3D divergence theorem examples

Background

• 3D divergence theorem
  • Flux in three dimensions
    • Divergence
    • Triple integrals

The divergence theorem (quick recap)

• $\mathbf{\text{F}}(x,y,z)$‍ is some three-dimensional vector field.
• $V$‍ is a three-dimensional volume (think of a blob in space).
• $S$‍ is the surface of $V$‍.
• $\widehat{\mathbf{\text{n}}}$‍ is a function which gives unit normal vectors on the surface of $S$‍.

Strategizing

Example 1: Surface integral through a cube.

Choose 1 answer:

Choose 1 answer:

• (Choice A)  

  ${\displaystyle {\int }_0^1{\int }_0^1{\int }_0^1\mathrm{\nabla }\times (2y\widehat{\mathbf{\text{i}}}+3y^2\widehat{\mathbf{\text{j}}}+4z\widehat{\mathbf{\text{k}}})dxdydz}$‍

  A

  ${\displaystyle {\int }_0^1{\int }_0^1{\int }_0^1\mathrm{\nabla }\times (2y\widehat{\mathbf{\text{i}}}+3y^2\widehat{\mathbf{\text{j}}}+4z\widehat{\mathbf{\text{k}}})dxdydz}$‍
• (Choice B)  

  ${\displaystyle {\int }_0^1{\int }_0^1{\int }_0^1\mathrm{\nabla }\cdot (2y\widehat{\mathbf{\text{i}}}+3y^2\widehat{\mathbf{\text{j}}}+4z\widehat{\mathbf{\text{k}}})dxdydz}$‍

  B

  ${\displaystyle {\int }_0^1{\int }_0^1{\int }_0^1\mathrm{\nabla }\cdot (2y\widehat{\mathbf{\text{i}}}+3y^2\widehat{\mathbf{\text{j}}}+4z\widehat{\mathbf{\text{k}}})dxdydz}$‍
• (Choice C)  

  ${\displaystyle 6{\int }_0^1{\int }_0^1\mathrm{\nabla }\times (2y\widehat{\mathbf{\text{i}}}+3y^2\widehat{\mathbf{\text{j}}}+4z\widehat{\mathbf{\text{k}}})dxdy}$‍

  C

  ${\displaystyle 6{\int }_0^1{\int }_0^1\mathrm{\nabla }\times (2y\widehat{\mathbf{\text{i}}}+3y^2\widehat{\mathbf{\text{j}}}+4z\widehat{\mathbf{\text{k}}})dxdy}$‍
• (Choice D)  

  ${\displaystyle 6{\int }_0^1{\int }_0^1\mathrm{\nabla }\cdot (2y\widehat{\mathbf{\text{i}}}+3y^2\widehat{\mathbf{\text{j}}}+4z\widehat{\mathbf{\text{k}}})dxdy}$‍

  D

  ${\displaystyle 6{\int }_0^1{\int }_0^1\mathrm{\nabla }\cdot (2y\widehat{\mathbf{\text{i}}}+3y^2\widehat{\mathbf{\text{j}}}+4z\widehat{\mathbf{\text{k}}})dxdy}$‍

Check

Answer

${\displaystyle {\int }_0^1{\int }_0^1{\int }_0^1\mathrm{\nabla }\cdot (2y\widehat{\mathbf{\text{i}}}+3y^2\widehat{\mathbf{\text{j}}}+4z\widehat{\mathbf{\text{k}}})dxdydz}$‍

The divergence theorem says that the surface integral of our vector field $(2y\widehat{\mathbf{\text{i}}}+3y^2\widehat{\mathbf{\text{j}}}+4z\widehat{\mathbf{\text{k}}})$‍ through the surface of the cube is the same as the triple integral of the divergence of that vector field throughout the cube.

The divergence is written like this:

$\mathrm{\nabla }\cdot (2y\widehat{\mathbf{\text{i}}}+3y^2\widehat{\mathbf{\text{j}}}+4z\widehat{\mathbf{\text{k}}})$‍

Describing the cube with a triple integral is relatively straightforward, since each variable $x$‍, $y$‍ and $z$‍ ranges from $0$‍ to $1$‍, so the appropriate triple integral structure just looks like this:

${\displaystyle {\int }_0^1{\int }_0^1{\int }_0^1\dots dxdydz}$‍

$\mathrm{\nabla }\cdot (2y\widehat{\mathbf{\text{i}}}+3y^2\widehat{\mathbf{\text{j}}}+4z\widehat{\mathbf{\text{k}}})=$‍

Check

Answer

If this feels unfamiliar, consider reviewing the article on divergence.

To compute divergence, think of the symbol $\mathrm{\nabla }$‍ as a vector of partial derivative operators:

${\displaystyle \mathrm{\nabla }=\left[\begin{array}{c}{\displaystyle \frac{\partial }{\partial x}}\\ \\ {\displaystyle \frac{\partial }{\partial y}}\\ \\ {\displaystyle \frac{\partial }{\partial z}}\end{array}\right]}$‍

Then take the "dot product" between this vector-ish thing and the vector-valued function whose divergence you are computing:

$\begin{array}{rl}& \mathrm{\nabla }\cdot (2y\widehat{\mathbf{\text{i}}}+3y^2\widehat{\mathbf{\text{j}}}+4z\widehat{\mathbf{\text{k}}})\\ \\ =& \left[\begin{array}{c}{\displaystyle \frac{\partial }{\partial x}}\\ \\ {\displaystyle \frac{\partial }{\partial y}}\\ \\ {\displaystyle \frac{\partial }{\partial z}}\end{array}\right]\cdot \left[\begin{array}{c}2y\\ \\ 3y^2\\ \\ 4z\end{array}\right]\\ \\ =& {\displaystyle \frac{\partial }{\partial x}}(2y)+{\displaystyle \frac{\partial }{\partial y}}(3y^2)+{\displaystyle \frac{\partial }{\partial z}}(4z)\\ \\ =& 0+6y+4\end{array}$‍

Example 2: Surface integral through a cylinder

${\displaystyle \mathrm{\nabla }\cdot \left[\begin{array}{c}{x}^3\\ y^3\\ x^3+y^3\end{array}\right]=}$‍

Check

Answer

$\begin{array}{rl}& \mathrm{\nabla }\cdot \left[\begin{array}{c}{x}^3\\ y^3\\ x^3+y^3\end{array}\right]\\ \\ =& \left[\begin{array}{c}{\displaystyle \frac{\partial }{\partial x}}\\ \\ {\displaystyle \frac{\partial }{\partial y}}\\ \\ {\displaystyle \frac{\partial }{\partial z}}\end{array}\right]\cdot \left[\begin{array}{c}{x}^3\\ \\ y^3\\ \\ x^3+y^3\end{array}\right]\\ \\ =& {\displaystyle \frac{\partial }{\partial x}}(x^3)+{\displaystyle \frac{\partial }{\partial y}}(y^3)+{\displaystyle \frac{\partial }{\partial z}}(x^3+y^3)\\ \\ =& 3x^2+3y^2+0\end{array}$‍

${\displaystyle {\iiint }_C(\mathrm{\nabla }\cdot \left[\begin{array}{c}{x}^3\\ y^3\\ x^3+y^3\end{array}\right])dV=}$‍

Check

Answer

First, just plug in the value you found from the last problem:

$\begin{array}{rl}& {\iiint }_C\mathrm{\nabla }\cdot \left[\begin{array}{c}{x}^3\\ y^3\\ x^3+y^3\end{array}\right]dV\\ \\ =& {\iiint }_{C}3(x^2+y^2)dV\end{array}$‍

Next, because we are integrating over a cylinder, it is very natural to integrate using cylindrical coordinates. For this problem, that involves three steps:

• Replace $dV$‍ with $rd\theta drdz$‍ (don't forget the little $r$‍ in front, this is what turns $d\theta$‍ into a unit of length).
• Replace $x^2+y^2$‍ with $r^2$‍.
• Set bounds which describe our cylinder. Luckily, all of these are constant:

  • $r$‍ ranges from $0$‍ to $3$‍.
    • $\theta$‍ ranges from $0$‍ to $2\pi$‍.
    • $z$‍ ranges from $0$‍ to $5$‍.

Applying all this to the integral, you get something that can be worked out:

$\begin{array}{rl}& \underset{\text{Plug in bounds}}{\underbrace{{\iiint }_C}}3\underset{r^2}{\underbrace{(x^2+y^2)}}\underset{rd\theta drdz}{\underbrace{dV}}\\ \\ & ={\int }_{z=0}^{z=5}{\int }_{r=0}^{r=3}{\int }_{\theta =0}^{\theta =2\pi }(3r^2)rd\theta drdz\\ \\ & =3\underset{\begin{array}{c}\text{Two of these integrals}\\ \text{treat }r\text{ as a constant.}\end{array}}{\underbrace{{\int }_{z=0}^{z=5}{\int }_{r=0}^{r=3}{\int }_{\theta =0}^{\theta =2\pi }{r}^{3}d\theta drdz}}\\ \\ & =3{\int }_{r=0}^{r=3}{r}^3{\int }_{z=0}^{z=5}\underset{=2\pi }{\underbrace{{\int }_{\theta =0}^{\theta =2\pi }d\theta }}dzdr\\ \\ & =3(2\pi ){\int }_{r=0}^{r=3}{r}^3\underset{=5}{\underbrace{{\int }_{z=0}^{z=5}dz}}dr\\ \\ & =3(2\pi )(5){\int }_{r=0}^{r=3}{r}^{3}dr\\ \\ & =3(2\pi )(5){\left[{\displaystyle \frac{1}{4}}{r}^4\right]}_{r=0}^{r=3}\\ \\ & =3(2\pi )(5)({\displaystyle \frac{1}{4}}{3}^4-{\displaystyle \frac{1}{4}}{0}^4)\\ \\ & ={\displaystyle \frac{3(2\pi )(5)(3^4)}{4}}\\ \\ & ={\displaystyle \frac{\mathrm{1,215}\pi }{2}}\end{array}$‍

Example 3: Surface area from a volume integral

Choose 1 answer:

Choose 1 answer:

• (Choice A)  

  $\mathbf{\text{F}}=\widehat{\mathbf{\text{n}}}$‍

  A

  $\mathbf{\text{F}}=\widehat{\mathbf{\text{n}}}$‍
• (Choice B)  

  $\mathbf{\text{F}}=\widehat{\mathbf{\text{n}}}\times (x\widehat{\mathbf{\text{i}}}+y\widehat{\mathbf{\text{j}}}+z\widehat{\mathbf{\text{k}}})$‍

  B

  $\mathbf{\text{F}}=\widehat{\mathbf{\text{n}}}\times (x\widehat{\mathbf{\text{i}}}+y\widehat{\mathbf{\text{j}}}+z\widehat{\mathbf{\text{k}}})$‍

Check

Answer

$\mathbf{\text{F}}=\widehat{\mathbf{\text{n}}}$‍

With this choice, $\mathbf{\text{F}}\cdot \widehat{\mathbf{\text{n}}}=\widehat{\mathbf{\text{n}}}\cdot \widehat{\mathbf{\text{n}}}$‍, and any unit vector dotted against itself equals $1$‍.

Choose 1 answer:

Choose 1 answer:

• (Choice A)  

  ${\displaystyle {\iiint }_B\mathrm{\nabla }\cdot \widehat{\mathbf{\text{n}}}dV}$‍

  A

  ${\displaystyle {\iiint }_B\mathrm{\nabla }\cdot \widehat{\mathbf{\text{n}}}dV}$‍
• (Choice B)  

  ${\displaystyle {\iiint }_B\mathrm{\nabla }\cdot (\widehat{\mathbf{\text{n}}}\cdot \widehat{\mathbf{\text{n}}})dV}$‍

  B

  ${\displaystyle {\iiint }_B\mathrm{\nabla }\cdot (\widehat{\mathbf{\text{n}}}\cdot \widehat{\mathbf{\text{n}}})dV}$‍

Check

Answer

The first choice is correct. In fact, the other choice doesn't even make sense, since $(\widehat{\mathbf{\text{n}}}\cdot \widehat{\mathbf{\text{n}}})$‍ gives a scalar valued function, and you can only take the divergence of vector-valued functions.

The surface integral we want to compute looks like this:

${\displaystyle {\iint }_{S}1d\mathrm{\Sigma }={\iint }_S(\widehat{\mathbf{\text{n}}}\cdot \widehat{\mathbf{\text{n}}})d\mathrm{\Sigma }}$‍

This is the flux of the unit normal vector function $\widehat{\mathbf{\text{n}}}$‍ through the surface. According to the divergence theorem, this is the same as integrating the divergence of that vector field within the volume of the sphere:

${\displaystyle {\iiint }_B\mathrm{\nabla }\cdot \widehat{\mathbf{\text{n}}}dV}$‍

Choose 1 answer:

Choose 1 answer:

• (Choice A)  

  ${\displaystyle \widehat{\mathbf{\text{n}}}(x,y,z)=(x+y+z)(\widehat{\mathbf{\text{i}}}+\widehat{\mathbf{\text{j}}}+\widehat{\mathbf{\text{k}}})}$‍

  A

  ${\displaystyle \widehat{\mathbf{\text{n}}}(x,y,z)=(x+y+z)(\widehat{\mathbf{\text{i}}}+\widehat{\mathbf{\text{j}}}+\widehat{\mathbf{\text{k}}})}$‍
• (Choice B)  

  ${\displaystyle \widehat{\mathbf{\text{n}}}(x,y,z)={\displaystyle \frac{1}{\sqrt{3}}}x\widehat{\mathbf{\text{i}}}+{\displaystyle \frac{1}{\sqrt{3}}}y\widehat{\mathbf{\text{j}}}+{\displaystyle \frac{1}{\sqrt{3}}}z\widehat{\mathbf{\text{k}}}}$‍

  B

  ${\displaystyle \widehat{\mathbf{\text{n}}}(x,y,z)={\displaystyle \frac{1}{\sqrt{3}}}x\widehat{\mathbf{\text{i}}}+{\displaystyle \frac{1}{\sqrt{3}}}y\widehat{\mathbf{\text{j}}}+{\displaystyle \frac{1}{\sqrt{3}}}z\widehat{\mathbf{\text{k}}}}$‍
• (Choice C)  

  ${\displaystyle \widehat{\mathbf{\text{n}}}(x,y,z)=x\widehat{\mathbf{\text{i}}}+y\widehat{\mathbf{\text{j}}}+z\widehat{\mathbf{\text{k}}}}$‍

  C

  ${\displaystyle \widehat{\mathbf{\text{n}}}(x,y,z)=x\widehat{\mathbf{\text{i}}}+y\widehat{\mathbf{\text{j}}}+z\widehat{\mathbf{\text{k}}}}$‍

Check

Answer

${\displaystyle \widehat{\mathbf{\text{n}}}(x,y,z)=x\widehat{\mathbf{\text{i}}}+y\widehat{\mathbf{\text{j}}}+z\widehat{\mathbf{\text{k}}}}$‍

The neat thing about the unit sphere centered at the origin is that if you take a vector pointing from the origin to a point on the sphere, then move it so that its tail is on that point, it will be normal to the sphere. This is easiest to draw and see on a circle, but the same principle applies for a sphere:

What's more, since the problem asks about a unit sphere, these vectors are all automatically unit vectors, so there's no need to normalize!

$\mathrm{\nabla }\cdot \widehat{\mathbf{\text{n}}}=$‍

Check

Answer

Plugging in ${\displaystyle \widehat{\mathbf{\text{n}}}(x,y,z)=x\widehat{\mathbf{\text{i}}}+y\widehat{\mathbf{\text{j}}}+z\widehat{\mathbf{\text{k}}}}$‍, here's what we get:

$\begin{array}{rl}& \mathrm{\nabla }\cdot \widehat{\mathbf{\text{n}}}\\ \\ & =\left[\begin{array}{c}{\displaystyle \frac{\partial }{\partial x}}\\ \\ {\displaystyle \frac{\partial }{\partial y}}\\ \\ {\displaystyle \frac{\partial }{\partial z}}\end{array}\right]\cdot \left[\begin{array}{c}x\\ \\ y\\ \\ z\end{array}\right]\\ \\ & ={\displaystyle \frac{\partial }{\partial x}}(x)+{\displaystyle \frac{\partial }{\partial y}}(y)+{\displaystyle \frac{\partial }{\partial z}}(z)\\ \\ & =1+1+1\\ \\ & =3\end{array}$‍

${\displaystyle {\iiint }_B\mathrm{\nabla }\cdot \widehat{\mathbf{\text{n}}}dV=}$‍

Check

Answer

Luckily, the expression $\mathrm{\nabla }\cdot \widehat{\mathbf{\text{n}}}$‍ turned out to be a constant, $3$‍. This means we are left with a volume integral:

$\begin{array}{rl}& {\iiint }_B\underset{=3}{\underbrace{\mathrm{\nabla }\cdot \widehat{\mathbf{\text{n}}}}}dV\\ \\ & ={\iiint }_B(3)dV\\ \\ & =3\underset{\text{Volume of unit ball}}{\underbrace{{\iiint }_{B}dV}}\\ \\ & =3{\displaystyle \frac{4}{3}}\pi \\ \\ & =4\pi \\ \end{array}$‍

And wouldn't you know it, that's the surface area of a unit sphere!

Summary

• The divergence theorem is useful whenever the interior volume of a region is easier to describe than its surface.
• It also helps if the divergence of the relevant vector field turns it into a simpler function.