Consider the vector field $F=<0,0,x+z>$,and the surface S which is the part of the plane $x+y+2z=4$ inside the first octant. Assume the unit normal vector n to S has positive third component.Use the Divergence Theorem to compute the flux of F across S by finding instead the flux of F across a different surface and an appropriate triple integral.

Here is my work: $$div(\mathbf F) = \frac{\partial 0}{\partial x} + \frac{\partial 0}{\partial y}+\frac{\partial}{\partial z}(x+z) = 1 $$ By divergence theorem, $$\int\int\int_Ddiv\mathbf FdV=\int\int_S(F\cdot n)d\sigma$$ $$\int\int\int_D1dV = the\;volume\;under\;"x+y+2z=4"$$ so I got $\frac {16}3$, but the correct answer is 16. Does anyone know where I did it wrong?


1 Answer 1


That is because applying Divergence Theorem will give you flux through the entire closed region which is a tetrahedron with $4$ surfaces enclosed by $x + y + 2z = 4$ in the first octant and the coordinate planes, including the XY plane between $(0, 0, 0), (4, 0, 0), (0, 4, 0)$ and similarly part of $YZ$ and $XZ$ planes.

But I think your question asks you to find flux only through the surface $x + y + 2z = 4$. So you need to subtract the flux through the other $3$ surfaces.

Given our vector field is $(0, 0, x + z)$, the flux through $XZ$ plane and $YZ$ planes will be zero as the unit normal vectors to these planes will have no $z$ component and dot product $\vec{F} \cdot \hat{n}$ will be zero. But that is not the case for $XY$ plane.

The outward pointing unit normal vector through the surface which is $XY$ plane is $(0, 0, -1)$.

Flux through this surface will be $\displaystyle \int_0^4 \int_0^{4-y} (0, 0, x + 0) \cdot (0, 0, -1) \, dx \, dy = -\frac{32}{3}$.

So the flux through surface ($S, x + y + 2z = 4$) will be the flux through the entire closed region minus the flux through other $3$ surfaces.

$$ = \frac{16}{3} - (-\frac{32}{3}) = 16$$.

  • $\begingroup$ @Boba did this help? Do you have any questions? $\endgroup$
    – Math Lover
    Dec 15, 2020 at 18:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .