Surface Integral: Compute g = xy over Triangle x+y+z=1, x,y,z>=0

[mit_hacker]

Homework Statement

(Q) Compute the surface integral g = xy over the triangle x+y+z=1,
x,y,z>=0.

Homework Equations

The Attempt at a Solution

The triangular region basically means that the region in consideration is a
plane and not a sphere, cylinder etc...

Therefore, we can let the region r be defined as r = xi + yj + (1-x-y)k.

thus, partial derivative of r w.r.t x = i-k and that w.r.t y = j-k.

their cross product comes to -i+j+k. the modulus of this is of course
sqrt(3).

So the double integral will be the double integral over region R of
sqrt(3)xydA.

the problem is, I have no idea as to how I can find the limits of this
double integral.

Please advice.

Thank-you very much for your time and effort!

 

[HallsofIvy]

Homework Statement

(Q) Compute the surface integral g = xy over the triangle x+y+z=1,
x,y,z>=0.

Homework Equations

The Attempt at a Solution

The triangular region basically means that the region in consideration is a
plane and not a sphere, cylinder etc...

Therefore, we can let the region r be defined as r = xi + yj + (1-x-y)k.

thus, partial derivative of r w.r.t x = i-k and that w.r.t y = j-k.

their cross product comes to -i+j+k. the modulus of this is of course
sqrt(3).

So the double integral will be the double integral over region R of
sqrt(3)xydA.

the problem is, I have no idea as to how I can find the limits of this
double integral.

Please advice.

Thank-you very much for your time and effort!

Draw a picture! You are told that x, y,z are all greater than or equal to 0 so you are restricted to the first octant. When x and y are both 0, what is z? When x and z are both 0, what is y? When y and z are both 0, what is x? Those three points are the vertices of the triangle cut off when the plane crosses the first octant. Since you have chosen to integrate over x and y, project that triangle down onto the xy-plane- just drop the z-coordinate of the vertices- and integrate over that region in the xy plane.

 

[mit_hacker]

Thanks!

I get it. Thanks a lot!