Find the mass of the surface of a triangel in 3d space

[MeMoses]

Homework Statement

Find the mass of the surface of the triangle with vertice (2,0,0), (0,2,0), (0,0,1) if the density is 4xz

Homework Equations

The Attempt at a Solution

I know I just need to take the integral of 4xy in the region of the triangle, but how do i set up the integral for the triangle? I am not used to triangles in 3d space, do I need to make a transformation or is there some easier way. I'm not really sure where to start on this one, but once i get the integral set up I supposed its easy.

 

[LCKurtz]

Write the equation of the plane containing those points. Then use that to figure out your dS element of surface area.

 

[MeMoses]

ok I got the plane: x+y+2z=2 or z = 1 - x/2 - y/2 and d/dx and d/dy both equal -1/2
So I should take the triple integral of sqrt(1+2(1/2)**2) with the limits
0<z<1
0<x<2-2z
0<y<2-2z
but this does not give me the value for area I got by double checking using the magnitude of the cross product of 2 of the lines, divided by 2, which was sqrt(6). I'm assuming I messeed up the limits but I can't quite see how at the moment.

 

[LCKurtz]

ok I got the plane: x+y+2z=2 or z = 1 - x/2 - y/2 and d/dx and d/dy both equal -1/2
So I should take the triple integral of sqrt(1+2(1/2)**2) with the limits
0<z<1
0<x<2-2z
0<y<2-2z
but this does not give me the value for area I got by double checking using the magnitude of the cross product of 2 of the lines, divided by 2, which was sqrt(6). I'm assuming I messeed up the limits but I can't quite see how at the moment.

You don't evaluate a surface integral as a triple integral. It is always a double integral. Since you have decided to solve for z as a function of x and y, you must express everything, including the area density in terms of x and y and integrate over the appropriate region in the xy plane.

 

[HallsofIvy]

I would set up a new coordinate system in the plane determined by those three points.

For example, a vector from the point (2, 0, 0) to (0, 2, 0) is <2, -2, 0> so the line from (2, 0, 0) to (0, 2, 0) can be written as x= 2- 2u, y= 2u, z= 0 and we can take u to be one of the parameters for this new coordianate system. We also get that 2x- 2y= C for any plane perpendicular to that line and, in particular, 2x- 2y= 0 is a plane perpendicular to that line containing the point (0, 0, 1). The line x= 2- 2u, y= 2u, z= 0 intersects that plane where 2(2- 2u)- 2(2u)= 4- 8u= 0 or u= 1/2. That is the point (1, 1, 0) and so the line through (1, 1, 0) and (0, 0, 1) is a line in that plane perpendicular to the line. A vector from (1, 1, 0) to (0, 0, 1) is <1, 1, -1> and so the line is given by x= v+ 1, y= v+ 1, z= -v and we can take v as the other parameter.

That is, the line from (2, 0, 0) to (0, 2, 0) is the u-axis, v= 0, with u going from 0 to 1. The point (2, 0, 0) is given in uv coordines as (0, 0, (0, 2, 0) by (1, 0), and (0, 0, 1) by (1/2, 1). Integrate the density function (changed to u,v coordinates, of course) over the triangle with those coordinates.