Line integral problems

pinsky · 2008-01-03 14:36:34

Hy!

So here is my problem.

I have to calculate a line integral from a curve C

integral (x+y-xyz)ds

where C is a border of a triangle whit peaks A(3,0,1) B(0,2,1) C(1,0,2)

My idea is to integrate separately for each line segment AB,BC,CA, but what i don't know is how to write those lines as functions and how to define borders of integration.

Off topic question, is there a free program which has mathematical simbols. I've seen people post formulas as jpg-s on this forum and they look fancy with integral sibols etc...

Tnx

Ricky · 2008-01-03 17:12:23

This is latex, and you can download it for your computer using MikTex on Windows, and a variety of programs on linux.

If you are doing line integrals, you should have already had a course in vector geometry which would go over line parametrization. However, here is a very quick overview.

Given two points (a, b, c) and (x, y, z), we wish to construct a line based upon t. So

C(0) = (a, b, c)
C(1) = (x, y, z)

We do this by splitting it up into three equations Cx(t), Cy(t), and Cz(t). We note that Cx(0) = a and Cx(1) = x. Since these are lines, we know the equation of any line is:

Cx(t) = mt + n

So since Cx(0) = a, we know that n = a. Since Cx(1) = x and Cx(1) = m + n = m + a, we get that m = (x - a). So in full, the equation is:

Cx(t) = (x - a)t + a

If you wish to check that this is the correct line, simply plug in Cx(0) and Cx(1) and make sure you get the right values.

This must be done for all pieces of the path (x, y, and z in this case, but it could be as many as n dimensions in general). Once this is done, you may use the fundamental theorem of line integrals:

Where C(t) is your path with the three parts, Cx(t), Cy(t), Cz(t). The notation may be a bit confusing at first, but you plug in Cx(t) into the first parameter of f (which would be your x variable), Cy(t) into the second parameter (replace y with Cy(t)) and so on.

Ricky · 2008-01-03 18:05:16

A good exercise to see how this is simply a generalization of every thing you learned in single variable calculus, take a simple example with:

f(x, y) = x
The path from (a, 0) to (b, 0)

You get that C(t) breaks down to:

Cx(t) = (b - a)t + a
Cy(t) = 0

So that:

On the other hand:

pinsky · 2008-01-04 04:26:56

Given two points (a, b, c) and (x, y, z), we wish to construct a line based upon t. So
C(0) = (a, b, c)
C(1) = (x, y, z)

Correct me if i'm wrong. You wrote points as solutions of the line function C(0) = (a,b,c). In my example i have three points and all i know is that i have to integrate over a triangle which has these points as peaks.

So since Cx(0) = a, we know that n = a. Since Cx(1) = x and Cx(1) = m + n = m + a, we get that m = (x - a). So in full, the equation is:
Cx(t) = (x - a)t + a

How do i get that if i don't know the value of the argument in C(x), why did u use 0 and 1?

Why did you define your limits of integration from 0 to 1?

The professor gave us a different formula for the line integral

Is that the same formula u are using?

Here is the whole task, so i you have the time perhaps you could solve just the part with the parametrization.

Calculate

where k is the border of a triangle with peaks A(3,0,1) B(0,2,1) C(1,0,2).

Thx for the latex hint. Cool thing.

pinsky · 2008-01-04 07:57:06

Figured out why

Ricky · 2008-01-04 17:03:14

Correct me if i'm wrong. You wrote points as solutions of the line function C(0) = (a,b,c). In my example i have three points and all i know is that i have to integrate over a triangle which has these points as peaks.

I thought you already understood this part when you said:

My idea is to integrate separately for each line segment AB,BC,CA, but what i don't know is how to write those lines as functions and how to define borders of integration.

But just to be clear, yes, break the problem into three separate line integrals. You start from one point on the triangle and integrate to the next point. Then start at this next point and go to the third. Finally, start at the third and go back to the first.

How do i get that if i don't know the value of the argument in C(x), why did u use 0 and 1?

Just by convention. If you can come up with a path with t going from your t_0 to t_1, then you can come up with an equivalent path that goes from 0 to 1. Also, integrating from 0 to 1 tends to be easier than anything else.

Is that the same formula u are using?

Yes.

Here is one line done for you. The side of the triangle going from the vertex A(3,0,1) to B(0,2,1).

x(t) = (0 - 3)t + 3 = -3t + 3
y(t) = (2 - 0)t + 0 = 2t
z(t) = (1 - 1)t + 1 = 1

Now we integrate:

pinsky · 2008-01-08 10:00:39

Why didn't you derivate first when getting dS? [x(t)']^2 [y(t)]^2 [z(t)']^2

Ricky · 2008-01-08 11:10:09

Cause sometimes I forget little things like that sometimes.

Good catch.

Edit: And that's supposed to be a "3t+3" not "3t+t"

Math Is Fun Forum

#1 2008-01-03 14:36:34

Line integral problems

#2 2008-01-03 17:12:23

Re: Line integral problems

#3 2008-01-03 18:05:16

Re: Line integral problems

#4 2008-01-04 04:26:56

Re: Line integral problems

#5 2008-01-04 07:57:06

Re: Line integral problems

#6 2008-01-04 17:03:14

Re: Line integral problems

#7 2008-01-08 10:00:39

Re: Line integral problems

#8 2008-01-08 11:10:09

Re: Line integral problems

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