Divergence Theorem Examples

Why This Matters

The Divergence Theorem connects what happens inside a region (a volume integral of divergence) to what happens on its boundary (a flux integral across a closed surface). It's one of the most useful tools in vector calculus because it lets you swap a difficult surface integral for an easier volume integral, or vice versa.

These examples cover the main geometric regions you'll encounter. For each one, the central question is the same: what symmetry does this region have, and which coordinate system matches it? Recognizing that quickly is what separates a clean solution from a messy one. Beyond choosing coordinates, you need to handle variable integration limits, surface orientation, and coordinate transformations. Each shape below highlights a different combination of these skills.

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Regions with Cartesian Symmetry

These shapes align naturally with Cartesian coordinates, featuring flat faces and integration limits that are either constant or linear. The key advantage is that bounds are simple to write down, making the triple integral straightforward to evaluate.

Solid Cube

• Constant integration limits on all six faces. For a cube with side length $a$, the volume integral is just $\int_0^a \int_0^a \int_0^a (\nabla \cdot \mathbf{F}) \, dV$ with no variable-dependent bounds.
• Direct surface integral comparison. Each face has an outward normal parallel to a coordinate axis ($\pm \hat{i}, \pm \hat{j}, \pm \hat{k}$), so computing flux face-by-face is clean.
• This makes the cube the go-to verification example: compute the volume integral and the surface integral independently, then confirm they're equal.

Rectangular Prism

• Generalized Cartesian region with dimensions $a \times b \times c$, maintaining constant limits $\int_0^a \int_0^b \int_0^c$.
• Separable integrals. When $\nabla \cdot \mathbf{F}$ factors into functions of single variables (e.g., $f(x)g(y)h(z)$), the volume integral becomes a product of three one-dimensional integrals. This is a huge computational shortcut.
• Functionally the same setup as the cube, just with unequal side lengths. It's a standard warm-up before tackling curved surfaces.

Tetrahedron

• Linear but variable bounds. A typical tetrahedron with vertices at the origin and on the coordinate axes gives limits like $z$ from $0$ to $1 - x - y$, $y$ from $0$ to $1 - x$, and $x$ from $0$ to $1$. You have to set these up carefully and choose a good order of integration.
• Four triangular faces make it manageable to verify flux through each surface individually, but parameterizing each face requires finding the outward normal to a plane.
• The tetrahedron tests mastery of dependent limits in Cartesian coordinates without introducing any curved boundaries.

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Regions with Spherical Symmetry

Spherical coordinates $(r, \theta, \phi)$ transform these regions into simple bounds. The volume element $dV = r^2 \sin\phi \, dr \, d\theta \, d\phi$ and radial symmetry often collapse complex vector fields into manageable expressions.

Spherical Shell

• Concentric sphere boundaries. The region $a \leq r \leq b$ has constant angular limits ($0 \leq \theta \leq 2\pi$, $0 \leq \phi \leq \pi$), so all the complexity lives in the radial direction.
• Radial vector fields simplify dramatically. For $\mathbf{F} = f(r)\hat{r}$, the divergence is $\nabla \cdot \mathbf{F} = \frac{1}{r^2}\frac{d}{dr}(r^2 f)$, which depends only on $r$. The angular integrals just contribute a factor of $4\pi$.
• Classic inverse-square law application. The field $\mathbf{F} = \frac{\hat{r}}{r^2}$ has $\nabla \cdot \mathbf{F} = 0$ everywhere except the origin. If the shell excludes the origin ($a > 0$), the volume integral of divergence is zero, meaning the flux into the inner sphere equals the flux out of the outer sphere.

Hemisphere

• Half-sphere with a planar cap. Bounded by $r \leq R$ and $0 \leq \phi \leq \frac{\pi}{2}$. The Divergence Theorem requires a closed surface, so you must include the flat circular disk at $z = 0$ as part of the boundary.
• Symmetry reduces angular integration. Many integrands vanish when integrated over the hemisphere due to odd-function cancellation in $\phi$ or $\theta$.
• Common exam setup: you're asked for flux through the curved surface alone. The strategy is to use the Divergence Theorem on the closed region (curved surface + disk), compute the volume integral, then subtract the flux through the disk.

Ellipsoid

• Stretched spherical symmetry. Defined by $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} \leq 1$. Spherical coordinates don't give clean bounds here, so you need a different approach.
• Jacobian scaling. The substitution $u = x/a$, $v = y/b$, $w = z/c$ maps the ellipsoid to a unit sphere with $dV = abc \, du \, dv \, dw$. After this transformation, you can use standard spherical coordinates on the unit sphere.
• Tests coordinate transformation skills. You need to understand how the divergence of the original field interacts with the scaled volume element. The factor of $abc$ in the Jacobian accounts for the stretching.

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Regions with Cylindrical Symmetry

Cylindrical coordinates $(r, \theta, z)$ are ideal when a region has rotational symmetry about one axis. The volume element $dV = r \, dr \, d\theta \, dz$ handles circular cross-sections efficiently.

Cylindrical Shell

• Annular cross-section bounded by $a \leq r \leq b$, $0 \leq \theta \leq 2\pi$, and $0 \leq z \leq h$. All limits are constant, making this the cylindrical analog of the rectangular prism.
• Rotational symmetry. If the vector field is independent of $\theta$, the $\theta$-integral factors out as $2\pi$, reducing the problem to a double integral in $r$ and $z$.
• Divergence in cylindrical coordinates: $\nabla \cdot \mathbf{F} = \frac{1}{r}\frac{\partial}{\partial r}(rF_r) + \frac{1}{r}\frac{\partial F_\theta}{\partial \theta} + \frac{\partial F_z}{\partial z}$. You need to have this memorized for these problems.

Cone

• Variable radius with height. A cone of height $h$ and base radius $R$ gives the region $0 \leq r \leq \frac{R}{h}z$, $0 \leq z \leq h$. The $r$-bound depends linearly on $z$.
• Order of integration matters. You can integrate $r$ first (from $0$ to $\frac{R}{h}z$) for each fixed $z$, or flip the order, but the $r$-first approach usually reads more naturally for a cone.
• The apex at $z = 0$ is a single point where the cross-section shrinks to zero. This doesn't cause problems for the integral itself, but if the divergence has singular behavior near the origin, you'll need to check convergence.

Paraboloid

• Curved boundary in the $r$-$z$ plane. A typical setup is $z = r^2$ (opening upward) capped by a plane $z = c$, giving the region $0 \leq r \leq \sqrt{c}$, $r^2 \leq z \leq c$.
• Choosing integration order. Integrating $z$ first (from $r^2$ to $c$) for fixed $r$ is often cleanest. The $r$-limits are then constant: $0$ to $\sqrt{c}$.
• Orientation matters. Whether the paraboloid opens upward or downward changes which surface is the "cap" and which is the curved part, affecting the outward normal direction on each piece.

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Regions with Complex Topology

These shapes challenge standard approaches due to non-convexity, holes, or unusual parameterizations. The Divergence Theorem still applies, but careful attention to surface orientation and closed boundaries is essential.

Torus

• Doughnut topology. The hole through the center means the surface has genus 1. Parameterization requires two angles: a toroidal angle $\theta$ (around the central axis) and a poloidal angle $\phi$ (around the tube).
• Standard parameterization: $x = (R + r\cos\phi)\cos\theta$, $y = (R + r\cos\phi)\sin\theta$, $z = r\sin\phi$, where $R$ is the distance from the center of the tube to the center of the torus (major radius) and $r$ is the tube radius (minor radius).
• Non-convex region. Outward normals point away from the center on the outer surface but toward the center on the inner surface. Despite this, the Divergence Theorem applies without modification since the torus still encloses a well-defined volume.

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Quick Reference Table

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Self-Check Questions

1. Which two regions both use cylindrical coordinates but differ in whether the $r$-bounds are constant or variable? What does this difference mean for setting up the integral?

2. For a radial vector field $\mathbf{F} = r^n \hat{r}$, which region would most naturally exploit the symmetry to simplify the divergence calculation? Why?

3. Compare the Tetrahedron and the Solid Cube: both use Cartesian coordinates, so what makes one more computationally challenging than the other?

4. If you need to verify the Divergence Theorem by computing both the volume integral and the surface integral independently, which region would you choose for the cleanest calculation? Justify your choice.

5. The Hemisphere requires including a flat circular base to form a closed surface. Why is this necessary for applying the Divergence Theorem, and how would you compute the flux through this base?