How do I parameterize a triangle for a surface integral?

[Benny]

Hi, I'm having problems evaluating a surface integral.

$$\int\limits_{}^{} {\int\limits_S^{} {xdS} }$$

where S is the triangle with vertices (1,0,0), (0,2,0) and (0,1,1).

I need to parameterise the triangle but I don't know how to.

I tried (x,y,z) = (1,0,0) + u[(0,2,0)-(1,0,0)] + v[(0,1,1)-(1,0,0)] = (1,0,0) + u(-1,2,0) + v(-1,1,1).

But that is a paramterisation for a plane. I can't figure this one out can someone please help me out?

 

[0rthodontist]

Your parametrization is correct, you just need to set a domain for u and v. You can see that u >= 0 and v >= 0 gives you an infinite region that shares 2 edges with the triangle, and shares the vertex (1, 0, 0). And if you know that then you can probably also see that 0 <= u <= 1 and 0 <= v <= 1 gives you a parallelogram with 3 of its vertices at (0, 2, 0), (1, 0, 0), and (0, 1, 1). So basically you want to restrict u and v a little bit more than that to narrow it down to just the triangle. You have to split the parallelogram "diagonally" so can you guess what the restriction should be? (how do you split a square diagonally?)

Another way to look at it is to translate the vertices of your triangle so that one of them is at the origin, then use linear algebra to find a transformation that takes a simple-to-integrate triangle to your translated triangle. Then add a translation to that transformation so that the simple-to-integrate triangle maps to the original triangle.

 

[Benny]

I used 0 <= u <= 1 and 0 <= v <= 1-u and it worked out but I don't really know why. If 'double' the triangle were a square then I can see how this could work but as far as I can see, doubling the triangle in the question would result in a parallelogram - the kind which isn't a square.

Or should I avoid trying to visualise what's going on when I calculate surface integrals? After all, once I choose the parameterisation I only need to find the range of the u and v values.

Also, how would one proceed to use linear algebra to produce the appropriate translation of the triangle? Thanks for help.

 

[siddharth]

Benny, why do you want to parametrise the triangle?

Isn't the surface integral much easier if you project it on the x-y plane and then integrate?

 

[HallsofIvy]

Of course, "projecting" onto the xy-plane is a parameterization! It is using x and y as the parameters.

Benny, you certainly should be able to write the equation of the plane in the form Ax+ By+ Cz= D and then solve for z: z= (1/C)(D- Ax- By).
Your parameterization would be x= x, y= y, z= (1/C)(D- Ax- By).

 

[siddharth]

Of course, "projecting" onto the xy-plane is a parameterization! It is using x and y as the parameters.

Yeah, I should have seen that.

 

[Benny]

Now that you mention it, I could just use the point-normal form for a plane to obtain the cartesian equation of the related plane. But I'm not sure how that helps me me calculate the surface integral over the triangle.

 

[HallsofIvy]

The equation of the plane is 2x+ y+ z= 2.

At this point there are two (equivalent) ways to go.
Projecting the plane down into the xy-plane, where z= 0, gives the region bounded by 2x+ y= 2, x= 0, and y= 0. The normal to the plane is the vector 2i+ j+ k. Since that already has coefficient of k= 1, it is already "normalized" so that its length gives the differential of area relative to that of the xy-plane: $||2i+ j+ k||= \sqrt{6}$ so the differential of surface area is $\sqrt{6}dxdy$.
[tex]\int xdS= \int_{x=0}^1\int_{y=0}^{2-2x} x\sqrt{6}dydx[/itex]

The other way to write the "position vector", xi+ yj+ zk, in terms of x and y as parameter: since 2x+ y+ z= 2, z= 2- 2x- y and xi+ yj+ zk= xi+ yj+ (2-2x-y)k. Differentiate that with respect to x: i-2k and with respect to y: j- k. Finally, form the "fundamental vector product" by taking the cross product of those. It should be no surprise that that is 2i+ j+ k again. Once again, the length of that gives the differential of surface area relative to the xy-plane, $\sqrt{6}dydx$.

(The differential of area of one plane region relative to another plane is always just a constant.)

 

[0rthodontist]

I used 0 <= u <= 1 and 0 <= v <= 1-u and it worked out but I don't really know why. If 'double' the triangle were a square then I can see how this could work but as far as I can see, doubling the triangle in the question would result in a parallelogram - the kind which isn't a square.

Or should I avoid trying to visualise what's going on when I calculate surface integrals? After all, once I choose the parameterisation I only need to find the range of the u and v values.

Also, how would one proceed to use linear algebra to produce the appropriate translation of the triangle? Thanks for help.

(Edit: removed incorrect statement)
Your transformation is an Affine transformation (a linear transformation followed by a translation), which means points go to points, edges go to edges. You know that your domain is some subset of the unit square in the u-v plane. What are the 3 points of your domain corresponding to the vertices of your triangle? Where are 2 of the edges of your triangle mapped from?

So, where should the third edge of the triangle be mapped from, and what's the domain? (it is what you said)

What I was suggesting earlier with linear algebra was just another way to come to the same conclusion. What was unclear?

 

[Benny]

It wasn't that what you said was unclear. I just couldn't think of a way to do it using your alternative suggestion.

 

[0rthodontist]

It wasn't that what you said was unclear. I just couldn't think of a way to do it using your alternative suggestion.

I pretty much said what you could do--consider the triangle translated so one vertex is at the origin, and map a simple triangle in the u, v plane to the translated triangle using a linear transformation. Can you find the vertices of your triangle when it is translated so one vertex is at the origin? For the triangle in the u,v plane just use what you were already using--the triangle with vertices (0,0), (1, 0), and (0, 1). Can you find a linear transformation that takes that triangle to the triangle you want (the one translated to the origin)?

Then translate the result of that mapping back to the original triangle (you know how?), so you have a linear transformation followed by a translation.

Also, do you understand the other way of looking at it--"edges to edges"?

 

[Benny]

I don't think I've ever done mappings of shapes such as triangles to other shapes. The most I can recall doing is determining matrices for rotations by considering actions on bases. I was also never too big on the geometric interpretation of linear transformations so I guess that's probably why I haven't been able to apply your suggestion. But I don't think that this question requires it though.