Integrating surface areas with dS and dA

[KnowledgeisPo]

Homework Statement

ζζs yz ds
S is the part of the plane x+y+z=1 that lies in the first Octant

Homework Equations

ζζs f(x,y,z) dS = ζζd f(x,y,g(x,y)) sqrt(1+(df/dx)²+(df/dy)²) dA

The Attempt at a Solution

turned dS into dA, and attempted many different regions. Always got an area that was 4-6 times too big.

Using double integrals, I'm limited in my abilities to evaluate all three variables.

edit: the sqrt comes to root(3). The relevant equations is one given in examples, but the actual examples don't ever seem to deal with a g(x,y) for z substitution. One example uses it, but instead of having a yz to integrate, it just has a y.

I integrate over the region 0<x<1 and 0<y<1, as proportionately instructed in examples, but that leaves the z variable unintegrated. I tried numerous equation substitutions also, and as I said they come out 4-6 times too big as for as close as I can get.

 

[Quinzio]

I integrate over the region 0<x<1 and 0<y<1

Do you mean you integrate on a square ?

as proportionately instructed in examples

The region over which to integrate is determined in this exercise, I don't care what other examples might say

 

[KnowledgeisPo]

Do you mean you integrate on a square ?

The region on the xz would be x=1-z, the region on the xy would be y=1-x, over the yz would be z=1-y. Since the surface area with equation, x+y+z=1, is a triangle flush against the xz and yz planes, I set my dA to be the region on the XY plane, in which x and y go from 0 to 1. Every time I set them to go to equations, I end up with a variable left over in the integrated area.

The region over which to integrate is determined in this exercise, I don't care what other examples might say

This exercise doesn't tell me the region to integrate. Proportionately was the 'attempted' region for this exercise that the examples instruct upon.

 

[KnowledgeisPo]

ζζs yz ds
S is the part of the plane x+y+z=1 that lies in the first Octant

Homework Equations

ζζs f(x,y,z) dS = ζζd f(x,y,g(x,y)) sqrt(1+(df/dx)²+(df/dy)²) dA

solution
sqrt(1+(df/dx)²+(df/dy)²)-- df/dx=-1, df/dy=-1 therefor sqrt(1+ (-1)² + (-1)²)= sqrt(3)
ζζs f(x,y,z) dS=ζζs yz dS=ζζd f(x,y,g(x,y)) sqrt(1+(df/dx)²+(df/dy)²) dA=(ζ0>1) (ζ0>1-y)yz sqrt(3)dz dy. Apparently, yz to be integrated is corresponding directly to the surface area to be integrated upon, thus prompting us to substitute yz for y(1-x-y).
Through proper integrations, the first integral will consist of a region aside from any Z parameters, leaving us to integrate a function of two variables instead of 3 (twice).

 

[HallsofIvy]

x+ y+ z= 1 meets the coordinate axes at (1, 0, 0), (0, 1, 0) and (0, 0, 1). The region on that plane in the first octant is the triangle with those vertices. Projecting down to the xy plane, gives a triangle with vertices (1, 0), (0, 1), and (0, 0). Your want to integrate your "dA" over that region. Since taking z= 0 changes the equation to x+ y= 1, your region of integration is the triangle with sides y= 1- x and the x and y axes. One way to do that would be to integrate with x going from 0 to 1 and, for each x, y going from 0 to 1- x.

As for getting the integrand, I would do it this way: the plane is x+ y+ z= 1 or z= 1- x- y. Using x and y as parameters, we can write the "position vector" of any point in that plane as r(x,y)= xi+ yj+ zk= xi+ yj+ (1- x- y)k.

The tangent vector are r_x= i- k and r_y= j- k. The cross product of those two vectors, i+ j+ k, is perpendicular to plane and the "differential of surface area" is given by its length, sqrt(3), times dxdy.

You want to integerate
int_{x=0}^1 int_{y=0}^{1- x} (yz) (sqrt{3}dydx)

Of course, z= 1- x- y so that is
sqrt(3) int_{x=0}^1 int_{y= 0}^{1- x} y(1- x- y) dydx

= sqrt(3) int_{x=0}^1 int_{y=0}^{1- x} y- xy- y^2 dy dx

 

[KnowledgeisPo]

x+ y+ z= 1 meets the coordinate axes at (1, 0, 0), (0, 1, 0) and (0, 0, 1). The region on that plane in the first octant is the triangle with those vertices. Projecting down to the xy plane, gives a triangle with vertices (1, 0), (0, 1), and (0, 0). Your want to integrate your "dA" over that region. Since taking z= 0 changes the equation to x+ y= 1, your region of integration is the triangle with sides y= 1- x and the x and y axes. One way to do that would be to integrate with x going from 0 to 1 and, for each x, y going from 0 to 1- x.

As for getting the integrand, I would do it this way: the plane is x+ y+ z= 1 or z= 1- x- y. Using x and y as parameters, we can write the "position vector" of any point in that plane as r(x,y)= xi+ yj+ zk= xi+ yj+ (1- x- y)k.

The tangent vector are r_x= i- k and r_y= j- k. The cross product of those two vectors, i+ j+ k, is perpendicular to plane and the "differential of surface area" is given by its length, sqrt(3), times dxdy.

You want to integerate
int_{x=0}^1 int_{y=0}^{1- x} (yz) (sqrt{3}dydx)

Of course, z= 1- x- y so that is
sqrt(3) int_{x=0}^1 int_{y= 0}^{1- x} y(1- x- y) dydx

= sqrt(3) int_{x=0}^1 int_{y=0}^{1- x} y- xy- y^2 dy dx

No joke. That's pretty much the way I explained it in the post above... Much appreciated if you were doing it while I was explaining the solution I got. I don't think I mentioned the cross product of the r_x and r_y vectors, but yeah, that's how you get the segment to integrate with the function to actually be integrated... Tricky stuff.

 

[HallsofIvy]

If you are referring to your post #4, I see nothing there about the triangle in the xy-plane with vertices (0, 0), (0, 1), and (1, 0) nor anything about the y integral having limits 0 and 1- x. And those are the crucial points.

 

[KnowledgeisPo]

That could be because you don't actually need to do it in the xy plane. And my integral was indeed defined from 0>z>1-y, and 0>y>1. If I was substituting z in the integral with z=1-x-y, then yes, I would want to integrate over the xy, and yes, if you wanted to point out a conceptual typo, you could explain the concept behind ζ0>1-y, and use that to say that the said typo of dz instead of dx meant I did not instruct on region integration at all.