# Line Integral: Int(yzdy) but

1. Feb 26, 2006

### Living_Dog

Hi all,

(This is part of a DJGriffiths, 3rd ed., problem: Prob. 1.28)

Line Integral: Int(yzdy) [lower limit = (1,1,0); upper limit = (1,1,1)] but y does not change and is supposed to be integrated, while z changes and is not integrated.

I have 2 questions:

[1] I take z out of the integral since the integral is only along the y direction,
z*Int(ydy) = zy^2/2 .

Now evaluate: ... but z is changing - is the line integral a fnc of z?

[2] Trying to evaluate the above "answer" I get z*y^2/2 (from y=1 to y=1) which means z*(1/2 - 1/2) = 0.

Edit: at first I thought that z (or y) can be expressed as a fnc of y (or z) but that is not the case for this line segment.

-LD

Last edited: Feb 26, 2006
2. Feb 26, 2006

### vaishakh

Try to express the change in z, or more appropriately z, with respect to y.

3. Feb 26, 2006

### HallsofIvy

Staff Emeritus
A line integral is taken along a specific path- it generally doesn't make sense to just give end points. Is this integral along the straight line between (1,1,0) and (1, 1, 1)??

Typically, in order to integrate a line integral on a particular path, you write parametric equations for the path, use that to put x, y, z, dx, dy, dz in terms of the parameter.

Assuming that you are integrating along the straight line between these points there is an obvious parametrization: x= 1, y= 1, z= t with 0<= t<= 1. If y= 1, what is dy?

4. Feb 26, 2006

### Living_Dog

The path is from the point x=1 y=1 z=0 to x=1 y=1 z=1, so y is fixed at 1. z changes from 0 to 1... I am sorry but I don't understand how there is a change in z wrt y when y is fixed at 1. (Quite seriously, it's not you, it's me.)

-LD

5. Feb 26, 2006

### Living_Dog

Yes it is. Sorry I missed that subtlety in my description.

The path is only the vertical line starting at ri = (1,1,0) and ending at rf = (1,1,1). It's so simple that I don't see how to write a parametric equation for x=1 y=1 and 0 <= z <= 1.

z = t?? This seems like I am only changing the name of the last variable from 'z' to 't'.

Since y is not changing along this line, then dy = 0.

I'm thick as a brick...

-LD

6. Feb 26, 2006

Int(yzdy) between (1,1,0) and (1, 1, 1)

You might want to take what I'm saying with a grain of salt, because it's been awhile since I've done calc (getting back into it right now). But I believe you can think of the integral as follows:

You have $\int yz\,dy$. This is of the form $\vec F \cdot \,d\vec r$. It may not seem like this, but $\vec F$ is actually $(0,yz,0)$ and since $d\vec r = (dx,dy,dz)$ then you get your integral above. The obvious paratmetrization is $\vec r(t) = ?$. How do you parametrize a line?

7. Feb 27, 2006

### Living_Dog

We are in the same boat when it comes to calculus.

From what I remember x, y, and z are all written in terms of some paramater, say 't'. E.g. x = t2 + 2, y = (3/2)t - 4, and z = 2t. Then it turns out that the line itself can be marked in units of this 't'. And so the equation of the line, which was in terms of x, y, and z, is now a line in 't'.

By asking me that I think I see the suggestion from vaishakh more clearly, namely: f(x,y,z) = yz. Now find a paramterization which will allow me to write both y and z in terms of this new parameter, 't'. Once I do that then I can convert yz into a function of 't' and then whatever the 't' parameterization is for y I imediately know dy and then I can do the integral over 't'.

Great! ... now what parameterization is there for y when it is constant from z=0 to z=1?? I think that's my problem, the integral is over y but y is not changing. Thus no parameterization is possible - y does not change along the line - so y != f(t).

thx for the help...
-LD

Last edited: Feb 27, 2006
8. Feb 27, 2006

### HallsofIvy

Staff Emeritus
Oh, I don't know, I've seen some pretty thick bricks!

But you have it now: $\int f(x,yz)dy= 0$ for any function f(x,y,z), any end values, if dy= 0!

9. Feb 27, 2006

### Living_Dog

oh,... thx man! :tongue2: I sometimes don't know I know the answer.

Also, thanks to FrogPad's poignant question I now understand parameterization MUCH better than ever before...

I'm starting to like this site more and more! woohoo! :rofl: (seriously - not the 'rofl' as the icon suggests. It's more like Scrooge in the original classic movie... you knowwhen he's as giddy as a drunken man... etc. ... sorry )

-LD