Question

Let S be the part of the plane 2x+2y+z=2 which lies in the first octant, oriented upward. Find the flux of the vector field F=4i+1j+2k across the surface S.

Answer 1

First of all, a normal vector to your plane is (4,2,1). By oriented "upward" I assume you mean that this is the positively oriented normal. Now, you needto make it a unit normal in order to calculate the flux:

n = (4,2,1)/||(4,2,1)|| = (4,2,1)/sqrt(21)

Now the flux is what you get when integrate F.n over your plane. Since both F and n are constant, you might as well just calculate F.n = (4+8+3)/sqrt(21) =15/sqrt(21). Thus the flux will be the area of the surface times 15/sqrt(21). So what is the area? Well the surface will the intersection of the plane withthe first octant, which will be a triangle, so in order to calculate its area, you can find it's vertices. To do this, set z, x, and y equal zero two at atime in the equation for the plane:

z = 2, so one vertex is (0,0,2)
2y = 2, so another vertex is (0,1,0)
4x = 2, so the last vertex is (1/2,0,0)

The area of this triangle is thus:

1/2*|(0,-1,2) x (1/2,-1,0)| = 1/2*|(2,1,1/2)| = sqrt(21)/4.

Therefore the flux is sqrt(21)/4*15/sqrt(21) = 15/4.