## 17Calculus - Divergence Theorem

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The Divergence Theorem is the three-dimensional version of the flux form of Green's Theorem and it relates the flow (or flux) through the boundary of a closed surface S to the divergence of the vector field through the volume Q. The surface S encloses the volume Q.

Divergence Theorem

If $$\vec{N}$$ is the unit normal vector directed outward from a solid region Q, then

$$\displaystyle{ \iint\limits_S { \vec{F} \cdot \vec{N} ~dS = \iiint\limits_Q { ( \nabla \vec{F} ) ~ dV } } }$$

Divergence Theorem in Detail

When you studied surface integrals, you learned how to calculate the flux integral $$\displaystyle{\iint_S{\vec{F} \cdot \vec{N} ~ dS} }$$ which is basically the flow through a surface S.
When you have a solid region bounded by a closed surface S, there is an easier way to calculate this integral, using the Divergence Theorem.

Basically what this Divergence Theorem says is that the flow or flux of the vector field $$\vec{F}$$ through the closed surface S, which is given by the integral $$\displaystyle{\iint_S{\vec{F} \cdot \vec{N} ~ dS} }$$ is equal to the divergence of $$\vec{F}$$ through the volume Q. This is given by the integral $$\displaystyle{ \iiint\limits_Q { ( \nabla \vec{F} ) ~ dV } }$$.

The required relationship between the surface S in the first integral and the volume Q in the second integral is that the volume Q must be bounded by the surface S.

Here is a great video explaining the Divergence Theorem.

### Dr Chris Tisdell - Divergence theorem of Gauss [12mins-20secs]

video by Dr Chris Tisdell

Usually one of the integrals is easier to set up and evaluate than the other. Most of the time the easier integral is the volume integral since surface integrals are more difficult to set up in general. But you will get a feel for this as you get more experience.

Before we go on, let's look at an example.

Divergence Theorem Example

Verify the Divergence Theorem for $$\vec{F}(x,y,z)=xy\hat{i}+z\hat{j}+(x+y)\hat{k}$$ where the surface consists of the bounding planes $$y=4$$, $$z=4-x$$ and the coordinate planes.

Verify the Divergence Theorem for $$\vec{F}(x,y,z)=xy\hat{i}+z\hat{j}+(x+y)\hat{k}$$ where the surface consists of the bounding planes $$y=4$$, $$z=4-x$$ and the coordinate planes.

Solution

Plot 1 - plotted with 3d Grapher

Verify the Divergence Theorem for $$\vec{F}(x,y,z)=xy\hat{i}+z\hat{j}+(x+y)\hat{k}$$ where the surface consists of the bounding planes $$y=4$$, $$z=4-x$$ and the coordinate planes.

When the problem says to verify the Divergence Theorem, it means to calculate both integrals and confirm that they are equal. Plot 1 shows the plane $$z-4-x$$. Basically, we have a square box 4 units on each side with the plane $$z=4-x$$ cutting it in half giving us a wedge.
We will start with the longer one first, i.e. the surface integral. This is longer since we have 5 sides to work with and, therefore, 5 integrals. Here is a list of the sides and the outward pointing normal vector for each side.

 1 xz-plane $$\vec{N}=-\hat{j}$$ 2 xz-plane at $$y=4$$ $$\vec{N}=\hat{j}$$ 3 xy-plane $$\vec{N}=-\hat{k}$$ 4 yz-plane $$\vec{N}=-\hat{i}$$ 5 $$z=4-x$$ plane $$\displaystyle{ \vec{N}=\frac{\hat{i}+\hat{k}}{\sqrt{2}} }$$

Note - We calculated the normal vector of side 5 using $$\displaystyle{ \vec{N}=\frac{\vec{\nabla}G}{\| \vec{\nabla}G \|} }$$ where $$G=z-4+x$$.
Now we set up all 5 integrals but we let you evaluate them.
Side 1 - $$\vec{F}\cdot\vec{N}=-z$$
In the plane $$y=0$$, $$0\leq x\leq 4$$ and $$0\leq z\leq 4-x$$ giving us $$\displaystyle{ \int_{0}^{4}{ \int_{0}^{4-x}{ -z~dz~dx } } = -32/3}$$

Side 2 - $$\vec{F}\cdot\vec{N}=z$$
In the plane $$y=4$$, the surface is the same as in side 1. So we have $$\displaystyle{ \int_{0}^{4}{ \int_{0}^{4-x}{ z~dz~dx } } = 32/3}$$

Side 3 - $$\vec{F}\cdot\vec{N}=-(x+y)$$
In the xy-plane, the surface is a square with sides of length 4. So our integral is $$\displaystyle{ \int_{0}^{4}{ \int_{0}^{4}{ -(x+y)~dy~dx } } =-64 }$$

Side 4 - $$\vec{F}\cdot\vec{N}=-xy$$
In the yz-plane, the surface is a square with sides of length 4. So our integral is $$\displaystyle{ \int_{0}^{4}{ \int_{0}^{4}{ -xy~dy~dz } } = 0 }$$ since $$x=0$$.

Side 5 - $$\vec{F}\cdot\vec{N}=(xy+x+y)/\sqrt{2}$$
This side will take a bit more work. The surface is at an angle, so $$dS = \sqrt{g_x^2+g_y^2+1}dA$$ where $$z=g(x,y)=4-x$$ so $$dS=\sqrt{2}dA$$ giving us an integral $$\displaystyle{ \int_{0}^{4}{ \int_{0}^{4}{ xy+x+y ~ dy ~ dx } } = 128 }$$

Now we sum up all the values of the integrals to get $$-32/3+32/3-64+0+128=64$$

Okay, now we will evaluate the volume integral.
$$div ~ \vec{F}=y$$ so the triple integral is $$\displaystyle{ \int_{0}^{4}{ \int_{0}^{4}{ \int_{0}^{4-x}{ y~dz~dy~dx } } } = 64 }$$

Notice how much easier the volume integral was to evaluate.

Note - As is usual in higher mathematics, we have left out a lot of detail in this solution. We do not expect you to be able to sit and look at this solution and visualize all the steps in your head. You need to get out a pencil and a piece of paper and fill in the details yourself in order to understand where everything comes from. Get used to doing this all the time with examples and solutions. If you do, then you will understand and be able to use calculus and higher math.

Divergence Theorem in Depth

Here are several full length video lectures that explain the divergence theorem in more depth. These are important to watch in order to understand what is really going on with this theorem and how to use it properly.

### Evans Lawrence - Lecture 34 - Divergence Theorem [22mins-22secs]

video by Evans Lawrence

video by MIT OCW

video by MIT OCW

### Divergence Theorem Proof Videos

Although not completely necessary to the use of the divergence theorem, going through the proof will help you understand in more detail how this theorem works.

### Khan Academy - Divergence Theorem Proof (5) [3mins-11secs]

Okay, you are now ready for some practice problems. After which, the next logical step is Stokes' Theorem, if you haven't already learned it.

Practice

Unless otherwise instructed, solve these problems using The Divergence Theorem.
- If you are asked to calculate the surface integral, you will usually evaluate the volume integral.
- Similarly, if you are asked to calculate the volume integral, you will usually evaluate the surface integral.

Using the Divergence Theorem, determine the flux of the vector field $$\vec{F} = \langle x^2,y^2,z^3 \rangle$$ across the surface bounded by the planes $$x=0, x=2, y=0, y=2, z=0, z=4.$$

Problem Statement

Using the Divergence Theorem, determine the flux of the vector field $$\vec{F} = \langle x^2,y^2,z^3 \rangle$$ across the surface bounded by the planes $$x=0, x=2, y=0, y=2, z=0, z=4.$$

320

Problem Statement

Using the Divergence Theorem, determine the flux of the vector field $$\vec{F} = \langle x^2,y^2,z^3 \rangle$$ across the surface bounded by the planes $$x=0, x=2, y=0, y=2, z=0, z=4.$$

Solution

### 1985 video

video by MIP4U

320

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Using the Divergence Theorem, determine the flux of the vector field $$\vec{F} = \langle x^2/2,yz,-xz \rangle$$ across the surface $$z = \sqrt{4-x^2-y^2}, z = 0$$.

Problem Statement

Using the Divergence Theorem, determine the flux of the vector field $$\vec{F} = \langle x^2/2,yz,-xz \rangle$$ across the surface $$z = \sqrt{4-x^2-y^2}, z = 0$$.

$$4\pi$$

Problem Statement

Using the Divergence Theorem, determine the flux of the vector field $$\vec{F} = \langle x^2/2,yz,-xz \rangle$$ across the surface $$z = \sqrt{4-x^2-y^2}, z = 0$$.

Solution

### 1986 video

video by MIP4U

$$4\pi$$

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Using the Divergence Theorem, compute the flux of $$\vec{F} = \langle x^4y, -2x^3y^2,z^2 \rangle$$ through the surface of the solid bounded by $$z=0, z=h$$ and $$x^2 + y^2 = R^2$$.

Problem Statement

Using the Divergence Theorem, compute the flux of $$\vec{F} = \langle x^4y, -2x^3y^2,z^2 \rangle$$ through the surface of the solid bounded by $$z=0, z=h$$ and $$x^2 + y^2 = R^2$$.

$$\pi h^2 R^2$$

Problem Statement

Using the Divergence Theorem, compute the flux of $$\vec{F} = \langle x^4y, -2x^3y^2,z^2 \rangle$$ through the surface of the solid bounded by $$z=0, z=h$$ and $$x^2 + y^2 = R^2$$.

Solution

### 1987 video

video by MIT OCW

$$\pi h^2 R^2$$

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Using the Divergence Theorem, calculate the surface integral for $$\vec{F} = \langle 4x^3z,4y^3z,3z^4 \rangle$$ where the surface of the solid is bounded by the hemispheres $$z = \sqrt{4-x^2-y^2}, z = \sqrt{1-x^2-y^2}$$ and the plane $$z=0$$.

Problem Statement

Using the Divergence Theorem, calculate the surface integral for $$\vec{F} = \langle 4x^3z,4y^3z,3z^4 \rangle$$ where the surface of the solid is bounded by the hemispheres $$z = \sqrt{4-x^2-y^2}, z = \sqrt{1-x^2-y^2}$$ and the plane $$z=0$$.

$$126\pi$$

Problem Statement

Using the Divergence Theorem, calculate the surface integral for $$\vec{F} = \langle 4x^3z,4y^3z,3z^4 \rangle$$ where the surface of the solid is bounded by the hemispheres $$z = \sqrt{4-x^2-y^2}, z = \sqrt{1-x^2-y^2}$$ and the plane $$z=0$$.

Solution

### 1988 video

video by MIP4U

$$126\pi$$

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Using the Divergence Theorem, calculate the surface integral for $$\vec{F} = \langle 3xy^2,xe^z,z^3 \rangle$$ where the solid is bounded by the cylinder $$y^2 + z^2 = 25$$ and the planes $$x=-4$$ and $$x=3$$.

Problem Statement

Using the Divergence Theorem, calculate the surface integral for $$\vec{F} = \langle 3xy^2,xe^z,z^3 \rangle$$ where the solid is bounded by the cylinder $$y^2 + z^2 = 25$$ and the planes $$x=-4$$ and $$x=3$$.

$$13125 \pi/2$$

Problem Statement

Using the Divergence Theorem, calculate the surface integral for $$\vec{F} = \langle 3xy^2,xe^z,z^3 \rangle$$ where the solid is bounded by the cylinder $$y^2 + z^2 = 25$$ and the planes $$x=-4$$ and $$x=3$$.

Solution

### 1989 video

video by MIP4U

$$13125 \pi/2$$

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Using the Divergence Theorem, calculate the surface integral for $$\vec{F} = \langle e^x\sin y, e^x\cos y, yz^2 \rangle$$ where the surface is the box bounded by the planes $$x=4, y=5, z=1$$ and the coordinate planes.

Problem Statement

Using the Divergence Theorem, calculate the surface integral for $$\vec{F} = \langle e^x\sin y, e^x\cos y, yz^2 \rangle$$ where the surface is the box bounded by the planes $$x=4, y=5, z=1$$ and the coordinate planes.

50

Problem Statement

Using the Divergence Theorem, calculate the surface integral for $$\vec{F} = \langle e^x\sin y, e^x\cos y, yz^2 \rangle$$ where the surface is the box bounded by the planes $$x=4, y=5, z=1$$ and the coordinate planes.

Solution

### 1990 video

video by MIP4U

50

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Suppose $$\vec{F}$$ is a vector field on $$\mathbb{R}^3\setminus(0,0,0)$$ with $$div \vec{F} = \sqrt{x^2+y^2+z^2}$$. Let $$S_1$$ be the sphere $$x^2+y^2+z^2=1$$, oriented outward. Let $$S_2$$ be the sphere $$x^2+y^2+z^2=4$$, oriented outward. Suppose $$\iint\limits_{S_1}{\vec{F}\cdot d\vec{S}} = 2\pi$$. Calculate $$\iint\limits_{S_2}{\vec{F}\cdot d\vec{S}}$$.

Problem Statement

Suppose $$\vec{F}$$ is a vector field on $$\mathbb{R}^3\setminus(0,0,0)$$ with $$div \vec{F} = \sqrt{x^2+y^2+z^2}$$. Let $$S_1$$ be the sphere $$x^2+y^2+z^2=1$$, oriented outward. Let $$S_2$$ be the sphere $$x^2+y^2+z^2=4$$, oriented outward. Suppose $$\iint\limits_{S_1}{\vec{F}\cdot d\vec{S}} = 2\pi$$. Calculate $$\iint\limits_{S_2}{\vec{F}\cdot d\vec{S}}$$.

$$17 \pi$$

Problem Statement

Suppose $$\vec{F}$$ is a vector field on $$\mathbb{R}^3\setminus(0,0,0)$$ with $$div \vec{F} = \sqrt{x^2+y^2+z^2}$$. Let $$S_1$$ be the sphere $$x^2+y^2+z^2=1$$, oriented outward. Let $$S_2$$ be the sphere $$x^2+y^2+z^2=4$$, oriented outward. Suppose $$\iint\limits_{S_1}{\vec{F}\cdot d\vec{S}} = 2\pi$$. Calculate $$\iint\limits_{S_2}{\vec{F}\cdot d\vec{S}}$$.

Solution

### 2216 video

$$17 \pi$$

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### Trig Formulas

The Unit Circle

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Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

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