,y,z and also in terms of t. So would this be correct?

x=t^2; y=t; z= 4t (for example)

and F(x,y,z) = x^2yz^4.

Then when I do the gradient of F, it would be:

2xyz^4i + x^2z^4j + x^2y4z^3k.
And then I would plug in the values of x,y and z in terms of t.
But I would not do the differentiation with respect to x=x(t).
For the (i) direction term I would use: 2xyz^4i = 2(t^2)(t)(4t)^4, instead of doing the derivative with respect to x=x(t) as well:
2(2t)(t)(4t)^4.

 

[Hurkyl]

AH, I see. I thought that when you did the gradient you had to do it with respcet to x,y,z and also in terms of t. So would this be correct?

Yep. The notation \grad F(x(t), y(t), z(t)) means that you take the gradient of F, evaluated at the point (x(t), y(t), z(t)). F itself doesn't care about t. It doesn't care about 'x', 'y', or 'z' either -- recall that

F(x,y,z) = x^2yz^4

means exactly the same thing as

F(p,q,r) = p^2qr^4
F(\alpha,\beta,\gamma) = \alpha^2\beta\gamma^4
and even
F(\spadesuit,\diamondsuit,\clubsuit) = \spadesuit^2\diamondsuit\clubsuit^4

The equation merely defines the function pointwise. In fact, you should strive to use different variables like this when you're confused. (IMHO, you should always do it, or at least until you're very comfortable with this abuse of notation)

 

[Cyrus]

Ive never seen the gradient of a royal flush before...interesting
Thanks for the help, Merry Christmas Hurkyl.

Cheers,

Cyrus Abdollahi

 

[Hurkyl]

They're my favorite TeX symbols. I knew a professor who liked to use smiley-faces for variables in his lectures, though!

 

[Cyrus]

Another question:

They say that:

f(x,y)= \int^{(x,y)}_{(a,b)} F*dr

Now normally, It is written as:

f(x,y) =\int_C F*dr

Is the reason that they used the endpoints as the upper and lower limit because for a conservative vector field it does not matter which curve you use? So they decided to replace the C by simply the end points?

 

[Cyrus]

I think my math book has an error but ill let you guys decide. It trys to prove the conservation of energy based on the gradient of a line integral. I will just quote it below:

"Now let's further assume that F is a conservative force field;that is, we can write F = \nabla f. In physics, the potential energy of an object at the point (x,y,z) is defined as P(x,y,z) = -f(x,y,z), so we have F = - \nabla P. Then by theorem 2 we have:

W= \int_c F*dr = - \int_c \nabla P *dr

= - [P(r(b))-P(r(a))]
= P(A)-P(B)

Comparing this equation with Equation 16, we see that

P(A)+K(A) = P(B) + K(B) "

First point:

Isint this incorrect by definition, isint F =- \nabla f, by definition, where f is the potential function!?

It seems like they considered this when they said that P(x,y,z)= -f(x,y,z),
but when they added ANOTHER negative sign by saying that F =- \nabla P, they just canceled out their correction by using the first negative sign?

Its ok for me to factor out a negative sign if I do the gradient, so let's say I rewrite F= - \nabla P, in terms of the potential energy function f. Then I know that if P(x,y,z)= -f(x,y,z), then its also true that
\nabla P(x,y,z) = - \nabla f(x,y,z). Well then if I sub this back into the line integral, in the end I now get:
F(B)-F(A)

Oh, if I write that in terms of the potential P(x,y,z)= -f(x,y,z), I still get the same anwser. I see now. All this typing for nothing. But does anyone see my point. Isint it STUPID and CONFUSING to add in another damn negative when the definition is F= - \nabla f to start with!? This just makes things overly complicated when they need not be.

 

[Cyrus]

Another question:

They say that:

f(x,y)= \int^{(x,y)}_{(a,b)} F*dr

Now normally, It is written as:

f(x,y) =\int_C F*dr

Is the reason that they used the endpoints as the upper and lower limit because for a conservative vector field it does not matter which curve you use? So they decided to replace the C by simply the end points?

Also, how is this true?

f(x,y)= \int^{(x,y)}_{(a,b)} F*dr

When you evaluate this at the limits won't you get an anwser of:

f(x,y)-f(a,b)?

shouldent it be written as?:

f(x,y) - f(a,b) = \int^{(x,y)}_{(a,b)} F*dr

 

[Hurkyl]

They say that:

What exactly are they saying? This equation could be anything from a constraint on x & y to the definition of f.

Is the reason that they used the endpoints as the upper and lower limit because for a conservative vector field it does not matter which curve you use? So they decided to replace the C by simply the end points?

I have seen that notation used that way before.

 

[Cyrus]

They said, right before that equation, "Let A(a,b) be a fixed point in D. We construct the desired potential function f by defining" ... and then the equation.

But none the less, what ever happened to the evaluation of the lower limit f(a,b)?

 

[Hurkyl]

Why would there be an "evaluation at the lower limit"? This integral is a function of x and y, and their statement is defining f to be that function.


(Incidentally, what is f(a, b)?)

 

[Cyrus]

Let me restate the question Hurkyl. I thought about it and I wonder if this is the reason why. f is a potential function. But the potential of a function is always relative to some fixed point. f(a,b), the lower limit of the integral is that fixed point. This means that f(a,b) is where we define the potential energy to be zero, because when you evaluate the integral f(a,b) to f(a,b) is going to integrate into zero. Similarly, i think when they write f(x,y), they really mean that f(x,y)-f(a,b), but they don't write the f(a,b), because it is implied that this is the potential energery relative to the point at A(a,b). Yes, no?

 

[Hurkyl]

Are you trying to apply the fundamental theorem of calculus? If not, why should there be an f(a,b) on the LHS?

And yes, f(a, b) ... which is the integral of F*dr from (a,b) to (a,b) ... is indeed 0.

 

[Cyrus]

Yes, I was trying to apply the fundamental theorem of calculus for line integrals.

It says that:

\int_c \nabla f*dr = f(r(b))- f(r(a)).

But \nabla f = F

so this is equivalent to saying:

\int_c F*dr = f(r(b)) - f(r(a)).

but in our case we have r(b) = (x,y) so f(r(b)=f(x,y)
and r(a) = (a,b) so f(r(a))= f(a,b)

and then we have

f(x,y) - f(a,b)

but when they write:

f(x,y)= \int^{(x,y)}_{(a,b)} F*dr what happened to -f(a,b) on the left hand side?

 

[krab]

Hi Cyrus: You seem to have gotten confused about the meanings of the symbols you are using. What could you possibly mean by something like r(b)=(x,y)? You seem to be going from a description of C being from (a,b) to (x,y), to one of going from a to b. That's completely different; don't get mixed up!

Go back to your first formula
\int_c \nabla f*dr = f(r(b))- f(r(a))
First, the gradient operator generates a vector, not a scalar. So your asterisk here is inappropriate. It should be a dot product between two vectors. IOW, dr is a vector as well.

Second, What is C? It is some curve in space, parameterized by some variable s. IOW, x=x(s) and y=y(s) so that x and y vary simultaneously as s in order to keep you on the curve C. (Aside: you can do this in some cases by just finding a function y=y(x), but that is not general and in particular does not work in 3D.)

Third, you can choose the start of the curve as s=0 (you are completely free to do so). You seem to use (x,y)=(a,b) as your starting point. That's fine; it just means x(0)=a and y(0)=b. But now be careful as to your end point. If you simply use (x,y), you can get confused because x and y are in a sense your dummy integration variables, and you'll be using them in two completely different senses. So don't do this unless you really know what you are doing.

Let us decide that curve C starts at (a,b) and ends at (u,v). Then we have
\int_C \nabla f\cdot d\vec{r}=f(u,v)-f(a,b)
Remembering that d\vec{r}=(dx,dy), we can show this directly:
\int_C \nabla f\cdot d\vec{r}= \int_C{\partial f\over\partial x}dx+{\partial f\over\partial y}dy=\int_C df=f(u,v)-f(a,b)

Or,
\int_{(a,b)}^{(u,v)} \nabla f\cdot d\vec{r}= f(u,v)-f(a,b)

In your text, they've switched from (u,v) back to (x,y). Technically this is fine, but as I said if they are also using with variables x and y as integration variables, it is confusing.

So we end up with
\int_{(a,b)}^{(x,y)} \nabla f\cdot d\vec{r}= f(x,y)-f(a,b)
So to answer your question, they've dropped f(a,b) because they've set the potentail at (a,b) to be zero. In physics, one is able to do this because none of the real observables that the potential function is used for depend on the value of the potential; they only depend on the difference in the potential at 2 different places, or on the derivatives of the potential. So they've pulled a fast one on you, but it doesn't have any physical effect.

Choosing where f=0 is like choosing where s=0; you are free to make this choice as long as you stick to it once it is made. Come to think of it, they are zero at the same location: f(a,b)=f(x(0),y(0)) with our chosen parameterization variable s.

 

[Cyrus]

Thanks Krab. At first I was wondering why they just neglected the f(a,b) term, but when I looked at it a second time, I realized it was a potential energy function, so I thought they might have ment that the value of the potential at the point (a,b) will be defined as zero. As for the dot, i dident know how to put a dot, so i used a star, sorry. And for the r(b)-r(a) notation, I got sloppy. In the book they use the definition of the fundamental theorem for line integrals as such:

\int_C \nabla f \cdot dr = f(r(b)) - f(r(a)).
Whats why I reused the a and b, i dident mean a and b from (a,b) as the lower and upper limits.

 

[Cyrus]

I will just quote what my book has written down; however, I cannot put the picture so I am sorry on that note.

"Proof Let A(a,b) be a fixed point in D. We construct the desired potential function f by defining

f(x,y) = \int^{(x,y)}_{(a,b)} F \cdot dr

for any point (x,y) in D. Since \int_c F \cdot dr is independent of path, it does not matter which path C from (a,b) to (x,y) is used to evaluate f(x,y). Since D is open, there exists a disk contained in D with center (x,y). Choose any point (x_1,y) in the disk with x_1 < x and let C consist of any path C_1 from (a,b) to (x_1,y) followed by the horizontal line segement C_2 from (x_1,y) to (x,y). (see figure 4.) Then

f(x,y)= \int_{C_1} F \cdot dr + \int_{C_2} F \cdot dr = \int^{(x_1,y)}_{(a,b)} F \cdot dr + \int_{C_2} F \cdot dr

Notice that the first integral of these integrals does not depend on x, so

\frac {\partial}{\partial x}f(x,y) = 0 + \frac {\partial}{\partial x} \int_{C_2} F \cdot dr

If we write F =Pi +Qj , then

\int_{C_2} F \cdot dr = \int_{C_2} Pdx + Qdy

On C_2 y is constant, so dy=0. Using t as the parameter, where x_1 <= t >= x, we have

\frac {\partial}{\partial x}f(x,y)= \frac {\partial}{\partial x} \int_{C_2} Pdx+Qdy= \frac {\partial}{\partial x} \int^x_{x_1} P(t,y)dt = P(x,y)

by part 1 of the fundamental theorem of calculus (see section 5.4). A similar argument, using a verticle line segment (see figure 5), shows that

\frac {\partial}{\partial y}f(x,y)= \frac {\partial}{\partial y} \int_{C_2} Pdx+Qdy= \frac {\partial}{\partial y} \int^y_{y_1} Q(x,t)dt = Q(x,y)

Thus F= Pi +Qj = \frac {\partial f}{\partial x}i +\frac {\partial f}{\partial y}j "-end quote

Whew that's a lot of latex.
Now for the questions. I think Krab cleared up my first question about the f(x,y) = \int^{(x,y)}_{(a,b)} F \cdot dr
buisness.

Second Question, when they do the derivative with respect to x, and eliminate the integral of curve C_1. I've never seen that done before. I thought that when you take a derivative, it is with respect to the variables in the function, but here they are taking a derivative with respect to variables in the integrand. Also, they say the first curve does not depend on x, but in a way it does. You want the line segment from the end of the first curve to the end point of the second curve to be horizontal, so depending on where x is in the region D, determines where the curve C1 will terminate in order to keep the second line segment horizontal.

Third and final question, in the last step of this:
\frac {\partial}{\partial x}f(x,y)= \frac {\partial}{\partial x} \int_{C_2} Pdx+Qdy= \frac {\partial}{\partial x} \int^x_{x_1} P(t,y)dt = P(x,y)

they seem to be puling that same stunt that they did with f(a,b). What ever happened to the lower limit? Why isint it written as P(x,y)-P(x1,y). We origionally defined the potential to be zero at (a,b), so the value of P(x1,y) cannot be zero. What is justifiying them in doing this? Or is it becuase they are starting off with a new curve, (the horizontal one), that they are allowed to redefine where the potential is zero, in this case, the start of the second curve. (It seems that in solving this, they made two important assumptions they never bothered to mention, and that is where they defined the zero potential point in the evalutation of each line segment.

 

[krab]

Second Question, when they do the derivative with respect to x, and eliminate the integral of curve C_1. I've never seen that done before. I thought that when you take a derivative, it is with respect to the variables in the function, but here they are taking a derivative with respect to variables in the integrand. Also, they say the first curve does not depend on x, but in a way it does. You want the line segment from the end of the first curve to the end point of the second curve to be horizontal, so depending on where x is in the region D, determines where the curve C1 will terminate in order to keep the second line segment horizontal.

Third and final question, in the last step of this:
\frac {\delta}{\delta x}f(x,y)= \frac {\delta}{\delta x} \int_{C_2} Pdx+Qdy= \frac {\delta}{\delta x} \int^x_{x_1} P(t,y)dt = P(x,y)

they seem to be puling that same stunt that they did with f(a,b). What ever happened to the lower limit? Why isint it written as P(x,y)-P(x1,y). We origionally defined the potential to be zero at (a,b), so the value of P(x1,y) cannot be zero. What is justifiying them in doing this? Or is it becuase they are starting off with a new curve, (the horizontal one), that they are allowed to redefine where the potential is zero, in this case, the start of the second curve. (It seems that in solving this, they made two important assumptions they never bothered to mention, and that is where they defined the zero potential point in the evalutation of each line segment.

My first comment is that you are using delta's could you change them to \partial? Makes it easier to read.

Your second point. You are forgetting one of the things I tried to teach you in my first post: (x,y) is the terminal point of the integral: it is NOT the integration variable. So when C1 ends at (x1,y), this integral is entirely independent of x. They are not taking derivatives wrt the integration variables. Integration variables are dummy variables; they are not the x and y of the end point of C.

Third point: what they did was perfectly correct. It is because you don't quite understand integrals. Do for your self the following exercise: Find
\int_1^x u^2du
Now take the derivative of this result wrt x. Notice the lower limit, since it is constant, does not matter when you take the derivative.

 

[Cyrus]

Oh, I see now. So you have to integrate FIRST, then take the derivative of the result of your integral second. That makes things work out better, thank you. That part had me puzzeled for a while. What I kept trying to do was work from left to right, doing the derivative first. But your way makes more sense. Now that you see the rest of my first question I asked you in context, does that assumption of f(a,b) still hold true, or do you see a different reason for not including the f(a,b) now that you see the rest of the work that goes with it?


\frac{U^3}{3} |^x_1
= \frac{x^3}{3} - \frac{1}{3}
and taking the derivative with respect to x gives you:
x^2
But you called x a dummy variable. Is it still a dummy variable after you we-write the equation in terms of u to terms of x? What qualifies something as being a dummy variable?


Thanks,

Cyrus

 

[krab]

But you called x a dummy variable. Is it still a dummy variable after you we-write the equation in terms of u to terms of x? What qualifies something as being a dummy variable?


Thanks,

Cyrus

In the example I gave, the dummy variable happens to be u (the integration variable). x is not the dummy variable there; it is the integration limit. You seem to think x only ever has one meaning. I see that your mind is still not flexible enough to do math easily. Keep at it though!

Exercises:
What is the dummy variable in each case?
a. \int_1^x f(u)du
b. \int_1^u f(x)dx
c. \int_1^x f(x)dx
d. \int_u^v f(y)dy

Ans: a: u. b: x. c: this is not a properly formed integral! x cannot be both a limit and a dummy variable. d: y.

 

[Cyrus]

Ah, Gotcha. Thanks Krab. I have a hard time remembering things. I know it just fine when I take the course, but a month or two later, I have a tough time recalling all the information. For the most part, I can do standard definite integrals with not much trouble, its when it comes down to the proofs and theorems of the sort that things turn south.

 

[Cyrus]

Im reading up on Green's Theorem Now, Just one question:

The book says:

" \int \int_d \frac{\partial P}{\partial y} da = \int^a_b \int^{g_2(x)}_{g_1(x)} \frac { \partial P}{\partial y} (x,y) dy dx = \int^a_b [P(x,g_2(x))- P(x,g_1(x)] dx

Just to be certain I am not making any false assumptions. What this is saying is that you have the partial of a function P(x,y) that varies in with x and y. So it says, take the partial derivative of P(x,y) with resepct to y, then integrate this with respect to y, (which basically gives you P(x,y) again. And now evaluate it at the limits of the integral. The difference between this and the other case, \frac {\partial} {\partial x} \int_{C_2} F \cdot dr is that unlike above, this one says to integate first, and take the derviative second, even though the notion has the partial term written in order first.

Oh yeah, and one other thing. I would assume that Greens theorem is only good for R^2? The reason being that it only involves vector fields that have an x,y component. What are some applications for this theorem? My first guess, seeing a line integral around a closed loop would be faradays law. But with faradays law, don't we have to associate a "sheet" when we do the integral, (that trick maxwell played when the problem with flux for a capacitor arose and test charge buisness). I don't see any sheet in Greens theorem except maybe for the one on the planar region.

 

[Cyrus]

Actually, looking back on it i have to say no I am wrong to myself. Faradays law had to do with flux through a sheet of a closed loop, but greens theorem has to do with a line integral and the tangential part in the same direction. So I guess the only real application is an easier way to solve an otherwise hard line integral. It has nothing to do with flux vectors going through the closed path.

 

[Cyrus]

From my book:

" As an aid to our memory, let's rewrite Equation 1 using operator notation. We introduced the vector operator \nabla ("del") as

\nabla = i \frac{\partial}{\partial x}+j \frac{\partial}{\partial y}+k \frac{\partial}{\partial z}

It has meaning when it operates on a scalar function to produce the gradient of f:

\nabla f = i \frac{\partial f}{\partial x}+j \frac{\partial f}{\partial y}+k \frac{\partial f}{\partial z}= \frac{\partial f}{\partial x}i + \frac{\partial f}{\partial y}j+ \frac{\partial f}{\partial z}k "

How come they define the differential operator with the i,j,k on the left hand side of the partial derivative fraction? Then they do the gradient of f, and they seem to move all the i,j,k's from the left side to the right side? Also, the book also uses the differnetail operator with i,j,k on the right side. Whats going on with this inconsistancy? Is it just carelessness?

 

[krab]

How come they define the differential operator with the i,j,k on the left hand side of the partial derivative fraction?

That's to emphasize that the partial derivatives do not operate on the i,j,k.

Then they do the gradient of f, and they seem to move all the i,j,k's from the left side to the right side?

That's just to put them back in the usual order that people are used to.