Surface Integral Over Tetrahedron

[Lancelot59]

I have to integrate this function:
$$f(x,y,z)=y+x$$

Over the region S which is a tetrahedron defined by points (0,0,0), (2,0,0), (0,2,0), (0,0,2).

So after I drew it out I saw that three of the faces were right up against the XZ, YZ, and XY planes. I'm getting stuck on parameterizing the faces.

I tried to do the face against the XZ plane. Using u and v I drew the plane and got a line going from (0,2) to (2,0) in the new U/V plane.

I got these as my functions for the line:
$$v=-u+2$$
$$u=-v+2$$

From there I parameterized it like so:
$$r(u,v)=(u,0,v)$$
$$r(u,v)=(u,0,-u+2)$$

with my normal being:
$$\vec{N}=T_{u} X T_{v} (0,-1,0)$$
$$\vec{N}=(1,0,-1) X (0,0,0)$$
$$||\vec{N}||=0$$

So then combing f and r:
$$f(r(u,v))=0+u=u$$

To get:
$$\int\int f(r(u,v)) * ||N(u,v)|| dA$$
$$\int\int u * 0 *dA$$

So I'm confused here. Does that mean that there is no surface integral on that face? Once I get the integrals over the rest of the faces, do I just add them up?

 

[LCKurtz]

I have to integrate this function:
$$f(x,y,z)=y+x$$

Over the region S which is a tetrahedron defined by points (0,0,0), (2,0,0), (0,2,0), (0,0,2).

So after I drew it out I saw that three of the faces were right up against the XZ, YZ, and XY planes. I'm getting stuck on parameterizing the faces.

I tried to do the face against the XZ plane. Using u and v I drew the plane and got a line going from (0,2) to (2,0) in the new U/V plane.

I got these as my functions for the line:
$$v=-u+2$$
$$u=-v+2$$

From there I parameterized it like so:
$$r(u,v)=(u,0,v)$$
$$r(u,v)=(u,0,-u+2)$$

You don't want to parameterize the line, you want to parametrize the triangular surface. Your original r(u,v) = (u,0,v) is correct because you need two variables. But you need to specify v goes from 0 to 2 -u or u goes from 0 to 2-v in the inner integral limits.

with my normal being:
$$\vec{N}=T_{u} X T_{v} (0,-1,0)$$

That would be r_u x r_v. (Why don't you just use x and z as the natural parameters? And for that matter observe the outward normal is -k.)

$$\vec{N}=(1,0,-1) X (0,0,0)$$
$$||\vec{N}||=0$$

Not sure what that is, but it's wrong.

So then combing f and r:
$$f(r(u,v))=0+u=u$$

To get:
$$\int\int f(r(u,v)) * ||N(u,v)|| dA$$
$$\int\int u * 0 *dA$$

So I'm confused here. Does that mean that there is no surface integral on that face? Once I get the integrals over the rest of the faces, do I just add them up?

Your dA should be ||(0,-1,0)|| du dv = du dv so your integral should be

$$\int_0^2\int_0^{2-v} u*1\, du dv$$

Onward! Only 3 more faces left.

 

[Lancelot59]

You don't want to parameterize the line, you want to parametrize the triangular surface. Your original r(u,v) = (u,0,v) is correct because you need two variables. But you need to specify v goes from 0 to 2 -u or u goes from 0 to 2-v in the inner integral limits.

Okay, that makes sense...

That would be r_u x r_v. (Why don't you just use x and z as the natural parameters? And for that matter observe the outward normal is -k.)

I like u and v...can't really say why.

How is that the normal? The face is up against the XZ plane.

Not sure what that is, but it's wrong.

That was supposed to be the normal...

Your dA should be ||(0,-1,0)|| du dv = du dv so your integral should be

$$\int_0^2\int_0^{2-v} u*1\, du dv$$

Onward! Only 3 more faces left.

Could be worse I guess.

 

[LCKurtz]

That would be r_u x r_v. (Why don't you just use x and z as the natural parameters? And for that matter observe the outward normal is -k.)

How is that the normal? The face is up against the XZ plane.

That was supposed to be the normal...

That's a woops! Obviously I meant -j, which is what I actually used as ||(0,-1,0)||dudv. And the = dudv in that equation is also a copy & paste typo.