Evaluating Flux Integrals: Plane x+y+z=1

[mathman44]

Homework Statement

Directly compute the flux integral \int\int(G*n dS) where D is the part of the plane x+y+z=1 in the first octant, oriented upwards.

G = -y-z-x

My attempt at solution:

normal vector, normalized, is 1/sqrt(3) * (1,1,1) and since z=1-y-x, the integral simplifies to

1/sqrt(3)*\int\int(-1 dS)

How do I evaluate this, without parametrization of the plane?

 

[tiny-tim]

Hi mathman44!

(have a square-root: √ and an integral: ∫ )

∫∫ dS is the area, so you need the area of an equilateral triangle of side … ?

 

[mathman44]

Doh!

Of side √2. Meaning the area is 1, and the flux is -1/√3... is that okay?

 

[tiny-tim]

Doh!

Of side √2. Meaning the area is 1 …

sin60º ≠ 1

 

[mathman44]

Ha, oops... so area is √3/2 and the flux is then -1/2.

 

[tiny-tim]

Ha, oops... so area is √3/2

Yup!

and the flux is then -1/2.

ah … now I didn't actually understand the original question …

G is presumably a vector, but what are its coordinates?

 

[mathman44]

G is -y(i) - z(j) - x(k)

so G dotted with n

= 1/√(3)*(1,1,1) * G = 1/√(3) * (-y-z-x)

But because x+y+z=1, z=1-x-y and so -y-z-x = -1, becoming:
1/√(3) * (-y-z-x) = 1/√(3) * (-1)

So then flux should be -1/√3 * dS, which is -1/√3 * √3/2 = -1/2

 

[tiny-tim]

G is -y(i) - z(j) - x(k)

ah! all is now clear!

Yes, -1/2 looks good.