Flux through a closed surface using the Divergence Theorem

Key takeaway: This is a classic Divergence Theorem setup: turn the surface flux into a triple integral over the solid, then look for symmetry before doing any heavy computation.

Use the Divergence Theorem:

$\iint_S \vec F\cdot d\vec S=\iiint_E (\nabla\cdot \vec F)\,dV,$

where $E$ is the solid bounded by the surface.

1) Compute the divergence

Given

$\vec F=\langle yx^2,\;xy^2-3z^4,\;x^3+y^2\rangle,$

we have

$$=\frac{\partial}{\partial x}(yx^2)+\frac{\partial}{\partial y}(xy^2-3z^4)+\frac{\partial}{\partial z}(x^3+y^2).$$

So,

$\nabla\cdot \vec F=2xy+2xy+0=4xy.$

2) Use symmetry of the solid

The region is the part of the sphere $x^2+y^2+z^2\le 16$ with

$y\le 0, \qquad z\le 0.$

That solid is symmetric with respect to $x\mapsto -x$.

Since the integrand is $4xy$, it is odd in $x$:

$4(-x)y=-4xy.$

Over a region symmetric in $x$, the triple integral cancels out:

$\iiint_E 4xy\,dV=0.$

3) Final result

$\iint_S \vec F\cdot d\vec S=0.$

You do not need to parameterize the spherical patch or the planar pieces separately once the divergence symmetry is noticed.

---

Pitfalls the pros know 👇 The main mistake here is integrating only over the spherical cap and forgetting that the closed surface also includes the coordinate-plane pieces. The Divergence Theorem applies only to the full closed boundary. Another common slip is expanding the divergence incorrectly; the $-3z^4$ term vanishes in the divergence because it is differentiated with respect to $y$.

What if the problem changes? If the region were not symmetric in $x$, then you would need to compute

$\iiint_E 4xy\,dV$

in spherical coordinates or by slicing. For a different vector field, the same method still works: find $\nabla\cdot\vec F$, then integrate over the enclosed solid. The closed-surface condition is what makes the theorem usable.

Tags: Divergence Theorem, flux integral, triple integral