0

Calculate the flux of the vector field $F(x, y, z) = (0, 0, z)$ up through the part of the plane $x + y + z = 1$ that lies in the f rst octant, by closing to the surface and using the divergence theorem.

Usually I'd have something to say but this time I don't even know where to begin. I have a plane right so what do I do now?

Ted Shifrin
  • 115,160
Need_MathHelp
  • 1,069

1 Answers1

2

Notice that the plane intersects the first quadrant at the points $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$. We create a closed volume with the desired surface on its boundary by considering the tetrahedron created by these vertices and the vertex at the origin, $(0,0,0)$. In set notation, this volume is given by $\Omega = \{(x,y,z)\in\mathbb{R}^3:x,y,z\geq 0\text{ and }x+y+z \leq 1\}$. The boundary of this region consists of four faces, one lying on each of the planes $x,y,z=0$ and the fourth being the the surface we are interested in. We label these surfaces $S_1$, $S_2$, $S_3$, and $S_4$, respectively.

The divergence theorm tells us that $$ \int_\Omega\nabla\cdot F dV = \int_{\partial\Omega} F\cdot \hat{n}dS. $$ We will compute each of these integrals, compare, then use them to evaluate the flux through $S_4$, which is what we want.

First notice that $\nabla\cdot F = 1$, so we have $$ \int_\Omega\nabla\cdot F dV = \int_\Omega dV = \int_0^1\int_0^{1-x}\int_0^{1-x-y} dzdydx =\frac{1}{6}. $$

We now compute the surface integral. We can decompose the boundary of $\Omega$ into integrals over the four surfaces defined earlier: $$ \int_{\partial\Omega} F\cdot \hat{n}dS = \int_{S_1} F\cdot \hat{n}dS+\int_{S_2} F\cdot \hat{n}dS+\int_{S_3} F\cdot \hat{n}dS+\int_{S_4} F\cdot \hat{n}dS. $$ Now, notice that $F\perp \hat{n}$ on both $S_1$ and $S_2$, so their dotproduct in the integrand is $0$. Additionally, $F=0$ on $S_3$ since $z=0$, so this integral is $0$ as well. We then have $$ \int_{\partial\Omega} F\cdot \hat{n}dS = \int_{S_4} F\cdot \hat{n}dS. $$

Combining this with the Divergence Theorem, we have

$$ \frac{1}{6} = \int_\Omega\nabla\cdot F dV = \int_{\partial\Omega} F\cdot \hat{n}dS = \int_{S_4} F\cdot \hat{n}dS, $$ so the flux through $S_4$ is equal to $1/6$.

whpowell96
  • 5,530
  • thanks for the help but what suggests that I'm dealing with a tetrahedron? When I intially was sketching the graph I got something that looks like a triangle. I'm asking because (0,0,0) is not on the plane. – Need_MathHelp May 07 '23 at 10:48
  • This was the same idea I suggested in my hint above. You need to create a closed surface with this triangle as one part of it. This is the most natural surface that comes to my mind. If I gave you a closed disk (circle and its interior), you might naturally think of a hemisphere or a paraboloid. – Ted Shifrin May 07 '23 at 15:04
  • @Need_MathHelp The most natural way to apply the divergence theorem in this context is to construct a closed volume with this triangular section of the plane as part of its boundary. This particular tetrahedron is one such shape and it probably results in the easiest computation – whpowell96 May 07 '23 at 16:46