# Flux integral

## 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?

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tiny-tim
Homework Helper
Hi mathman44!

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

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

Doh!

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

tiny-tim
Homework Helper
Doh!

Of side √2. Meaning the area is 1 …
sin60º ≠ 1

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

tiny-tim
Homework Helper
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?

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