How do you intergrate a triangle ?

[Haths]

Basically. How would I even begin to go about doing the surface integral of a triangle bound by the co-ordinates;

(1,0,0)
(0,1,0)
(0,0,1)

This is part of a larger question to compute;

$$\int \int \vec{F} \cdot d \vec{r}$$

Around the surface using stokes theorum, the scalar multiplier comes out as 2x+2y+2z so I'm not worried about that, but this last section of the question has dumfounded me completely. I might guess that I am being asked to develop a new co-ordinate referance point, and set of functions to describe it, but that appears like a hell of a lot of work for very little output.

Unless I am meant to cheat at this point and say; area of triangle 1/2 ab sin (theta) plug in sqrt(3) for a and b, to get an area of sqrt(3)/2 then multiply that by the scalar value to get a total of 3sqrt(3).

Yet that sounds very wrong.

Cheers,
Haths

 

[belliott4488]

Basically. How would I even begin to go about doing the surface integral of a triangle bound by the co-ordinates;

(1,0,0)
(0,1,0)
(0,0,1)

You could just "scan" across the triangle, that is, integrate horizontal lines across the face of the triangle from 0 to 1 in z, where the length of each line of a function of z. Just add 'em up. You'll have to figure out how to express the length of each line as a function of z, but I'll bet you can do that!

 

[Haths]

Now that would work if I was only interested in the area, but I'm interested in the curl of the vector field. The path around the triangle A-B-C etc. is in the same direction or opposite direction to the 'rotation' of the field. The scalar quanta from before points in the positive r direction, which via the right hand rule means that the field is 'rotating' counter clockwise.

As you see the intergration needs to be done with some respect to this path somehow, and that's where the limits come in.

Haths

 

[gabbagabbahey]

What is your original vector function? Have you tried a change of coordinates, where one coordinate is parallel to the base of the triangle, the second is parallel to the height of the triangle , and the third is orthogonal to the first two?