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The Divergence Theorem

Explain the meaning of the divergence theorem.

Overview of Theorems

Before examining the divergence theorem, it is helpful to begin with an overview of the versions of the Fundamental Theorem of Calculus we have discussed:

  1. The Fundamental Theorem of Calculus:
    \[{\int }_{a}^{b}{f}^{'}(x)dx=f(b)-f(a).\]
    This theorem relates the integral of derivative \({f}^{'}\) over line segment \([a,b]\) along the x-axis to a difference of \(f\) evaluated on the boundary.
  2. The Fundamental Theorem for Line Integrals:
    \[{\int }_{C}∇f\cdot d\text{r}=f({P}_{1})-f({P}_{0}),\]
    where \({P}_{0}\) is the initial point of C and \({P}_{1}\) is the terminal point of C. The Fundamental Theorem for Line Integrals allows path C to be a path in a plane or in space, not just a line segment on the x-axis. If we think of the gradient as a derivative, then this theorem relates an integral of derivative \(∇f\) over path C to a difference of \(f\) evaluated on the boundary of C.
  3. Green’s theorem, circulation form:
    \[{∬}_{D}({Q}_{x}-{P}_{y})dA={\int }_{C}\text{F}\cdot d\text{r}.\]
    Since \({Q}_{x}-{P}_{y}=\text{curl}\ \text{F}\cdot \text{k}\) and curl is a derivative of sorts, Green’s theorem relates the integral of derivative curlF over planar region D to an integral of F over the boundary of D.
  4. Green’s theorem, flux form:
    \[{∬}_{D}({P}_{x}+{Q}_{y})dA={\int }_{C}\text{F}\cdot \text{N}ds.\]
    Since \({P}_{x}+{Q}_{y}=\text{div}\ \text{F}\) and divergence is a derivative of sorts, the flux form of Green’s theorem relates the integral of derivative divF over planar region D to an integral of F over the boundary of D.
  5. Stokes’ theorem:
    \[{∬}_{S}\text{curl}\ \text{F}\cdot d\text{S}={\int }_{C}\text{F}\cdot d\text{r}.\]
    If we think of the curl as a derivative of sorts, then Stokes’ theorem relates the integral of derivative curlF over surface S (not necessarily planar) to an integral of F over the boundary of S.

Stating the Divergence Theorem

The divergence theorem follows the general pattern of these other theorems. If we think of divergence as a derivative of sorts, then the divergence theorem relates a triple integral of derivative divF over a solid to a flux integral of F over the boundary of the solid. More specifically, the divergence theorem relates a flux integral of vector field F over a closed surface S to a triple integral of the divergence of F over the solid enclosed by S.

Recall that the flux form of Green’s theorem states that \({∬}_{D}\text{div}\ \text{F}dA={\int }_{C}\text{F}\cdot \text{N}ds.\) Therefore, the divergence theorem is a version of Green’s theorem in one higher dimension.

The proof of the divergence theorem is beyond the scope of this text. However, we look at an informal proof that gives a general feel for why the theorem is true, but does not prove the theorem with full rigor. This explanation follows the informal explanation given for why Stokes’ theorem is true.

Condensed — the full section is in OpenStax Calculus Volume 3.

Using the Divergence Theorem

The divergence theorem translates between the flux integral of closed surface S and a triple integral over the solid enclosed by S. Therefore, the theorem allows us to compute flux integrals or triple integrals that would ordinarily be difficult to compute by translating the flux integral into a triple integral and vice versa.

Example

Try it.

Calculate the surface integral \({∬}_{S}\text{F}\cdot d\text{S},\) where S is cylinder \({x}^{2}+{y}^{2}=1,0\le z\le 2,\) including the circular top and bottom, and \(\text{F}=〈\frac{{x}^{3}}{3}+yz,\frac{{y}^{3}}{3}-\text{sin}(xz),z-x-y〉.\)

Solution

We could calculate this integral without the divergence theorem, but the calculation is not straightforward because we would have to break the flux integral into three separate integrals: one for the top of the cylinder, one for the bottom, and one for the side. Furthermore, each integral would require parameterizing the corresponding surface, calculating tangent vectors and their cross product, and using .

By contrast, the divergence theorem allows us to calculate the single triple integral \({∭}_{E}\text{div}\ \text{F}dV,\) where E is the solid enclosed by the cylinder. Using the divergence theorem and converting to cylindrical coordinates, we have

\[\begin{array}{ll}{∬}_{s}\text{F}\cdot d\text{S} & ={∭}_{E}\text{div}\ \text{F}\ dV \\ & ={∭}_{E}({x}^{2}+{y}^{2}+1)dV \\ & ={\int }_{0}^{2\pi }{\int }_{0}^{1}{\int }_{0}^{2}({r}^{2}+1)r\ dz\ dr\ d\theta \\ & =\frac{3}{2}{\int }_{0}^{2\pi }d\theta =3\pi .\end{array}\]

illustrates a remarkable consequence of the divergence theorem. Let S be a piecewise, smooth closed surface and let F be a vector field defined on an open region containing the surface enclosed by S. If F has the form \(\text{F}=〈f(y,z),g(x,z),h(x,y)〉,\) then the divergence of F is zero. By the divergence theorem, the flux of F across S is also zero. This makes certain flux integrals incredibly easy to calculate. For example, suppose we wanted to calculate the flux integral \({∬}_{S}\text{F}\cdot d\text{S}\) where S is a cube and

Condensed — the full section is in OpenStax Calculus Volume 3.

Application to Electrostatic Fields

The divergence theorem has many applications in physics and engineering. It allows us to write many physical laws in both an integral form and a differential form (in much the same way that Stokes’ theorem allowed us to translate between an integral and differential form of Faraday’s law). Areas of study such as fluid dynamics, electromagnetism, and quantum mechanics have equations that describe the conservation of mass, momentum, or energy, and the divergence theorem allows us to give these equations in both integral and differential forms.

One of the most common applications of the divergence theorem is to electrostatic fields. An important result in this subject is Gauss’ law. This law states that if S is a closed surface in electrostatic field E, then the flux of E across S is the total charge enclosed by S (divided by an electric constant). We now use the divergence theorem to justify the special case of this law in which the electrostatic field is generated by a stationary point charge at the origin.

If \((x,y,z)\) is a point in space, then the distance from the point to the origin is \(r=\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}.\) Let \({\text{F}}_{r}\) denote radial vector field \({\text{F}}_{r}=\frac{1}{{r}^{2}}〈\frac{x}{r},\frac{y}{r},\frac{z}{r}〉.\) The vector at a given position in space points in the direction of unit radial vector \(〈\frac{x}{r},\frac{y}{r},\frac{z}{r}〉\) and is scaled by the quantity \(1\text{/}{r}^{2}.\) Therefore, the magnitude of a vector at a given point is inversely proportional to the square of the vector’s distance from the origin. Suppose we have a stationary charge of q Coulombs at the origin, existing in a vacuum. The charge generates electrostatic field E given by

\[\text{E}=\frac{q}{4\pi {\epsilon }_{0}}{\text{F}}_{r},\]

where the approximation \({\epsilon }_{0}=8.854\ \times \ {10}^{-12}\) farad (F)/m is an electric constant. (The constant \({\epsilon }_{0}\) is a measure of the resistance encountered when forming an electric field in a vacuum.) Notice that E is a radial vector field similar to the gravitational field described in . The difference is that this field points outward whereas the gravitational field points inward. Because

\[\text{E}=\frac{q}{4\pi {\epsilon }_{0}}{\text{F}}_{r}=\frac{q}{4\pi {\epsilon }_{0}}(\frac{1}{{r}^{2}}〈\frac{x}{r},\frac{y}{r},\frac{z}{r}〉),\]

Notice that since the divergence of \({\text{F}}_{r}\) is zero and E is \({\text{F}}_{r}\) scaled by a constant, the divergence of electrostatic field E is also zero (except at the origin).

Condensed — the full section is in OpenStax Calculus Volume 3.

Key Concepts

  • The divergence theorem relates a surface integral across 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, and is therefore a higher dimensional version of the Fundamental Theorem of Calculus.
  • The divergence theorem can be used to transform a difficult flux integral into an easier triple integral and vice versa.
  • The divergence theorem can be used to derive Gauss’ law, a fundamental law in electrostatics.

The Divergence Theorem

For the following exercises, use a computer algebraic system (CAS) and the divergence theorem to evaluate surface integral \({∬}_{S}^{}\text{F}\cdot \text{N}ds\) for the given choice of F and the boundary surface S. For each closed surface, assume N is the outward unit normal vector.

For the following exercises, use a CAS along with the divergence theorem to compute the net outward flux for the fields across the given surfaces S.

For the following exercises, use a CAS and the divergence theorem to compute the net outward flux for the vector fields across the boundary of the given regions D.

For the following exercises, Fourier’s law of heat transfer states that the heat flow vector F at a point is proportional to the negative gradient of the temperature; that is, \(\text{F}=\text{-}k∇T,\) which means that heat energy flows hot regions to cold regions. The constant \(k>0\) is called the conductivity, which has metric units of joules per meter per second-kelvin or watts per meter-kelvin. A temperature function for region D is given. Use the divergence theorem to find net outward heat flux \({∬}_{S}\text{F}\cdot \text{N}dS=\text{-}k{∬}_{S}∇T\cdot \text{N}dS\) across the boundary S of D, where \(k=1.\)

The Divergence Theorem

This section relies heavily on understanding flux integrals as well as the meaning of the divergence of a vector field from and triple integrals from .

Introduction

In we examined vector fields to consider how the strength of a vector field changes in different regions. In particular, we developed the divergence of a vector field as a local measurement (based on density) of how the strength of the vector field changes. In particular, we did this by looking at the flux of the vector field through a closed path in two dimensions and then generalized these ideas to higher dimensions.

In , we measured how much of a vector field flowed through a section of a surface in three dimensions as a generalization of our argument from . In this section, we will connect the ideas of flux of a vector field through a closed surface in three dimensions and the divergence of that vector field.

Recalling the surfaces we studied in , where we applied , notice that all of these surfaces had the property that they had a boundary along which we calculated circulation. It turns out that giving a precise definition for boundary is challenging. For our purposes, however, you might think of a boundary curve of a surface as being a curve along which you could walk with your arms stretched out on either side with one arm lying over the surface and the other arm not lying over the surface as you walk. In this section, we will study closed surfaces in three dimensions, which are those surfaces without boundary. In , none of the surfaces were closed, because each had a boundary curve. On the other hand, in , we show two more surfaces that are not closed, as demonstrated by the magenta curves marking the edge where the surface ends.

In fact, the yellow cylinder has two edges where the surface ends (does not meet itself). The surfaces in are closed because there is no edge to the surface. Note that the cylinder has been filled in with a cap at top and bottom (plotted in gray and green, respectively) to become a closed surface.

Closed surfaces can be used to define the boundary of a volume in space. If we have the top and bottom on our cylinder, we have a well-defined volume of space, in that we know which points are inside the volume and which are outside of the volume. With different top and bottom surfaces, the enclosed volume would be different. illustrates three different ways to complete the cylindrical surface into a closed surface.

The preview activity showed how the flux through a closed surface can be subdivided into the flux through surfaces which combine to be the closed surface (with orientation switches corresponding to additive inverse). The net flux through the closed surface measures the net amount of the vector field that is created or destroyed on the interior of the closed surface.

Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.

The Divergence Theorem

The divergence of a vector field was developed as a measurement of the density with which the strength of vector field is changing. In three dimensions, the divergence measures the density per unit volume in which the vector field is being created or destroyed. This means that if we integrate the divergence of a vector field over a volume of space, we will get the net amount of the vector field that is created or destroyed in that particular volume of space. Since the net amount of a vector field that is created or destroyed in a volume of space is the same as the net flux of the vector field through the closed surface that is the boundary of that volume, we have a correspondence between a triple integral of the divergence of a vector field on the interior of a closed surface and the flux integral of the vector field over the closed surface.

The preview activity and the discussion before the statement of the have hopefully given you some intuition as to why the theorem is true. The ideas should also seem similar to the manner in which we approached and . In the next example, we will verify the Divergence Theorem for a particular case.

The next activity walks you through evaluating both the flux integrals necessary to calculate the flux directly and the triple integral given in the for a specific vector field and closed surface.

The next activity asks you to compute flux in some circumstances where it may be wise to apply .

Activity

Find the flux of the vector field \(\vF = \langle 3x^2+y^5,5+e^{z^3},z\rangle\) through the surface of the solid cube \(Q\) in \(\R^3\) with \(-2\leq x\leq 2\), \(-2\leq y\leq 2\), and \(-2\leq z\leq 2\).

Find the flux of the vector field \(\vG = \langle x^3,y^3,z^3\rangle\) through surface consisting of the top half of sphere of radius \(3\) centered at the origin and the disc of radius \(3\) in the \(xy\)-plane (centered at the origin).

Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

  1. Verify the divergence theorem for vector field \(\text{F}=〈x-y,x+z,z-y〉\) and surface S that consists of cone \({x}^{2}+{y}^{2}={z}^{2},0\le z\le 1,\) and the circular top of the cone (see the following figure). Assume this surface is oriented outward.

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    Let E be the solid cone enclosed by S. To verify the theorem for this example, we show that \({∭}_{E}\text{div}\ \text{F}dV=\underset{S}{∬}\text{F}\cdot d\text{S}\) by calculating each integral separately.

    To compute the triple integral, note that \(\text{div}\ \text{F}={P}_{x}+{Q}_{y}+{R}_{z}=2,\) and therefore the triple integral is

    \[\begin{array}{ll}{∭}_{E}\text{div}\ \text{F}dV & =2{∭}_{E}dV \\ & =2(\text{volume of}\ E).\end{array}\]

    The volume of a right circular cone is given by \(\pi {r}^{2}\frac{h}{3}.\) In this case, \(h=r=1.\) Therefore,

    \[{∭}_{E}\text{div}\ \text{F}dV=2(\text{volume of}\ E)=\frac{2\pi }{3}.\]

    To compute the flux integral, first note that S is piecewise smooth; S can be written as a union of smooth surfaces. Therefore, we break the flux integral into two pieces: one flux integral across the circular top of the cone and one flux integral across the remaining portion of the cone. Call the circular top \({S}_{1}\) and the portion under the top \({S}_{2}.\) We start by calculating the flux across the circular top of the cone. Notice that \({S}_{1}\) has parameterization

    \[\text{r}(u,v)=〈u\ \text{cos}\ v,u\ \text{sin}\ v,1〉,0\le u\le 1,0\le v\le 2\pi .\]

    Then, the tangent vectors are \({\text{t}}_{u}=〈\text{cos}\ v,\text{sin}\ v,0〉\) and \({\text{t}}_{v}=〈\text{-}u\ \text{cos}\ v,u\ \text{sin}\ v,0〉.\) Therefore, the flux across \({S}_{1}\) is

    \[\begin{array}{ll}{∬}_{{S}_{1}}\text{F}\cdot d\text{S} & ={\int }_{0}^{1}{\int }_{0}^{2\pi }\text{F}(\text{r}(u,v))\cdot ({\text{t}}_{u}\ \times \ {\text{t}}_{v})dA \\ & ={\int }_{0}^{1}{\int }_{0}^{2\pi }〈u\ \text{cos}\ v-u\ \text{sin}\ v,u\ \text{cos}\ v+1,1-u\ \text{sin}\ v〉\cdot 〈0,0,u〉dvdu \\ & ={\int }_{0}^{1}{\int }_{0}^{2\pi }u-{u}^{2}\text{sin}\ v\ dvdu=\pi .\end{array}\]

    We now calculate the flux over \({S}_{2}.\) A parameterization of this surface is

    \[\text{r}(u,v)=〈u\ \text{cos}\ v,u\ \text{sin}\ v,u〉,0\le u\le 1,0\le v\le 2\pi .\]

    The tangent vectors are \({\text{t}}_{u}=〈\text{cos}\ v,\text{sin}\ v,1〉\) and \({\text{t}}_{v}=〈\text{-}u\ \text{sin}\ v,u\ \text{cos}\ v,0〉,\) so the cross product is

    \[{\text{t}}_{u}\ \times \ {\text{t}}_{v}=〈\text{-}u\ \text{cos}\ v,\text{-}u\ \text{sin}\ v,u〉.\]

    Notice that the negative signs on the x and y components induce the inward orientation of the cone. Since the surface is oriented outward, we use vector \({\text{t}}_{v}\ \times \ {\text{t}}_{u}=〈u\ \text{cos}\ v,u\ \text{sin}\ v,\text{-}u〉\) in the flux integral. The flux across \({S}_{2}\) is then

    \[\begin{array}{ll}{∬}_{{S}_{2}}\text{F}\cdot d\text{S} & ={\int }_{0}^{1}{\int }_{0}^{2\pi }\text{F}(\text{r}(u,v))\cdot ({\text{t}}_{v}\ \times \ {\text{t}}_{u})dA \\ & ={\int }_{0}^{1}{\int }_{0}^{2\pi }〈u\ \text{cos}\ v-u\ \text{sin}\ v,u\ \text{cos}\ v+u,u-\text{sin}\ v〉\cdot 〈u\ \text{cos}\ v,u\ \text{sin}\ v,\text{-}u〉 \\ & ={\int }_{0}^{1}{\int }_{0}^{2\pi }{u}^{2}{\text{cos}}^{2}v+2{u}^{2}\text{sin}\ v-{u}^{2}dvdu=-\frac{\pi }{3}.\end{array}\]

    The total flux across S is

    \[{∬}_{{S}_{2}}\text{F}\cdot d\text{S}={∬}_{{S}_{1}}\text{F}\cdot d\text{S}+{∬}_{{S}_{2}}\text{F}\cdot d\text{S}=\frac{2\pi }{3}={∭}_{E}\text{div}\ \text{F}dV,\]

    and we have verified the divergence theorem for this example.

  2. Verify the divergence theorem for vector field \(\text{F}(x,y,z)=〈x+y+z,y,2x-y〉\) and surface S given by the cylinder \({x}^{2}+{y}^{2}=1,0\le z\le 3\) plus the circular top and bottom of the cylinder. Assume that S is oriented outward.

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    Both integrals equal \(6\pi .\)

  3. Calculate the surface integral \({∬}_{S}\text{F}\cdot d\text{S},\) where S is cylinder \({x}^{2}+{y}^{2}=1,0\le z\le 2,\) including the circular top and bottom, and \(\text{F}=〈\frac{{x}^{3}}{3}+yz,\frac{{y}^{3}}{3}-\text{sin}(xz),z-x-y〉.\)

    အဖြေကို ဖော်ပြပါ

    We could calculate this integral without the divergence theorem, but the calculation is not straightforward because we would have to break the flux integral into three separate integrals: one for the top of the cylinder, one for the bottom, and one for the side. Furthermore, each integral would require parameterizing the corresponding surface, calculating tangent vectors and their cross product, and using .

    By contrast, the divergence theorem allows us to calculate the single triple integral \({∭}_{E}\text{div}\ \text{F}dV,\) where E is the solid enclosed by the cylinder. Using the divergence theorem and converting to cylindrical coordinates, we have

    \[\begin{array}{ll}{∬}_{s}\text{F}\cdot d\text{S} & ={∭}_{E}\text{div}\ \text{F}\ dV \\ & ={∭}_{E}({x}^{2}+{y}^{2}+1)dV \\ & ={\int }_{0}^{2\pi }{\int }_{0}^{1}{\int }_{0}^{2}({r}^{2}+1)r\ dz\ dr\ d\theta \\ & =\frac{3}{2}{\int }_{0}^{2\pi }d\theta =3\pi .\end{array}\]

  4. Use the divergence theorem to calculate flux integral \({∬}_{S}\text{F}\cdot d\text{S},\) where S is the boundary of the box given by \(0\le x\le 2,1\le y\le 4,0\le z\le 1,\) and \(\text{F}=〈{x}^{2}+yz,y-z,2x+2y+2z〉\) (see the following figure).

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    30

  5. Let \(\text{v}=〈-\frac{y}{z},\frac{x}{z},0〉\) be the velocity field of a fluid. Let C be the solid cube given by \(1\le x\le 4,2\le y\le 5,1\le z\le 4,\) and let S be the boundary of this cube (see the following figure). Find the flow rate of the fluid across S.

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    The flow rate of the fluid across S is \({∬}_{S}\text{v}\cdot d\text{S}.\) Before calculating this flux integral, let’s discuss what the value of the integral should be. Based on , we see that if we place this cube in the fluid (as long as the cube doesn’t encompass the origin), then the rate of fluid entering the cube is the same as the rate of fluid exiting the cube. The field is rotational in nature and, for a given circle parallel to the xy-plane that has a center on the z-axis, the vectors along that circle are all the same magnitude. That is how we can see that the flow rate is the same entering and exiting the cube. The flow into the cube cancels with the flow out of the cube, and therefore the flow rate of the fluid across the cube should be zero.

    To verify this intuition, we need to calculate the flux integral. Calculating the flux integral directly requires breaking the flux integral into six separate flux integrals, one for each face of the cube. We also need to find tangent vectors, compute their cross product, and use . However, using the divergence theorem makes this calculation go much more quickly:

    \[\begin{array}{ll}{∬}_{S}\text{v}\cdot d\text{S} & ={∭}_{C}\text{div}(\text{v})dV \\ & ={∭}_{C}0\ dV=0.\end{array}\]

    Therefore the flux is zero, as expected.

  6. Let \(\text{v}=〈\frac{x}{z},\frac{y}{z},0〉\) be the velocity field of a fluid. Let C be the solid cube given by \(1\le x\le 4,2\le y\le 5,1\le z\le 4,\) and let S be the boundary of this cube (see the following figure). Find the flow rate of the fluid across S.

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    \(9\ \text{ln}(16)\)

  7. Verify that the divergence of \({\text{F}}_{r}\) is zero where \({\text{F}}_{r}\) is defined (away from the origin).

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    Since \(r=\sqrt{{x}^{2}+{y}^{2}+{z}^{2}},\) the quotient rule gives us

    \[\begin{array}{ll}\frac{∂}{∂x}(\frac{x}{{r}^{3}}) & =\frac{∂}{∂x}(\frac{x}{{({x}^{2}+{y}^{2}+{z}^{2})}^{3\text{/}2}}) \\ & =\frac{{({x}^{2}+{y}^{2}+{z}^{2})}^{3\text{/}2}-x[\frac{3}{2}{({x}^{2}+{y}^{2}+{z}^{2})}^{1\text{/}2}2x]}{{({x}^{2}+{y}^{2}+{z}^{2})}^{3}} \\ & =\frac{{r}^{3}-3{x}^{2}r}{{r}^{6}}=\frac{{r}^{2}-3{x}^{2}}{{r}^{5}}.\end{array}\]

    Similarly,

    \[\frac{∂}{∂y}(\frac{y}{{r}^{3}})=\frac{{r}^{2}-3{y}^{2}}{{r}^{5}}\ \text{and}\ \frac{∂}{∂z}(\frac{z}{{r}^{3}})=\frac{{r}^{2}-3{z}^{2}}{{r}^{5}}.\]

    Therefore,

    \[\begin{array}{ll}\text{div}\ {\text{F}}_{r} & =\frac{{r}^{2}-3{x}^{2}}{{r}^{5}}+\frac{{r}^{2}-3{y}^{2}}{{r}^{5}}+\frac{{r}^{2}-3{z}^{2}}{{r}^{5}} \\ & =\frac{3{r}^{2}-3({x}^{2}+{y}^{2}+{z}^{2})}{{r}^{5}} \\ & =\frac{3{r}^{2}-3{r}^{2}}{{r}^{5}}=0.\end{array}\]

  8. Suppose we have four stationary point charges in space, all with a charge of 0.002 Coulombs (C). The charges are located at \((0,1,1),(1,1,4),(-1,0,0),\ \text{and}\ (-2,-2,2).\) Let E denote the electrostatic field generated by these point charges. If S is the sphere of radius 2 oriented outward and centered at the origin, then find \({∬}_{S}\text{E}\cdot d\text{S}.\)

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    According to Gauss’ law, the flux of E across S is the total charge inside of S divided by the electric constant. Since S has radius 2, notice that only two of the charges are inside of S: the charge at \((0,1,1)\) and the charge at \((-1,0,0).\) Therefore, the total charge encompassed by S is 0.004 and, by Gauss’ law,

    \[{∬}_{S}\text{E}\cdot d\text{S}=\frac{0.004}{8.854\ \times \ {10}^{-12}}\approx 4.518\ \times \ {10}^{9}\ \text{V-m}.\]

  9. Work the previous example for surface S that is a sphere of radius 4 centered at the origin, oriented outward.

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    \(\approx 6.777\ \times \ {10}^{9}\)

  10. [T] \(\text{F}(x,y,z)=x\text{i}+y\text{j}+z\text{k};\) S is the surface of cube \(0\le x\le 1,0\le y\le 1,0

  11. [T] \(\text{F}(x,y,z)=(\text{cos}\ yz)\text{i}+{e}^{xz}\text{j}+3{z}^{2}\text{k}\text{;}\) S is the surface of hemisphere \(z=\sqrt{4-{x}^{2}-{y}^{2}}\) together with disk \({x}^{2}+{y}^{2}\le 4\) in the xy-plane.

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    \({\int }_{S}^{}\text{F}\cdot \text{N}ds=24\pi \approx 75.3982\)

  12. [T] \(\text{F}(x,y,z)=({x}^{2}+{y}^{2}-{x}^{2})\text{i}+{x}^{2}y\text{j}+3z\text{k};S\) is the surface of the unit cube \(0\le x\le 1,0\le y\le 1,0\le z\le 1\) excluding the face \(z=0\).

  13. [T] \(\text{F}(x,y,z)=x\text{i}+y\text{j}+z\text{k}\text{;}\) S is the surface of the solid bounded by the parabola \(z={x}^{2}+{y}^{2}\\) and the plane \(z=9\).

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    \({\int }_{S}^{}\text{F}\cdot \text{N}ds=\frac{243\pi }{2}\approx 381.704\)

  14. [T] \(\text{F}(x,y,z)={x}^{2}\text{i}+{y}^{2}\text{j}+{z}^{2}\text{k}\text{;}\) S is the surface of sphere \({x}^{2}+{y}^{2}+{z}^{2}=4.\)

  15. [T] \(\text{F}(x,y,z)=x\text{i}+y\text{j}+({z}^{2}-1)\text{k}\text{;}\) S is the surface of the solid bounded by cylinder \({x}^{2}+{y}^{2}=4\) and planes \(z=0\ \text{and}\ z=1.\)

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    \({\int }_{S}^{}\text{F}\cdot \text{N}ds=12\pi \approx 37.6991\)

  16. [T] \(\text{F}(x,y,z)=x{y}^{2}\text{i}+y{z}^{2}\text{j}+{x}^{2}z\text{k}\text{;}\) S is the surface of the solid bounded above by sphere \(\rho =2\) and below by cone \(\phi =\frac{\pi }{4}\) in spherical coordinates. (Think of S as the surface of an “ice cream cone.”)

  17. [T] \(\text{F}(x,y,z)={x}^{3}\text{i}+{y}^{3}\text{j}+3{a}^{2}z\text{k}\ \text{(constant}\ a>0)\text{;}\) S is the surface of the solid bounded by cylinder \({x}^{2}+{y}^{2}={a}^{2}\) and planes \(z=0\ \text{and}\ z=1.\)

    အဖြေကို ဖော်ပြပါ

    \({\int }_{S}^{}\text{F}\cdot \text{N}ds=\frac{9\pi {a}^{4}}{2}\)

  18. [T] Surface integral \({∬}_{S}\text{F}\cdot d\text{S},\) where S is the surface of the solid bounded by paraboloid \(z={x}^{2}+{y}^{2}\) and plane \(z=4,\) and \(\text{F}(x,y,z)=(x+{y}^{2}{z}^{2})\text{i}+(y+{z}^{2}{x}^{2})\text{j}+(z+{x}^{2}{y}^{2})\text{k}\)

  19. Use the divergence theorem to calculate surface integral \({∬}_{S}\text{F}\cdot d\text{S},\) where \(\text{F}(x,y,z)=({e}^{{y}^{2}})\text{i}+(y+\text{sin}({z}^{2}))\text{j}+(z-1)\text{k}\) and S is the surface of the solid bounded by the sphere \({x}^{2}+{y}^{2}+{z}^{2}=1\text{,}\) and below by the plane \(z=0\).

    အဖြေကို ဖော်ပြပါ

    \({∬}_{S}\text{F}\cdot d\text{S}=\frac{4\pi }{3}\)

  20. Use the divergence theorem to calculate surface integral \({∬}_{S}\text{F}\cdot ds,\) where \(\text{F}(x,y,z)={x}^{4}\text{i}-{x}^{3}{z}^{2}\text{j}+4x{y}^{2}z\text{k}\) and \(S\) is the surface bounded by cylinder \({x}^{2}+{y}^{2}=1\) and planes \(z=x+2\) and \(z=0.\)

  21. Use the divergence theorem to calculate surface integral \({∬}_{S}\text{F}\cdot d\text{S}\) when \(\text{F}(x,y,z)={x}^{2}{z}^{3}\text{i}+2xy{z}^{3}\text{j}+x{z}^{4}\text{k}\) and S is the surface of the box with vertices \((\pm 1,\pm 2,\pm 3).\)

    အဖြေကို ဖော်ပြပါ

    \({∬}_{S}\text{F}\cdot d\text{S}=0\)

  22. Use the divergence theorem to calculate surface integral \({∬}_{S}\text{F}\cdot d\text{S}\) when \(\text{F}(x,y,z)=z\ {\text{tan}}^{-1}({y}^{2})\text{i}+{z}^{3}\text{ln}({x}^{2}+1)\text{j}+z\text{k}\) and S is the surface of the solid bounded by the paraboloid \({x}^{2}+{y}^{2}+z=2\) and the plane \(z=1\).

  23. [T] Use a CAS and the divergence theorem to calculate flux \({∬}_{S}\text{F}\cdot d\text{S},\) where \(\text{F}(x,y,z)=({x}^{3}+{y}^{3})\text{i}+({y}^{3}+{z}^{3})\text{j}+({z}^{3}+{x}^{3})\text{k}\) and S is a sphere with center (0, 0, 0) and radius 2.

    အဖြေကို ဖော်ပြပါ

    \({∬}_{S}\text{F}\cdot d\text{S}=\frac{384\pi }{5}\approx 241.2743\)

  24. Use the divergence theorem to compute the value of flux integral \({∬}_{S}\text{F}\cdot d\text{S},\) where \(\text{F}(x,y,z)=({y}^{3}+3x)\text{i}+(xz+y)\text{j}+[z+{x}^{4}\text{cos}({x}^{2}y)]\text{k}\) and S is the surface of the solid bounded by \({x}^{2}+{y}^{2}=1,x\ge 0,y\ge 0,\ \text{and}\ 0\le z\le 1.\)

  25. Use the divergence theorem to compute flux integral \({∬}_{S}\text{F}\cdot d\text{S},\) where \(\text{F}(x,y,z)=y\text{j}-z\text{k}\) and S consists of the union of paraboloid \(y={x}^{2}+{z}^{2},0\le y\le 1,\) and disk \({x}^{2}+{z}^{2}\le 1,y=1,\) oriented outward. What is the flux through just the paraboloid?

    အဖြေကို ဖော်ပြပါ

    \({∬}_{D}\text{F}\cdot d\text{S}=\) Net flux \(=0\); flux through the paraboloid \(=-\pi\)

  26. Use the divergence theorem to compute flux integral \({∬}_{S}\text{F}\cdot d\text{S},\) where \(\text{F}(x,y,z)=x+y\text{j}+{z}^{4}\text{k}\) and S is a part of cone \(z=\sqrt{{x}^{2}+{y}^{2}}\) beneath top plane \(z=1,\) oriented downward.

  27. Use the divergence theorem to calculate surface integral \({∬}_{S}\text{F}\cdot d\text{S}\) for \(\text{F}(x,y,z)={x}^{4}\text{i}-{x}^{3}{z}^{2}\text{j}+4x{y}^{2}z\text{k},\) where S is the surface inside the cylinder \({x}^{2}+{y}^{2}=1\) between the planes \(z=x+2\ \text{and}\ z=0.\)

    အဖြေကို ဖော်ပြပါ

    \({∬}_{S}\text{F}\cdot d\text{S}=\frac{2\pi }{3}\)

  28. Consider \(\text{F}(x,y,z)={x}^{2}\text{i}+xy\text{j}+(z+1)\text{k}.\) Let E be the solid enclosed by paraboloid \(z=4-{x}^{2}-{y}^{2}\) and plane \(z=0\) with normal vectors pointing outside E. Compute flux F across the boundary of E using the divergence theorem.

  29. [T] \(\text{F}=〈x,-2y,3z〉;\) S is sphere \(\{(x,y,z):{x}^{2}+{y}^{2}+{z}^{2}=6\}.\)

    အဖြေကို ဖော်ပြပါ

    \(16\sqrt{6}\pi\)

  30. [T] \(\text{F}=〈x,2y,z〉;\) S is the boundary of the tetrahedron in the first octant formed by plane \(x+y+z=1.\)

  31. [T] \(\text{F}=〈y-2x,{x}^{3}-y,{y}^{2}-z〉;\) S is sphere \(\{(x,y,z):{x}^{2}+{y}^{2}+{z}^{2}=4\}.\)

    အဖြေကို ဖော်ပြပါ

    \(-\frac{128\pi }{3}\)

  32. [T] \(\text{F}=〈x,y,z〉;\) S is the surface of paraboloid \(z=4-{x}^{2}-{y}^{2},\) for \(z\ge 0,\) plus its base in the xy-plane.

  33. [T] \(\text{F}=〈z-x,x-y,2y-z〉;\) D is the region between spheres of radius 2 and 4 centered at the origin.

    အဖြေကို ဖော်ပြပါ

    \(-224\pi \approx -703.7168\)

  34. [T] \(\text{F}=\frac{\text{r}}{||\text{r}||}=\frac{〈x,y,z〉}{\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}};\) D is the region between spheres of radius 1 and 2 centered at the origin.

  35. [T] \(\text{F}=〈{x}^{2},\text{-}{y}^{2},{z}^{2}〉;\) D is the region in the first octant between planes \(z=4-x-y\) and \(z=2-x-y.\)

    အဖြေကို ဖော်ပြပါ

    20

  36. Let \(\text{F}(x,y,z)=2x\text{i}-3xy\text{j}+x{z}^{2}\text{k}.\) Use the divergence theorem to calculate \({∬}_{S}\text{F}\cdot d\text{S},\) where S is the surface of the cube with corners at \((0,0,0),(1,0,0),(0,1,0),\) \((1,1,0),(0,0,1),(1,0,1),(0,1,1),\ \text{and}\ (1,1,1),\) oriented outward.

  37. Use the divergence theorem to find the outward flux of field \(\text{F}(x,y,z)=({x}^{3}-3y)\text{i}+(2yz+1)\text{j}+xyz\text{k}\) through the cube bounded by planes \(x=\pm 1,y=\pm 1,\ \text{and}\ z=\pm 1.\)

    အဖြေကို ဖော်ပြပါ

    \({∬}_{S}\text{F}\cdot d\text{S}=8\)

  38. Let \(\text{F}(x,y,z)=2x\text{i}-3y\text{j}+5z\text{k}\) and let S be hemisphere \(z=\sqrt{9-{x}^{2}-{y}^{2}}\) together with disk \({x}^{2}+{y}^{2}\le 9\) in the xy-plane. Use the divergence theorem to calculate \({∬}_{S}F⋅dS\).

  39. Evaluate \({∬}_{S}\text{F}\cdot \text{N}dS,\) where \(\text{F}(x,y,z)={x}^{2}\text{i}+xy\text{j}+{x}^{3}{y}^{3}\text{k}\) and S is the surface consisting of all faces of the tetrahedron bounded by plane \(x+y+z=1\) and the coordinate planes, with outward unit normal vector N.

    အဖြေကို ဖော်ပြပါ

    \({∬}_{S}\text{F}\cdot \text{N}dS=\frac{1}{8}\)

  40. Find the net outward flux of field \(\text{F}=〈bz-cy,cx-az,ay-bx〉\) across any smooth closed surface in \({\text{R}}^{3},\) where a, b, and c are constants.

Symbols used here

\int f(x)\,dx,\ \int_a^b
integral
Antiderivative (indefinite) or signed area from a to b (definite).
\sqrt{x},\ \sqrt[n]{x}
square root, n-th root
The non-negative number whose square (n-th power) is x.
\pi
pi
Ratio of a circle's circumference to its diameter, 3.14159…
\theta
theta
The usual name for an angle.
\varepsilon,\ \delta
epsilon, delta
Small positive tolerances in the definition of a limit.
i
imaginary unit
i² = −1.
\leq,\ \geq
less/greater than or equal
Inequalities that allow equality; < and > exclude it.
\frac{\partial f}{\partial x}
partial derivative
Derivative with respect to x, holding the other variables fixed.
\nabla f
gradient (nabla, del)
Vector of partial derivatives; points uphill.
\iint,\ \oint
double / contour integral
Integral over a region of the plane; integral around a closed curve.
\mathbf{v},\ \vec{v}
vector
A quantity with magnitude and direction; a column of numbers.
\mathbf{u} \cdot \mathbf{v},\ \|\mathbf{v}\|
dot product, norm
Σ u_i v_i; the length of v, √(v·v).

How to: The Divergence Theorem

  1. Explain the meaning of the divergence theorem.
  2. Use the divergence theorem to calculate the flux of a vector field.
  3. Apply the divergence theorem to an electrostatic field.
  4. The divergence theorem relates a surface integral across closed surface
  5. The divergence theorem can be used to transform a difficult flux integral into an easier triple integral and vice versa.
  6. The divergence theorem can be used to derive Gauss’ law, a fundamental law in electrostatics.

Questions people ask

What is a partial derivative?

The ordinary derivative with respect to one variable while every other variable is frozen — the slope of the surface in one coordinate direction.

What does the gradient point at?

Uphill: the direction of steepest increase, with length equal to that steepest slope. It is perpendicular to the level curves.

သင့်ရဲ့ကိုယ်ပိုင်စမ်းသပ်

Parts of this page are adapted from Boelkins et al., Active Calculus Multivariable (CC BY-SA 4.0), OpenStax Calculus Volume 3 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

ပိုပြီး Multivariable Calculus

Functions of several variablesPartial derivatives and the gradientOptimisation in several variablesDouble and triple integralsVector fields, line integrals and the big theorems