Divergence Theorem (Defined w/ Step-by-Step Examples!)

Did you know that we can see the divergence theorem at work in some of our tastiest desserts?

Jenn (B.S., M.Ed.) of Calcworkshop® teaching divergence theorem

Jenn, Founder Calcworkshop^®, 15+ Years Experience (Licensed & Certified Teacher)

Imagine making a light and airy cream puff or éclair for dessert. The pate a choux dough is baked to perfection and is ready to be filled with some yummy pastry cream.

But let’s assume that the cream puff is rigid and does not expand when filled with cream.

What will happen?

Some of the pastry cream will eventually leak out of the puff.

And this is the idea behind the divergence theorem!

Divergence Theorem

Simply stated, the divergence theorem, also called Gauss’ theorem, states that the total expansion of a fluid or gas (or pastry cream) inside a closed surface is equal to the fluid escaping the closed surface.

Remember, a vector field $\vec{F}$ represents the flux (flow) of a fluid, and the divergence of $\vec{F}$ is the gathering (sink) or dispersing (source) of the fluid inside some region.

This means we can turn our surface integral (flux integral) into a triple integral over the volume enclosed by the surface.

Formula

Let $E$ be a simple solid region and let $S$ be the boundary surface of $E$ , with a positive outward orientation. And let $\vec{F}$ be a vector field whose component functions have continuous partial derivatives on an open region that contains $E$ . Then

$\iint_{S} \vec{F} \cdot d \vec{S} = ∭_{E} div \vec{F} d V$

Divergence Theorem Geometric Interpretation

Therefore, that divergence theorem represents the net rate of outward flux per unit volume and plays a significant importance to the field of mathematics and engineering, in particular, electrostatics and fluid dynamics.

Cool!

Also, it’s important to note that if there is net flow “out of” the closed surface (i.e., $d i v \vec{F} > 0$ ), the integral is positive. Thus, if there is a net flow “into” the closed surface (i.e., $d i v \vec{F} < 0$ ), the integral is negative, just as we would expect.

Example

Let’s look at an example.

Evaluate the surface integral using the divergence theorem $∭_{D} div \vec{F} d V$ if $\vec{F} (x, y, z) = ⟨ x, y, z - 1 ⟩$ where $D$ is the region bounded by the hemisphere $0 \leq z \leq \sqrt{16 - x^{2} - y^{2}}$ .

First, we will calculate $d i v \vec{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}$ .

$div \vec{F} = \frac{\partial}{\partial x} [x] + \frac{\partial}{\partial y} [y] + \frac{\partial}{\partial z} [z - 1] = 1 + 1 + 1 = 3$

Next, we will find our limit bounds. Because we are told that a hemisphere bounds our region, we should use spherical coordinates.

Bounded Region Hemisphere

$\begin{aligned} 0 \leq z \leq \sqrt{16 - x^{2} - y^{2}} \\ 0 \leq z^{2} \leq 16 - x^{2} - y^{2} \\ 0 \leq \underset{ρ^{2}}{\underset{⏟}{x^{2} + y^{2} + z^{2}}} \leq 16 \\ 0 \leq ρ^{2} \leq 16 \\ 0 \leq ρ \leq 4 \end{aligned}$

And since $D$ is a hemisphere is has $360^{\circ}$ rotation around the xy-plane such that $0 \leq θ \leq 2 π$ , and the angle from the top of the hemisphere (positive z-axis) to the bottom of the plane is $90^{\circ}$ we know that $0 \leq ϕ \leq \frac{π}{2}$

Now we’re ready to evaluate our triple integral knowing that $d i v \vec{F} = 3$ , $0 \leq ρ \leq 4$ , $0 \leq θ \leq 2 π$ , and $0 \leq ϕ \leq \frac{π}{2}$ . And we must not forget the Jacobian determinate, $ρ^{2} \sin ϕ$ , for spherical coordinates!

$∭_{D} div \vec{F} d V = \int_{0}^{\frac{π}{2}} \int_{0}^{2 π} \int_{0}^{4} (3) (p^{2} \sin ϕ) d ρ d θ d ϕ = 256 π$

Easy, right!

Recap

So, let’s quickly recap some important concepts about Gauss’s divergence theorem.

The divergence theorem tells us that the flux of $\vec{F}$ across a surface is found by integrating the divergence of $\vec{F}$ over the region enclosed by the surface.
The divergence theorem equates a surface integral across a closed surface $S$ to a triple integral over the solid enclosed by $S$ .
The divergence theorem is a higher dimensional version of the flux form of Green’s theorem.

Nice.

And I bet the next time you shake a can of soda, pump air into a basketball or eat an éclair, cream puff, or jelly-filled donut, you’ll think of divergence theorem and smile!

Alright, let’s dive right into our lesson.

Video Tutorial w/ Full Lesson & Detailed Examples (Video)

Divergence-Theorem-Example

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