# Line integral

1. Mar 3, 2010

### nothGing

what is the different between line integral and surface integral?
If we parameterize curve by x=t , y=t , what is the range of t ? Is it 0<= t <=1? why?

2. Mar 3, 2010

### HallsofIvy

A "line integral" is along a one-dimensional line and a "surface integral" is over a two dimensional surface.

There are no bounds. if you are talking about a portion of a curve, say from $(x_0, y_0)$ to $(x_1, y_1)$, the range of t is from whatever value of t gives $x_0$ and $y_0$ to whatever value of t gives $x_1$ and $y_1$.

3. Mar 3, 2010

### nothGing

Erm.. ok..
Let say, integral ( x^2 + y + z)ds , the line segment of curve is from (0,0,0) to (1,2,1).
We parameterize curve C by x=t , y=2t , z=t.
As you said, t should be between 0 and 2.
but according to my reference book, 0<=t<=1.
Why?

4. Mar 5, 2010

### elibj123

HallsofIvy said that the range of t is such that your parametrization will agree with your end points. (Of course, you first need to check that your parametrization does describe a line, in this case it does).

With your paramterization, its 0<=t<=1. But I could choose:

x=t/2, y=t, z=t/2

Which still describe a line with the same orientation, instead now t must range from 0 to 2 to give the desired segment.

5. Mar 5, 2010

### Redbelly98

Staff Emeritus
When t=1, x,y,z is the final endpoint. Try plugging t=1 into the forumulas for x,y,z to convince yourself of this.

6. Mar 5, 2010

### HallsofIvy

No, I did NOT say that! I said " if you are talking about a portion of a curve, say from (x0, y0) to (x0, y0) , the range of t is from whatever value of t gives x0 and y0 to whatever value of t gives x1 and y1" . Since this is in three dimensions, we need to include z0 and z1 also.

We must have x(t)= t= 0, y(t)= 2t= 0, and z(t)= t= 0. Obviously t= 0 satisfies all of those.

We must also have x(t)= t= 1, y(t)= 2t= 2, and z(t)= t= 1. Obviously t= 1, NOT t= 2, satisfies all of those.

7. Apr 15, 2010

### nothGing

HallsofIvy, ya.. i missunderstand already.. :p
Well, now i understand it.. thanks to elibj123,Redbelly98, HallsofIvy for helping..