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**Chemist****Member**- Registered: 2005-12-12
- Posts: 35

I just can't figure out what's the difference b/w the following two methods for solving line integrals.

In the book, they say for evaluating a line integral for a function f(x,y,z) = x - 3y[sup]2[/sup] + z over the line segment C joining the origin and the point (1,1,1) , we must 1st determine the parametric equations in terms of a parameter t.

r = t i + t j + t k 0 ≤ t ≤ 1 ( smooth parameter , 1st derivative is cont and never zero )

they give the line integral form : ∫ f(x,y,z)ds = ∫ f(t,t,t) |v(t)| dt 0 ≤ t ≤ 1

In class, however, this method is never used.

We start from ∫ F.dr ( both vectors ) where F = vector field = M i + N j + O k

∫ F.Tds ( T is the tangent unit vector, F.T----> dot product ) = ∫ F.dr = ∫ F.drdt /dt = ∫ ( M dx/dt + N dy/dt + O dz/dt ) dt

We've used this form to determine the line integral in all the problems we have encountered so far. If I try the 1st way, the answer is not always the same for the 2nd, actually, rarely is it the same. I tried to see the common point between the two:

∫ F.Tds = ∫ F.dr = ∫ F.dr dt / dr = ∫ F.V(t) dt , here obviously V is a velocity vector.

so, ∫ F.V dt = ∫ F |V(t)| cosθ dt ( θ b/w F and V )

In the book , they're using ∫ f(t,t,t) |V(t)| dt , so perhaps they assume that V is parallel to F ? Why would they do that, and what's the significance of either method?

Thanks in advance

"Fundamentally one will never be able to renounce abstraction."

Werner Heisenberg

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

The main point of below is that you must do different things when your function goes from R³⇒R and R³⇒ R³

For a ** Path** integral, you have the integral over a path c = (x(t), y(t), z(t)) of a function f(x, y, z), where f goes from R^3 -> R.

To integrate this, you integrate from a to b (the start and end points of your path) of f(c(t)) ||c'(t)|| dt, where ||c'(t)|| is the length of the derivative of c(t).

But a ** Line** integral is completely different (sort of...). Here, F goes from R^3 -> R^3, not to just R. Then you need to do:

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

If you tried to use both formulas on the same problem, then you either:

Tried to integrate a vector

or

Tried to take the dot product of a scalar.

Of course, neither makes sense.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**Chemist****Member**- Registered: 2005-12-12
- Posts: 35

Thanks a lot Ricky, I know that a line integral is used to compute the work or circulation depending on what the vector field is, but what's the path integral used for?

Tried to integrate a vector

or

Tried to take the dot product of a scalar.

Of course, neither makes sense.

Nah , I actually changed the scalar function to a vector field ... why can't we do that? Can't every scalar function have its vector components too?

"Fundamentally one will never be able to renounce abstraction."

Werner Heisenberg

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**Chemist****Member**- Registered: 2005-12-12
- Posts: 35

If we assume that the total differential form of a vector field is exact, the line integral is always dependent on the path, can we relate this to a conservative field? That is, a line integral exists only for conservative vector fields, which means if a vector field has a line integral over a certain path ( does it make a difference if it's differentiable or not here? ) , it must have a scalar potential function. Does this make any sense? I'm just wondering ...

"Fundamentally one will never be able to renounce abstraction."

Werner Heisenberg

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Nah , I actually changed the scalar function to a vector field ... why can't we do that? Can't every scalar function have its vector components too?

Well, take the example: f(x, y, z) = x + y + z

That's from R³⇒R. You could define it from R³⇒R³ by:

f(x, y, z) = (x + y + z, 0, 0)

But these are two completely different things with completely different properties. They look the same, but they aren't.

Is this what you are talking about?

Thanks a lot Ricky, I know that a line integral is used to compute the work or circulation depending on what the vector field is, but what's the path integral used for?

They are pretty much the same exact thing. That is, the both "sum up" the value of a function along a certain path. But one sums up vector values, such as a force, and one sums up constant values, such as heat or density.

If you have f(x, y, z) as the density of a wire at a point in 3d space, and c(t) be the path of that wire. Then the path integral will mass of that wire.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

The second integral you mentioned is mainly for computing work done by conservative force. As you can see, the integrated is a dot product of work on a small piece F.dr, where F and dr are both vectors.

The first integral can be used to integrate f(t,t,t) |v(t)|, f(t,t,t) is displacement from origin, and if v(t) stands for velocity, then|v(t)| stands for speed, or the magnitude of velocity. i cannot figure out what the product means.

So you'd better use the second defaultly.

So far, i haven't encountered the situation to use your first integral.

**X'(y-Xβ)=0**

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**Chemist****Member**- Registered: 2005-12-12
- Posts: 35

Well, take the example: f(x, y, z) = x + y + z

That's from R³⇒R. You could define it from R³⇒R³ by:

f(x, y, z) = (x + y + z, 0, 0)

I'm not sure what this is, what's the 0,0 for?

But these are two completely different things with completely different properties. They look the same, but they aren't.

Is this what you are talking about?

No , I was simply saying that if we have a scalar function f(x,y,z) = x +y +z , we can change it to a vector field or to its total differential form. This is what I did in calculating the path integral of a function f, I changed it to its vector form and calculated the scalar product.

They are pretty much the same exact thing. That is, the both "sum up" the value of a function along a certain path. But one sums up vector values, such as a force, and one sums up constant values, such as heat or density.

If you have f(x, y, z) as the density of a wire at a point in 3d space, and c(t) be the path of that wire. Then the path integral will mass of that wire.

Thanks for the explanation.

If we assume that the total differential form of a vector field is exact, the line integral is always dependent on the path, can we relate this to a conservative field? That is, a line integral exists only for conservative vector fields, which means if a vector field has a line integral over a certain path ( does it make a difference if it's differentiable or not here? ) , it must have a scalar potential function. Does this make any sense? I'm just wondering ...

I think I may be right here. I was just solving some problems, it turns out that whenever the total diff form of a vector field is exact ( rotor = 0 ), the line integral is independent of the path, and I can simply compute it F(B) - F(A) if we assume the line integral is from A to B across an arc. This really makes the computation easy for a line integral.

But if the total diff form is exact, continuous and its 1st derivative is continuous too, the line integral in a closed domain would be zero. Why's that?

And can anyone prove Green-Rieman's formula? I have an idea but I think it's incorrect.

"Fundamentally one will never be able to renounce abstraction."

Werner Heisenberg

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**Chemist****Member**- Registered: 2005-12-12
- Posts: 35

George,Y wrote:

The second integral you mentioned is mainly for computing work done by conservative force. As you can see, the integrated is a dot product of work on a small piece F.dr, where F and dr are both vectors.

The first integral can be used to integrate f(t,t,t) |v(t)|, f(t,t,t) is displacement from origin, and if v(t) stands for velocity, then|v(t)| stands for speed, or the magnitude of velocity. i cannot figure out what the product means.

So you'd better use the second defaultly.

So far, i haven't encountered the situation to use your first integral.

Thanks for the info , I think I got the difference between the two. And yes |V(t)| is the velocity magnitude or speed.

"Fundamentally one will never be able to renounce abstraction."

Werner Heisenberg

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**Chemist****Member**- Registered: 2005-12-12
- Posts: 35

We can obtain Green's theorem area formula by going backward :

A = ∫ ∫ dydx = ∫ ∫ ( 1/2 + 1/2 ) dydx = 1/2 ∫ ( xdy - ydx )

, ofcourse the domain is closed.

But it seems sometimes we can compute the area directly from A = ∫ xdy

Example:

Determine the area b/w y=x[sup]2[/sup] and y = x across C[sup]+[/sup]

A = 1/2 ( ∫ xdy - ydx )

I divided the line integral into arc OA and a straight line OA where A (1,1), the point of intersection of the parabola and the straight line, and O is the origin ofcourse.

I chose the parametirc equations,

For [OA] : y = x , x=x 0 ≤ x ≤ 1

For arc OA : x=y[sup]2[/sup] , y=y 1 ≤ y ≤ 0

I[sub]1[/sub] = 1/2 ∫ (x-x)dx = 0 ----> for [OA]

I[sub]2[/sub] = 1/2 ∫ (y[sup]2[/sup] - 2y[sup]2[/sup])dy = 1/6 ---> for arc OA

A = I[sub]1[/sub] + I[sub]2[/sub] = 1/6

Now notice that the area could have been calculated directly from:

A = ∫ xdy ( line int across C[sup]+[/sup] ) = ∫ xdy ( across [OA] ) + ∫ xdy ( across arc OA )

Using the same variables, A = 1/6

The second way is quicker, but I can't seem to figure out when to use it.

"Fundamentally one will never be able to renounce abstraction."

Werner Heisenberg

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

to compute the area enclosed by an ellipse.

**X'(y-Xβ)=0**

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**Chemist****Member**- Registered: 2005-12-12
- Posts: 35

You mean the area A = 1/2 ∫ ( xdy - ydx ) can be simplified to A = ∫ xdy in the case of an ellipse? It was correct for the closed region between x=y[sup]2[/sup] and y=x. There was no ellipse there ...

"Fundamentally one will never be able to renounce abstraction."

Werner Heisenberg

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