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16.8: The Divergence Theorem - Mathematics


Learning Objectives

  • Explain the meaning of the divergence theorem.
  • Use the divergence theorem to calculate the flux of a vector field.
  • Apply the divergence theorem to an electrostatic field.

We have examined several versions of the Fundamental Theorem of Calculus in higher dimensions that relate the integral around an oriented boundary of a domain to a “derivative” of that entity on the oriented domain. In this section, we state the divergence theorem, which is the final theorem of this type that we will study. The divergence theorem has many uses in physics; in particular, the divergence theorem is used in the field of partial differential equations to derive equations modeling heat flow and conservation of mass. We use the theorem to calculate flux integrals and apply it to electrostatic fields.

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 vecs abla f cdot dvecs 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 ( abla f) over path (C) to a difference of (f) evaluated on the boundary of (C).
  3. Green’s theorem, circulation form: [iint_D (Q_x - P_y),dA = int_C vecs F cdot dvecs r.] Since (Q_x - P_y = ext{curl } vecs F cdot mathbf{hat k}) and curl is a derivative of sorts, Green’s theorem relates the integral of derivative curl (vecs F) over planar region (D) to an integral of (vecs F) over the boundary of (D).
  4. Green’s theorem, flux form: [iint_D (P_x + Q_y),dA = int_C vecs F cdot vecs N , dS.] Since (P_x + Q_y = ext{div }vecs F) and divergence is a derivative of sorts, the flux form of Green’s theorem relates the integral of derivative div (vecs F) over planar region (D) to an integral of (vecs F) over the boundary of (D).
  5. Stokes’ theorem: [iint_S curl , vecs F cdot dvecs S = int_C vecs F cdot dvecs r.] If we think of the curl as a derivative of sorts, then Stokes’ theorem relates the integral of derivative curl (vecs F) over surface (S) (not necessarily planar) to an integral of (vecs 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 div (vecs F) over a solid to a flux integral of (vecs F) over the boundary of the solid. More specifically, the divergence theorem relates a flux integral of vector field (vecs F) over a closed surface (S) to a triple integral of the divergence of (vecs F) over the solid enclosed by (S).

The Divergence Theorem

Let (S) be a piecewise, smooth closed surface that encloses solid (E) in space. Assume that (S) is oriented outward, and let (vecs F) be a vector field with continuous partial derivatives on an open region containing (E) (Figure (PageIndex{1})). Then

[iiint_E ext{div }vecs F , dV = iint_S vecs F cdot dvecs S. label{divtheorem}]

Recall that the flux form of Green’s theorem states that

[ iint_D ext{div }vecs F , dA = int_C vecs F cdot vecs 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.

Proof

Let (B) be a small box with sides parallel to the coordinate planes inside (E) (Figure (PageIndex{2a})). Let the center of (B) have coordinates ((x,y,z)) and suppose the edge lengths are (Delta x, , Delta y), and (Delta z). (Figure (PageIndex{1b})). The normal vector out of the top of the box is (mathbf{hat k}) and the normal vector out of the bottom of the box is (-mathbf{hat k}). The dot product of (vecs F = langle P, Q, R angle) with (mathbf{hat k}) is (R) and the dot product with (-mathbf{hat k}) is (-R). The area of the top of the box (and the bottom of the box) (Delta S) is (Delta x Delta y).

The flux out of the top of the box can be approximated by (R left(x,, y,, z + frac{Delta z}{2} ight) ,Delta x ,Delta y) (Figure (PageIndex{2c})) and the flux out of the bottom of the box is (- R left(x,, y,, z - frac{Delta z}{2} ight) ,Delta x ,Delta y). If we denote the difference between these values as (Delta R), then the net flux in the vertical direction can be approximated by (Delta R, Delta x ,Delta y). However,

[Delta R ,Delta x ,Delta y = left(frac{Delta R}{Delta z} ight) ,Delta x ,Delta y Delta z approx left(frac{partial R}{partial z} ight) ,Delta V. onumber]

Therefore, the net flux in the vertical direction can be approximated by (left(frac{partial R}{partial z} ight)Delta V). Similarly, the net flux in the (x)-direction can be approximated by (left(frac{partial P}{partial x} ight),Delta V) and the net flux in the (y)-direction can be approximated by (left(frac{partial Q}{partial y} ight),Delta V). Adding the fluxes in all three directions gives an approximation of the total flux out of the box:

[ ext{Total flux }approx left(frac{partial P}{partial x} + frac{partial Q}{partial y} + frac{partial R}{partial z} ight) Delta V = ext{div }vecs F ,Delta V. onumber]

This approximation becomes arbitrarily close to the value of the total flux as the volume of the box shrinks to zero.

The sum of ( ext{div }vecs F ,Delta V) over all the small boxes approximating (E) is approximately (iiint_E ext{div }vecs F ,dV). On the other hand, the sum of ( ext{div }vecs F ,Delta V) over all the small boxes approximating (E) is the sum of the fluxes over all these boxes. Just as in the informal proof of Stokes’ theorem, adding these fluxes over all the boxes results in the cancelation of a lot of the terms. If an approximating box shares a face with another approximating box, then the flux over one face is the negative of the flux over the shared face of the adjacent box. These two integrals cancel out. When adding up all the fluxes, the only flux integrals that survive are the integrals over the faces approximating the boundary of (E). As the volumes of the approximating boxes shrink to zero, this approximation becomes arbitrarily close to the flux over (S).

(Box)

Example (PageIndex{1}): Verifying the Divergence Theorem

Verify the divergence theorem for vector field (vecs F = langle x - y, , x + z, , z - y angle) and surface (S) that consists of cone (x^2 + y^2 = z^2, , 0 leq z leq 1), and the circular top of the cone (see the following figure). Assume this surface is positively oriented.

Solution

Let (E) be the solid cone enclosed by (S). To verify the theorem for this example, we show that

[iiint_E ext{div } vecs F ,dV = iint_S vecs F cdot dvecs S onumber]

by calculating each integral separately.

To compute the triple integral, note that ( ext{div } vecs F = P_x + Q_y + R_z = 2), and therefore the triple integral is

[ egin{align*} iiint_E ext{div } vecs F , dV &= 2 iiint_E dV [4pt] &= 2 , (volume , of , E). end{align*}]

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

[iiint_E ext{div } vecs F ,dV = 2 , (volume , of , E) = frac{2pi}{3}. onumber]

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

[vecs r(u,v) = langle u , cos v, , u , sin v, , 1 angle, , 0 leq u leq 1, , 0 leq v leq 2pi. onumber]

Then, the tangent vectors are (vecs t_u = langle cos v, , sin v, , 0 angle ) and (vecs t_v = langle -u , sin v, , u , cos v, 0 angle ). Therefore, the flux across (S_1) is

[ egin{align*} iint_{S_1} vecs F cdot dvecs S &= int_0^1 int_0^{2pi} vecs F (vecs r ( u,v)) cdot (vecs t_u imes vecs t_v) , dA [4pt] &= int_0^1 int_0^{2pi} langle u , cos v - u , sin v, , u , cos v + 1, , 1 - u , sin v angle cdot langle 0,0,u angle , dv, du [4pt] &= int_0^1 int_0^{2pi} u - u^2 sin v , dv du [4pt] &= pi. end{align*}]

We now calculate the flux over (S_2). A parameterization of this surface is

[vecs r(u,v) = langle u , cos v, , u , sin v, , u angle, , 0 leq u leq 1, , 0 leq v leq 2pi. onumber]

The tangent vectors are (vecs t_u = langle cos v, , sin v, , 1 angle ) and (vecs t_v = langle -u , sin v, , u , cos v, 0 angle ), so the cross product is

[vecs t_u imes vecs t_v = langle - u , cos v, , -u , sin v, , u angle. onumber]

Notice that the negative signs on the (x) and (y) components induce the negative (or inward) orientation of the cone. Since the surface is positively oriented, we use vector (vecs t_v imes vecs t_u = langle u , cos v, , u , sin v, , -u angle) in the flux integral. The flux across (S_2) is then

[ egin{align*} iint_{S_2} vecs F cdot dvecs S &= int_0^1 int_0^{2pi} vecs F ( vecs r ( u,v)) cdot (vecs t_u imes vecs t_v) , dA [4pt] &= int_0^1 int_0^{2pi} langle u , cos v - u , sin v, , u , cos v + u, , u , - usin v angle cdot langle u , cos v, , u , sin v, , -u angle,dv,du [4pt] &= int_0^1 int_0^{2pi} u^2 cos^2 v + 2u^2 sin v - u^2 ,dv,du [4pt] &= -frac{pi}{3} end{align*}]

The total flux across (S) is

[iint_{S} vecs F cdot dvecs S = iint_{S_1}vecs F cdot dvecs S + iint_{S_2} vecs F cdot dvecs S = frac{2pi}{3} = iiint_E ext{div } vecs F ,dV, onumber]

and we have verified the divergence theorem for this example.

Exercise (PageIndex{1})

Verify the divergence theorem for vector field (vecs F (x,y,z) = langle x + y + z, , y, , 2x - y angle) and surface (S) given by the cylinder (x^2 + y^2 = 1, , 0 leq z leq 3) plus the circular top and bottom of the cylinder. Assume that (S) is positively oriented.

Hint

Calculate both the flux integral and the triple integral with the divergence theorem and verify they are equal.

Answer

Both integrals equal (6pi).

​​Recall that the divergence of continuous field (vecs F) at point (P) is a measure of the “outflowing-ness” of the field at (P). If (vecs F) represents the velocity field of a fluid, then the divergence can be thought of as the rate per unit volume of the fluid flowing out less the rate per unit volume flowing in. The divergence theorem confirms this interpretation. To see this, let (P) be a point and let (B_{ au}) be a ball of small radius (r) centered at (P) (Figure (PageIndex{3})). Let (S_{ au}) be the boundary sphere of (B_{ au}). Since the radius is small and (vecs F) is continuous, ( ext{div }vecs F(Q) approx ext{div }vecs F(P)) for all other points (Q) in the ball. Therefore, the flux across (S_{ au}) can be approximated using the divergence theorem:

Exercise (PageIndex{2})

Use the divergence theorem to calculate flux integral [iint_S vecs F cdot dvecs S, onumber] where (S) is the boundary of the box given by (0 leq x leq 2, , 0 leq y leq 4, , 0 leq z leq 1) and (vecs F = langle x^2 + yz, , y - z, , 2x + 2y + 2z angle ) (see the following figure).

Hint

Calculate the corresponding triple integral.

Answer

40

Example (PageIndex{3}): Applying the Divergence Theorem

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

Solution

The flow rate of the fluid across (S) is (iint_S vecs v cdot dvecs S). Before calculating this flux integral, let’s discuss what the value of the integral should be. Based on Figure (PageIndex{4}), 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. However, using the divergence theorem makes this calculation go much more quickly:

[ egin{align*} iint_S vecs v cdot dvecs S &= iiint_C ext{div }vecs v , dV [4pt]
&= iiint_C 0 , dV = 0.end{align*}]

Therefore the flux is zero, as expected.

Exercise (PageIndex{3})

Let (vecs v = leftlangle frac{x}{z}, , frac{y}{z}, , 0 ight angle) be the velocity field of a fluid. Find the flow rate of the fluid across (S).

Hint

Use the divergence theorem and calculate a triple integral

Answer

(9 , ln (16))

Example illustrates a remarkable consequence of the divergence theorem. Let (S) be a piecewise, smooth closed surface and let (vecs F) be a vector field defined on an open region containing the surface enclosed by (S). If (vecs F) has the form (F = langle f (y,z), , g(x,z), , h(x,y) angle), then the divergence of (vecs F) is zero. By the divergence theorem, the flux of (vecs 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 (iint_S vecs F cdot dvecs S) where (S) is a cube and

[vecs F = langle sin (y) , e^{yz}, , x^2z^2, , cos (xy) , e^{sin x} angle.]

Calculating the flux integral directly would be difficult, if not impossible, using techniques we studied previously. At the very least, we would have to break the flux integral into six integrals, one for each face of the cube. But, because the divergence of this field is zero, the divergence theorem immediately shows that the flux integral is zero.

We can now use the divergence theorem to justify the physical interpretation of divergence that we discussed earlier. Recall that if (vecs F) is a continuous three-dimensional vector field and (P) is a point in the domain of (vecs F), then the divergence of (vecs F) at (P) is a measure of the “outflowing-ness” of (vecs F) at (P). If (vecs F) represents the velocity field of a fluid, then the divergence of (vecs F) at (P) is a measure of the net flow rate out of point (P) (the flow of fluid out of (P) less the flow of fluid in to (P)). To see how the divergence theorem justifies this interpretation, let (B_{ au}) be a ball of very small radius r with center (P), and assume that (B_{ au}) is in the domain of (vecs F). Furthermore, assume that (B_{ au}) has a positive, outward orientation. Since the radius of (B_{ au}) is small and (vecs F) is continuous, the divergence of (vecs F) is approximately constant on (B_{ au}). That is, ifv (P') is any point in (B_{ au}), then ( ext{div } vecs F(P) approx ext{div } vecs F(P')). Let (S_{ au}) denote the boundary sphere of (B_{ au}). We can approximate the flux across (S_{ au}) using the divergence theorem as follows:

[egin{align*} iint_{S_{ au}} vecs F cdot dvecs S &= iiint_{B_{ au}} ext{div }vecs F , dV [4pt]
&approx iiint_{B_{ au}} ext{div } vecs F (P) , dV [4pt]
&= ext{div } vecs F (P) , V(B_{ au}). end{align*}]

As we shrink the radius (r) to zero via a limit, the quantity ( ext{div }vecs F (P) , V(B_{ au})) gets arbitrarily close to the flux. Therefore,

[ ext{div }vecs F(P) = lim_{ au ightarrow 0} frac{1}{V(B_{ au})} iint_{S_{ au}} vecs F cdot dvecs S]

and we can consider the divergence at (P) as measuring the net rate of outward flux per unit volume at (P). Since “outflowing-ness” is an informal term for the net rate of outward flux per unit volume, we have justified the physical interpretation of divergence we discussed earlier, and we have used the divergence theorem to give this justification.

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 (vecs E), then the flux of (vecs 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 (vecs F_{ au}) denote radial vector field (vecs F_{ au} = dfrac{1}{ au^2} leftlangle dfrac{x}{ au}, , dfrac{y}{ au}, , dfrac{z}{ au} ight angle ).The vector at a given position in space points in the direction of unit radial vector (leftlangle dfrac{x}{ au}, , dfrac{y}{ au}, , dfrac{z}{ au} ight angle ) and is scaled by the quantity (1/ au^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 (vecs E) given by

[vecs E = dfrac{q}{4pi epsilon_0}vecs F_{ au},]

where the approximation (epsilon_0 = 8.854 imes 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 (vecs E) is a radial vector field similar to the gravitational field described in [link]. The difference is that this field points outward whereas the gravitational field points inward. Because

[vecs E = dfrac{q}{4pi epsilon_0}vecs F_{ au} = dfrac{q}{4pi epsilon_0}left(dfrac{1}{ au^2} leftlangle dfrac{x}{ au}, , dfrac{y}{ au}, , dfrac{z}{ au} ight angle ight),]

we say that electrostatic fields obey an inverse-square law. That is, the electrostatic force at a given point is inversely proportional to the square of the distance from the source of the charge (which in this case is at the origin). Given this vector field, we show that the flux across closed surface (S) is zero if the charge is outside of (S), and that the flux is (q/epsilon_0) if the charge is inside of (S). In other words, the flux across S is the charge inside the surface divided by constant (epsilon_0). This is a special case of Gauss’ law, and here we use the divergence theorem to justify this special case.

To show that the flux across (S) is the charge inside the surface divided by constant (epsilon_0), we need two intermediate steps. First we show that the divergence of (vecs F_{ au}) is zero and then we show that the flux of (vecs F_{ au}) across any smooth surface (S) is either zero or (4pi). We can then justify this special case of Gauss’ law.

Example (PageIndex{4}): The Divergence of (F_{ au}) is Zero

Verify that the divergence of (vecs F_{ au}) is zero where (vecs F_{ au}) is defined (away from the origin).

Solution

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

[ egin{align*} dfrac{partial}{partial x} left( dfrac{x}{ au^3} ight) &= dfrac{partial}{partial x} left( dfrac{x}{(x^2+y^2+z^2)^{3/2}} ight) [4pt]
&= dfrac{(x^2+y^2+z^2)^{3/2} - xleft[dfrac{3}{2} (x^2+y^2+z^2)^{1/2}2x ight]}{(x^2+y^2+z^2)^3} [4pt]
&= dfrac{ au^3 -3x^2 au}{ au^6} = dfrac{ au^2 - 3x^2}{ au^5}. end{align*}]

Similarly,

[dfrac{partial}{partial y} left( dfrac{y}{ au^3} ight) = dfrac{ au^2 - 3y^2}{ au^5} , and , dfrac{partial}{partial z} left( dfrac{z}{ au^3} ight) = dfrac{ au^2 - 3z^2}{ au^5}. onumber ]

Therefore,

[ egin{align*} ext{div } vecs F_{ au} &= dfrac{ au^2 - 3x^2}{ au^5} + dfrac{ au^2 - 3y^2}{ au^5} + dfrac{ au^2 - 3z^2}{ au^5} [4pt]
&= dfrac{3 au^2 - 3(x^2+y^2+z^2)}{ au^5} [4pt]
&= dfrac{3 au^2 - 3 au^2}{ au^5} = 0. end{align*}]

Notice that since the divergence of (vecs F_{ au}) is zero and (vecs E) is (vecs F_{ au}) scaled by a constant, the divergence of electrostatic field (vecs E) is also zero (except at the origin).

Flux across a Smooth Surface

Let (S) be a connected, piecewise smooth closed surface and let (vecs F_{ au} = dfrac{1}{ au^2} leftlangle dfrac{x}{ au}, , dfrac{y}{ au}, , dfrac{z}{ au} ight angle). Then,

[iint_S vecs F_{ au} cdot dvecs S = egin{cases}0, & ext{if }S ext{ does not encompass the origin} 4pi, & ext{if }S ext{ encompasses the origin.} end{cases}]

In other words, this theorem says that the flux of (vecs F_{ au}) across any piecewise smooth closed surface (S) depends only on whether the origin is inside of (S).

Proof

The logic of this proof follows the logic of [link], only we use the divergence theorem rather than Green’s theorem.

First, suppose that (S) does not encompass the origin. In this case, the solid enclosed by (S) is in the domain of (vecs F_{ au}), and since the divergence of (vecs F_{ au}) is zero, we can immediately apply the divergence theorem and find that [iint_S vecs F cdot dvecs S ] is zero.

Now suppose that (S) does encompass the origin. We cannot just use the divergence theorem to calculate the flux, because the field is not defined at the origin. Let (S_a) be a sphere of radius a inside of (S) centered at the origin. The outward normal vector field on the sphere, in spherical coordinates, is

[vecs t_{phi} imes vecs t_{ heta} = langle a^2 cos heta , sin^2 phi, , a^2 sin heta , sin^2 phi, , a^2 sin phi , cos phi angle]

(see [link]). Therefore, on the surface of the sphere, the dot product (vecs F_{ au} cdot vecs N) (in spherical coordinates) is

[ egin{align*} vecs F_{ au} cdot vecs N &= left langle dfrac{sin phi , cos heta}{a^2}, , dfrac{sin phi , sin heta}{a^2}, , dfrac{cos phi}{a^2} ight angle cdot langle a^2 cos heta , sin^2 phi, a^2 sin heta , sin^2 phi, , a^2 sin phi , cos phi angle [4pt]
&= sin phi ( langle sin phi , cos heta, , sin phi , sin heta, , cos phi angle cdot langle sin phi , cos heta, sin phi , sin heta, , cos phi angle ) [4pt]
&= sin phi. end{align*}]

The flux of (vecs F_{ au}) across (S_a) is

[iint_{S_a} vecs F_{ au} cdot vecs N dS = int_0^{2pi} int_0^{pi} sin phi , dphi , d heta = 4pi.]

Now, remember that we are interested in the flux across (S), not necessarily the flux across (S_a). To calculate the flux across (S), let (E) be the solid between surfaces (S_a) and (S). Then, the boundary of (E) consists of (S_a) and (S). Denote this boundary by (S - S_a) to indicate that (S) is oriented outward but now (S_a) is oriented inward. We would like to apply the divergence theorem to solid (E). Notice that the divergence theorem, as stated, can’t handle a solid such as (E) because (E) has a hole. However, the divergence theorem can be extended to handle solids with holes, just as Green’s theorem can be extended to handle regions with holes. This allows us to use the divergence theorem in the following way. By the divergence theorem,

[ egin{align*} iint_{S-S_a} vecs F_{ au} cdot dvecs S &= iint_S vecs F_{ au} cdot dvecs S - iint_{S_a} vecs F_{ au} cdot dvecs S [4pt]
&= iiint_E ext{div } vecs F_{ au} , dV [4pt]
&= iiint_E 0 , dV = 0. end{align*}]

Therefore,

[iint_S vecs F_{ au} cdot dvecs S = iint_{S_a} vecs F_{ au} cdot dvecs S = 4pi, onumber]

and we have our desired result.

(Box)

Now we return to calculating the flux across a smooth surface in the context of electrostatic field (vecs E = dfrac{q}{4pi epsilon_0} vecs F_{ au} ) of a point charge at the origin. Let (S) be a piecewise smooth closed surface that encompasses the origin. Then

[ egin{align*} iint_S vecs E cdot dvecs S &= iint_S dfrac{q}{4pi epsilon_0} vecs F_{ au} cdot dvecs S[4pt]
&= dfrac{q}{4pi epsilon_0} iint_S vecs F_{ au} cdot dvecs S [4pt]
&= dfrac{q}{epsilon_0}. end{align*}]

If (S) does not encompass the origin, then

[iint_S vecs E cdot dvecs S = dfrac{q}{4pi epsilon_0} iint_S vecs F_{ au} cdot dvecs S = 0. onumber]

Therefore, we have justified the claim that we set out to justify: the flux across closed surface (S) is zero if the charge is outside of (S), and the flux is (q/epsilon_0) if the charge is inside of (S).

This analysis works only if there is a single point charge at the origin. In this case, Gauss’ law says that the flux of (vecs E) across (S) is the total charge enclosed by (S). Gauss’ law can be extended to handle multiple charged solids in space, not just a single point charge at the origin. The logic is similar to the previous analysis, but beyond the scope of this text. In full generality, Gauss’ law states that if (S) is a piecewise smooth closed surface and (Q) is the total amount of charge inside of (S), then the flux of (vecs E) across (S) is (Q/epsilon_0).

Example (PageIndex{5}): Using Gauss’ law

Suppose we have four stationary point charges in space, all with a charge of 0.002 Coulombs (C). The charges are located at ((0,0,1), , (1,1,4), (-1,0,0)), and ((-2,-2,2)). Let (vecs 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

[iint_S vecs E cdot dvecs S. onumber]

Solution

According to Gauss’ law, the flux of (vecs 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,

[iint_S vecs E cdot dvecs S = dfrac{0.004}{8.854 imes 10^{-12}} approx 4.418 imes 10^9 , V - m. onumber]

Exercise (PageIndex{4})

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

Hint

Use Gauss’ law.

Answer

(approx 6.777 imes 10^9)

divergence theorem
a theorem used to transform a difficult flux integral into an easier triple integral and vice versa
Gauss’ law
if S is a piecewise, smooth closed surface in a vacuum and (Q) is the total stationary charge inside of (S), then the flux of electrostatic field (vecs E) across (S) is (Q/epsilon_0)
inverse-square law
the electrostatic force at a given point is inversely proportional to the square of the distance from the source of the charge

Divergence theorem

The divergence theorem (also called Gauss's theorem or Gauss-Ostrogradsky theorem) is a theorem which relates the flux of a vector field through a closed surface to the vector field inside the surface. The theorem states that the outward flux of a vector field through a closed surface is equal to the triple integral of the divergence of the vector field inside the surface.

What this theorem actually states is the physical fact that in the absence of the creation or destruction of matter, the density within a region of space can change only by having it flow into or away from the region through its boundary.

The theorem is very applicable in different areas of physics, among others electrostatics and fluid dynamics. Another very important application of the theorem is that several physical laws can be written in both differential and integral form, see, for instance, Gauss' law for an application.


Vector Calculus - 1

For a scalar function (x, y, z) = x 2 + 3y 2 + 2z 2 , the directional derivative at the point P( 1, 2, -1) is the direction of a vector is

We have,


So, the direction derivative in the direction of

Use the divergence theorem the value of where, S is any closed surface enclosing volume V.


where, is an outward drawn unit normal vector to S.

Find the value of

So, from vector identity

If then the value of div at the point (1, 1, -1) will be

Apply Stoke&rsquos theorem, the value of where C is the boundary of the triangle with vertices (2, 0, 0), (0, 3, 0) and (0, 0, 6) is

Taking projection on three planes, we note that the surface S consists of three triangles, &Delta OAB in XT- plane, &Delta OBC in TZ-plane and &Delta OAC in XZ-plane. Using two point formula, the equation of the line AB, BC, CA are respectively 3x + 2y = 6 , 2y + z = 6 , 3x + z = 6


So, by Stake&rsquos theorem


The line integral from the origin to the point P( 1,1,1) is

cannot be determined without specifying the path


So, potential function of

So, line integral of the vector from point (0, 0, 0) to (1, 1, 1) is

If are to arbitrary vectors with magnitudes a and b respectively, will be equal to

Cross checking from option (a),


which is correct answer.


Contents

The fundamental theorem of calculus states that the integral of a function f over the interval [a, b] can be calculated by finding an antiderivative F of f :

Stokes' theorem is a vast generalization of this theorem in the following sense.

  • By the choice of F , dF / dx = f(x) . In the parlance of differential forms, this is saying that f(x) dx is the exterior derivative of the 0-form, i.e. function, F : in other words, that dF = fdx . The general Stokes theorem applies to higher differential forms ω instead of just 0-forms such as F .
  • A closed interval [a, b] is a simple example of a one-dimensional manifold with boundary. Its boundary is the set consisting of the two points a and b . Integrating f over the interval may be generalized to integrating forms on a higher-dimensional manifold. Two technical conditions are needed: the manifold has to be orientable, and the form has to be compactly supported in order to give a well-defined integral.
  • The two points a and b form the boundary of the closed interval. More generally, Stokes' theorem applies to oriented manifolds M with boundary. The boundary ∂M of M is itself a manifold and inherits a natural orientation from that of M . For example, the natural orientation of the interval gives an orientation of the two boundary points. Intuitively, a inherits the opposite orientation as b , as they are at opposite ends of the interval. So, "integrating" F over two boundary points a , b is taking the difference F(b) − F(a) .

In even simpler terms, one can consider the points as boundaries of curves, that is as 0-dimensional boundaries of 1-dimensional manifolds. So, just as one can find the value of an integral ( f dx = dF ) over a 1-dimensional manifold ( [a, b] ) by considering the anti-derivative ( F ) at the 0-dimensional boundaries ( <a, b >), one can generalize the fundamental theorem of calculus, with a few additional caveats, to deal with the value of integrals ( dω ) over n -dimensional manifolds ( Ω ) by considering the antiderivative ( ω ) at the (n − 1) -dimensional boundaries ( ∂Ω ) of the manifold.

So the fundamental theorem reads:

Let Ω be an oriented smooth manifold with boundary of dimension n and let α be a smooth n -differential form that is compactly supported on Ω . First, suppose that α is compactly supported in the domain of a single, oriented coordinate chart <U, φ> . In this case, we define the integral of α over Ω as

More generally, the integral of α over Ω is defined as follows: Let <ψi> be a partition of unity associated with a locally finite cover <Ui, φi> of (consistently oriented) coordinate charts, then define the integral

The generalized Stokes theorem reads:

Let M be a smooth manifold. A (smooth) singular k -simplex in M is defined as a smooth map from the standard simplex in R k to M . The group Ck(M, Z) of singular k -chains on M is defined to be the free abelian group on the set of singular k -simplices in M . These groups, together with the boundary map, ∂ , define a chain complex. The corresponding homology (resp. cohomology) group is isomorphic to the usual singular homology group Hk(M, Z) (resp. the singular cohomology group H k (M, Z) ), defined using continuous rather than smooth simplices in M .

On the other hand, the differential forms, with exterior derivative, d , as the connecting map, form a cochain complex, which defines the de Rham cohomology groups H k
dR (M, R) .

Differential k -forms can be integrated over a k -simplex in a natural way, by pulling back to R k . Extending by linearity allows one to integrate over chains. This gives a linear map from the space of k -forms to the k th group of singular cochains, C k (M, Z) , the linear functionals on Ck(M, Z) . In other words, a k -form ω defines a functional

  1. closed forms, i.e., = 0 , have zero integral over boundaries, i.e. over manifolds that can be written as ∂∑cMc , and
  2. exact forms, i.e., ω = , have zero integral over cycles, i.e. if the boundaries sum up to the empty set: ∑cMc = ∅ .

De Rham's theorem shows that this homomorphism is in fact an isomorphism. So the converse to 1 and 2 above hold true. In other words, if <ci> are cycles generating the k th homology group, then for any corresponding real numbers, <ai> , there exist a closed form, ω , such that

Stokes' theorem on smooth manifolds can be derived from Stokes' theorem for chains in smooth manifolds, and vice versa. [13] Formally stated, the latter reads: [14]

Theorem (Stokes' theorem for chains) — If c is a smooth k -chain in a smooth manifold M , and ω is a smooth (k − 1) -form on M , then

To simplify these topological arguments, it is worthwhile to examine the underlying principle by considering an example for d = 2 dimensions. The essential idea can be understood by the diagram on the left, which shows that, in an oriented tiling of a manifold, the interior paths are traversed in opposite directions their contributions to the path integral thus cancel each other pairwise. As a consequence, only the contribution from the boundary remains. It thus suffices to prove Stokes' theorem for sufficiently fine tilings (or, equivalently, simplices), which usually is not difficult.

The formulation above, in which Ω is a smooth manifold with boundary, does not suffice in many applications. For example, if the domain of integration is defined as the plane region between two x -coordinates and the graphs of two functions, it will often happen that the domain has corners. In such a case, the corner points mean that Ω is not a smooth manifold with boundary, and so the statement of Stokes' theorem given above does not apply. Nevertheless, it is possible to check that the conclusion of Stokes' theorem is still true. This is because Ω and its boundary are well-behaved away from a small set of points (a measure zero set).

A version of Stokes' theorem that allows for roughness was proved by Whitney. [15] Assume that D is a connected bounded open subset of R n . Call D a standard domain if it satisfies the following property: There exists a subset P of ∂D , open in ∂D , whose complement in ∂D has Hausdorff (n − 1) -measure zero and such that every point of P has a generalized normal vector. This is a vector v(x) such that, if a coordinate system is chosen so that v(x) is the first basis vector, then, in an open neighborhood around x , there exists a smooth function f(x2, …, xn) such that P is the graph < x1 = f(x2, …, xn) > and D is the region <x1 : x1 < f(x2, …, xn) > . Whitney remarks that the boundary of a standard domain is the union of a set of zero Hausdorff (n − 1) -measure and a finite or countable union of smooth (n − 1) -manifolds, each of which has the domain on only one side. He then proves that if D is a standard domain in R n , ω is an (n − 1) -form which is defined, continuous, and bounded on DP , smooth on D , integrable on P , and such that dω is integrable on D , then Stokes' theorem holds, that is,

The study of measure-theoretic properties of rough sets leads to geometric measure theory. Even more general versions of Stokes' theorem have been proved by Federer and by Harrison. [16]

The general form of the Stokes theorem using differential forms is more powerful and easier to use than the special cases. The traditional versions can be formulated using Cartesian coordinates without the machinery of differential geometry, and thus are more accessible. Further, they are older and their names are more familiar as a result. The traditional forms are often considered more convenient by practicing scientists and engineers but the non-naturalness of the traditional formulation becomes apparent when using other coordinate systems, even familiar ones like spherical or cylindrical coordinates. There is potential for confusion in the way names are applied, and the use of dual formulations.

Kelvin–Stokes theorem Edit

This is a (dualized) (1 + 1)-dimensional case, for a 1-form (dualized because it is a statement about vector fields). This special case is often just referred to as Stokes' theorem in many introductory university vector calculus courses and is used in physics and engineering. It is also sometimes known as the curl theorem.

The classical Kelvin–Stokes theorem relates the surface integral of the curl of a vector field over a surface Σ in Euclidean three-space to the line integral of the vector field over its boundary. It is a special case of the general Stokes theorem (with n = 2 ) once we identify a vector field with a 1-form using the metric on Euclidean 3-space. The curve of the line integral, ∂Σ , must have positive orientation, meaning that ∂Σ points counterclockwise when the surface normal, n , points toward the viewer.

One consequence of the Kelvin–Stokes theorem is that the field lines of a vector field with zero curl cannot be closed contours. The formula can be rewritten as:

Theorem — Suppose F = (P(x,y,z), Q(x,y,z), R(x,y,z)) is defined in a region with smooth surface Σ and has continuous first-order partial derivatives. Then

where P , Q , and R are the components of F , and ∂Σ is the boundary of the region Σ .

Green's theorem Edit

Green's theorem is immediately recognizable as the third integrand of both sides in the integral in terms of P , Q , and R cited above.

In electromagnetism Edit

Two of the four Maxwell equations involve curls of 3-D vector fields, and their differential and integral forms are related by the Kelvin–Stokes theorem. Caution must be taken to avoid cases with moving boundaries: the partial time derivatives are intended to exclude such cases. If moving boundaries are included, interchange of integration and differentiation introduces terms related to boundary motion not included in the results below (see Differentiation under the integral sign):

The above listed subset of Maxwell's equations are valid for electromagnetic fields expressed in SI units. In other systems of units, such as CGS or Gaussian units, the scaling factors for the terms differ. For example, in Gaussian units, Faraday's law of induction and Ampère's law take the forms: [17] [18]

respectively, where c is the speed of light in vacuum.

Divergence theorem Edit

is a special case if we identify a vector field with the (n − 1) -form obtained by contracting the vector field with the Euclidean volume form. An application of this is the case F = fc where c is an arbitrary constant vector. Working out the divergence of the product gives


Handbook of Numerical Methods for Hyperbolic Problems

2 On the Lax–Wendroff Theorem

FVMs seem to be perfectly suited to the conservation or divergence form of partial differential equations since they appear automatically in conservation form, and this form is important per se as we know from the now classical result of Lax and Wendroff (1960) . This result stated that if a FDM in conservation form converges at all, then the limit function is indeed a weak solution of the partial differential equation. In its original form the Lax–Wendroff theorem required a uniform grid in one space dimension and other technical assumptions like continuous fluxes and L ∞ -boundedness of the sequence of discrete numerical solutions when the mesh parameter tends to zero. It is, however, not clear a priori that this type of theorem extends to FVMs on non-Cartesian or triangular grids. Kröner et al. (1996) have shown that the theorem holds for two-dimensional polygonal meshes under some restrictive conditions on the mesh, cp. their conditions (2.3) and (2.4) in Kröner et al. (1996) . Godlewski and Raviart (1996 , p. 375 ff.) have relaxed the conditions for a triangular mesh. Even more general meshes are considered by Elling (2007) and his Lax–Wendroff-type theorem is fairly general indeed.

The importance of the Lax–Wendroff theorem is clearly understood if one studies the failure of convergence and strange behaviour if nonconservative numerical discretisation methods are applied to conservation laws (see Hou and LeFloch, 1994 ).


16.8: The Divergence Theorem - Mathematics

Day and Time: MTWR 4:30 pm - 6:05 pm
Location: 417 Mathematics

Instructor: Pak-Hin Lee
Email: phlee "at" math.columbia.edu
Office: 408 Mathematics
Office Hours: Monday 2 pm to 4 pm, or by appointment

Teaching Assistant: Alvin Peizhe Li
Email: pl2488 "at" columbia.edu
Help Room: 406 Mathematics
Help Room Hours: Wednesday 7 pm to 10 pm, Thursday 7 pm to 10 pm
(For more information about the Help Room, see below.)

Textbook: James Stewart, Calculus: Early Transcendentals, 7th Edition
(More information about the textbook is available here. Please make sure you have access to the corrrect edition when working on the homework problems.)

Course Description: Double and triple integrals. Change of variables. Line and surface integrals. Grad, div, and curl. Vector integral calculus: Green's theorem, divergence theorem, Stokes' theorem. (Chapters 15 and 16 of Stewart.)
Prerequisites: Calculus I, II, III. (Chapters 1 to 14 of Stewart, but 9 and 11 are of little relevance.) Integration techniques are especially important, so brush up on them early on!

  • July 6 (Mon): First day of class
  • July 6 (Mon): Last day of registration
  • July 10 (Fri): Last day of late registration
  • July 10 (Fri): Last day to drop a course or withdraw for full refund
  • July 20 (Mon): Midterm Exam 1
  • August 3 (Mon): Midterm Exam 2
  • August 3 (Mon): Last day to drop a course or change to pass/fail
  • August 13 (Thu): Last day of class (Final Exam)

Grading

Homework: There will be ten written problem sets, due at the end of class every Tuesday and Thursday (except on the second and last days of class). You may either turn them in to me in class, or leave them in the homework dropbox (located outside of 408 Mathematics) by 6:15 pm. The majority of the problems will be taken from the textbook, with a few extra problems. Solutions will be posted after the homework is due. No late submission will be accepted, but the lowest homework score will be dropped. Collaboration is encouraged, but you must write up your own solutions independently. You must cite any resources you consult other than the textbook (Google, Wikipedia, Help Room, classmates, etc) failure to do so is considered plagiarism.

Examinations: There will be two midterms (20% each) and one final (40%) held in class. The topics to be covered in each exam will be announced later. If you have a conflict with any of the exam dates, you must contact me at least one week in advance for alternative arrangements. If you are unable to take the exam because of a medical problem, you must go to the health center and get a note from them, and contact me as soon as you can. The use of books, cell phones, calculators, or notes of any sort is not permitted in any of the exams.

CourseWorks: All course materials (homework, solutions, etc.) and grades will be posted on CourseWorks. Important announcements will also be emailed via CourseWorks.

Plagiarism: Any work plagiarized from outside sources or between classmates will receive no credit and potentially result in disciplinary actions.

Resources

Help Room: The Help Room is open 9 am to 10 pm Monday to Friday. The full schedule is available here. Feel free to seek help from any other TA's who are on duty, if you cannot make Alvin's hours.

WebAssign: We will not be using WebAssign. If you want to use it for practice, please refer here.

Disability Services: In order to receive disability-related academic accommodations, students must first be registered with the Disability Services (DS). More information on the DS registration process is available online here. Registered students must present an accommodation letter to the instructor before exam or other accommodations can be provided. Students who have, or think they may have, a disability are invited to contact DS for a confidential discussion.


State the Divergence Theorem.

Q: Describe the set of real numbers satisfying Ix - 31 = Ix - 21 + 1 as a half-infinite interval.

A: Click to see the answer

Q: Can you please solve these three sub questions and show all of the steps

A: Click to see the answer

Q: Find the differential d5y of the function y=sin2x.

A: To Determine: Find the 5th differential of the given function y = sin(2x)

Q: Each of the differential equations in Problems 31 through 36 is of two different types considered in.

A: Click to see the answer

Q: Find the nth-order derivative of the function y=arctanx at the point x=0.

A: Click to see the answer

Q: a 1000-liter (L) tank contains 500 L of w. ater with a salt concentration of 10 g/L. Water with a sa.

A: Click to see the answer

Q: Bounded by the cylinder x 2 + y 2 = 1 and the planes y= z x =0 z= 0 in the first octant

A: Click to see the answer

Q: In Exerc R. evaluate the double integral over the given region /| (6y? 2x) dA, R: 0 < x< 1, 0 < y< 2.

A: Since you have asked multiple questions in a single request, we will be answering only the 1st quest.

Q: Test for symmetry with respect to each axis and to the origin y = x /(x2 + 1)

A: To test symmetry of a function about x axis, replace y with -y in the given function y axis, replac.


16.8: The Divergence Theorem - Mathematics

Multivariable Calculus
Math 2110Q § 003
Fall 2017

As stated in the syllabus, every student is required to submit a short paper on some topic that extends the topics discussed in the lecture.
Specifically, the paper can be submitted alone or with one other student.
The paper must be single spaced and between two and five complete pages in length.
The paper can be submitted for review or final grading at any point during the regular semester but will not be accepted during exam week.
The paper will recieve a letter grade (A - F) upon final submission.

  • Applications of the cross and dot products to physics
  • The cross product and an n-dimensional Pythagorean theorem
  • Limits and continuity for functions of several variables
  • Quadratic approximations and Taylor series for functions of two variables
  • Linearization in higher dimensions
  • Differentiability of multi-variable functions
  • When does Clairaut's theorem not apply?
  • The method of Lagrange multipliers finding extrema relative to constraints - Bailey Skrtich and Kevin Spichiger
  • The Jacobian and change of variables for a functions of two variables
  • Applications of multiple integrals
  • Fubini's theorem and when it fails - Xinba Li
  • TNB Frames and curvature for space curves
  • Deriving Keplar's laws of planetary motion
  • Using parametric equations to connect straight tracks of railroad
  • A history of calculus
  • The projects given in Stewart
    • Section 14.4
    • Section 14.7
    • Section 14.8
    • Section 15.6
    • Section 15.8
    • Section 16.8

    Feel free to use your favorite word processor for typing this paper as many have built in "drag and drop" mathematics. If you want to use something to similar to my preferred way of producing type mathematics I recommend using LaTeX. Specifically, trough the online platform Overleaf. While the learning curve can be steep, this quickly becomes the fastest and prettiest way to typeset mathematics. An example can be found here. For determining how to do something in the language of LaTeX, a quick Google search usually is the answer. Detexify can help you determine the commands for symbols you may need.


    State the Divergence Theorem.

    Q: Describe the set of real numbers satisfying Ix - 31 = Ix - 21 + 1 as a half-infinite interval.

    A: Click to see the answer

    Q: Can you please solve these three sub questions and show all of the steps

    A: Click to see the answer

    Q: Find the differential d5y of the function y=sin2x.

    A: To Determine: Find the 5th differential of the given function y = sin(2x)

    Q: Each of the differential equations in Problems 31 through 36 is of two different types considered in.

    A: Click to see the answer

    Q: Find the nth-order derivative of the function y=arctanx at the point x=0.

    A: Click to see the answer

    Q: a 1000-liter (L) tank contains 500 L of w. ater with a salt concentration of 10 g/L. Water with a sa.

    A: Click to see the answer

    Q: Bounded by the cylinder x 2 + y 2 = 1 and the planes y= z x =0 z= 0 in the first octant

    A: Click to see the answer

    Q: In Exerc R. evaluate the double integral over the given region /| (6y? 2x) dA, R: 0 < x< 1, 0 < y< 2.

    A: Since you have asked multiple questions in a single request, we will be answering only the 1st quest.

    Q: Test for symmetry with respect to each axis and to the origin y = x /(x2 + 1)

    A: To test symmetry of a function about x axis, replace y with -y in the given function y axis, replac.


    Q: O Problem 1 (Exercise 4.3.38) Sketch the phase portrait with the given eigenvalues and eigenvectors.

    Q: The equation below gives parametric equations and parameter intervals for the motion of a particle i.

    A: Given, x=5t-3 . (1) and, y=25t2 . (2)

    Q: A = Find the eigenvalues and the associated eigenvectors of A.

    A: Click to see the answer

    Q: Approximate the derinite Integral using the Trapezoldal Rule and Simpson Ruie with these results wit.

    A: According to the trapezoidal rule of integration, the integration ∫abfxdx is approximately equal to .

    Q: The set is a basis for a subspace W. Use the Gram-Schmidt process to produce an orthogonal basis for.

    A: The solution is given as

    Q: A small resort is situated on an island off a part of the coast of Mexico that has a perfectly strai.

    A: Let P be the shore point Let R be the resort Let S be the point where the pipe gets the water.

    Q: (a) Using laws of logarithms, write the expression below using sums and/or differences of logarithmi.

    A: Given below the detailed solution

    Q: Please show the inductive base step, the inductive hypothesis, and the inductive step. Prove that fo.

    A: Click to see the answer

    Q: Subnet Calculat. 5 Best Subnet Cal. The following graph shows at least one complete cycle of the gra.


    Watch the video: Μαθήματα Ανάλυσης Γ Λυκείου - 19. Συνέχεια συνάρτησης β (January 2022).