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13. VECTOR CALCULUS. VECTOR CALCULUS. 13.9 The Divergence Theorem. In this section, we will learn about: The Divergence Theorem for simple solid regions, and its applications in electric fields and fluid flow. INTRODUCTION.

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Vector calculus

13

VECTOR CALCULUS


Vector calculus

VECTOR CALCULUS

13.9

The Divergence Theorem

In this section, we will learn about:

The Divergence Theorem for simple solid regions,

and its applications in electric fields and fluid flow.


Introduction

INTRODUCTION

  • In Section 12.5, we rewrote Green’s Theorem in a vector version as: where C is the positively oriented boundary curve of the plane region D.


Introduction1

INTRODUCTION

Equation 1

  • If we were seeking to extend this theorem to vector fields on , we might make the guess that where S is the boundary surface of the solid region E.


Divergence theorem

DIVERGENCE THEOREM

  • It turns out that Equation 1 is true, under appropriate hypotheses, and is called the Divergence Theorem.


Divergence theorem1

DIVERGENCE THEOREM

  • Notice its similarity to Green’s Theorem and Stokes’ Theorem in that:

    • It relates the integral of a derivative of a function (div F in this case) over a region to the integral of the original function F over the boundary of the region.


Divergence theorem2

DIVERGENCE THEOREM

  • At this stage, you may wish to review the various types of regions over which we were able to evaluate triple integrals in Section 11.6


Simple solid region

SIMPLE SOLID REGION

  • We state and prove the Divergence Theorem for regions E that are simultaneously of types 1, 2, and 3.

  • We call such regions simple solid regions.

    • For instance, regions bounded by ellipsoids or rectangular boxes are simple solid regions.


Simple solid regions

SIMPLE SOLID REGIONS

  • The boundary of E is a closed surface.

  • We use the convention, introduced in Section 12.7, that the positive orientation is outward.

    • That is, the unit normal vector n is directed outward from E.


The divergence theorem

THE DIVERGENCE THEOREM

  • Let:

    • E be a simple solid region and let S be the boundary surface of E, given with positive (outward) orientation.

    • F be a vector field whose component functions have continuous partial derivatives on an open region that contains E.

  • Then,


The divergence theorem1

THE DIVERGENCE THEOREM

  • Thus, the Divergence Theorem states that:

    • Under the given conditions, the flux of Facross the boundary surface of E is equal to the triple integral of the divergence of Fover E.


Gauss s theorem

GAUSS’S THEOREM

  • The Divergence Theorem is sometimes called Gauss’s Theorem after the great German mathematician Karl Friedrich Gauss (1777–1855).

    • He discovered this theorem during his investigation of electrostatics.


Ostrogradsky s theorem

OSTROGRADSKY’S THEOREM

  • In Eastern Europe, it is known as Ostrogradsky’s Theorem after the Russian mathematician Mikhail Ostrogradsky (1801–1862).

    • He published this result in 1826.


The divergence theorem2

THE DIVERGENCE THEOREM

Proof

  • Let F = P i + Q j + R k

    • Then,

    • Hence,


The divergence theorem3

THE DIVERGENCE THEOREM

Proof

  • If n is the unit outward normal of S, then the surface integral on the left side of the Divergence Theorem is:


The divergence theorem4

THE DIVERGENCE THEOREM

Proof—Eqns. 2-4

  • So, to prove the theorem, it suffices to prove these equations:


The divergence theorem5

THE DIVERGENCE THEOREM

Proof

  • To prove Equation 4, we use the fact that E is a type 1 region: where D is the projection of Eonto the xy-plane.


The divergence theorem6

THE DIVERGENCE THEOREM

Proof

  • By Equation 6 in Section 11.6, we have:


The divergence theorem7

THE DIVERGENCE THEOREM

Proof—Equation 5

  • Thus, by the Fundamental Theorem of Calculus,


The divergence theorem8

THE DIVERGENCE THEOREM

Proof

  • The boundary surface S consists of three pieces:

    • Bottom surface S1

    • Top surface S2

    • Possibly a vertical surface S3, which lies above the boundary curve of D(It might happen that S3 doesn’t appear, as in the case of a sphere.)


The divergence theorem9

THE DIVERGENCE THEOREM

Proof

  • Notice that, on S3, we have k ∙n = 0, because k is vertical and n is horizontal.

    • Thus,


The divergence theorem10

THE DIVERGENCE THEOREM

Proof—Equation 6

  • Thus, regardless of whether there is a vertical surface, we can write:


The divergence theorem11

THE DIVERGENCE THEOREM

Proof

  • The equation of S2 is z = u2(x, y), (x, y) D, and the outward normal n points upward.

    • So, from Equation 10 in Section 12.7 (with F replaced by Rk), we have:


The divergence theorem12

THE DIVERGENCE THEOREM

Proof

  • On S1, we have z =u1(x, y).

  • However, here, n points downward.

    • So, we multiply by –1:


The divergence theorem13

THE DIVERGENCE THEOREM

Proof

  • Therefore, Equation 6 gives:


The divergence theorem14

THE DIVERGENCE THEOREM

Proof

  • Comparison with Equation 5 shows that:

    • Equations 2 and 3 are proved in a similar manner using the expressions for E as a type 2 or type 3 region, respectively.


The divergence theorem15

THE DIVERGENCE THEOREM

  • Notice that the method of proof of the Divergence Theorem is very similar to that of Green’s Theorem.


Divergence

DIVERGENCE

Example 1

  • Find the flux of the vector field F(x, y, z) = z i + y j + x k over the unit sphere x2 + y2 + z2 = 1

    • First, we compute the divergence of F:


Divergence1

DIVERGENCE

Example 1

  • The unit sphere S is the boundary of the unit ball B given by: x2 + y2 + z2≤ 1

    • So, the Divergence Theorem gives the flux as:


Divergence2

DIVERGENCE

Example 2

  • Evaluate where:

    • F(x, y, z) = xy i + (y2 + exz2) j + sin(xy)k

    • S is the surface of the region E bounded by the parabolic cylinder z = 1 – x2and the planesz = 0, y = 0, y + z = 2


Divergence3

DIVERGENCE

Example 2

  • It would be extremely difficult to evaluate the given surface integral directly.

    • We would have to evaluate four surface integrals corresponding to the four pieces of S.

    • Also, the divergence of F is much less complicated than F itself:


Divergence4

DIVERGENCE

Example 2

  • So, we use the Divergence Theorem to transform the given surface integral into a triple integral.

    • The easiest way to evaluate the triple integral is to express E as a type 3 region:


Divergence5

DIVERGENCE

Example 2

  • Then, we have:


Divergence6

DIVERGENCE

Example 2


Unions of simple solid regions

UNIONS OF SIMPLE SOLID REGIONS

  • The Divergence Theorem can also be proved for regions that are finite unions of simple solid regions.

    • The procedure is similar to the one we used in Section 12.4 to extend Green’s Theorem.


Unions of simple solid regions1

UNIONS OF SIMPLE SOLID REGIONS

  • For example, let’s consider the region E that lies between the closed surfaces S1 and S2, where S1 lies inside S2.

    • Let n1 and n2 be outward normals of S1 and S2.


Unions of simple solid regions2

UNIONS OF SIMPLE SOLID REGIONS

  • Then, the boundary surface of E is: S = S1S2Its normal n is given by: n = –n1 on S1n = n2 on S2


Unions of simple solid rgns

UNIONS OF SIMPLE SOLID RGNS.

Equation 7

  • Applying the Divergence Theorem to S, we get:


Applications electric field

APPLICATIONS—ELECTRIC FIELD

  • Let’s apply this to the electric field (Example 5 in Section 12.1): where S1 is a small sphere with radius a and center the origin.


Applications electric field1

APPLICATIONS—ELECTRIC FIELD

  • You can verify that div E = 0 (Exercise 23).

  • Thus, Equation 7 gives:


Applications electric field2

APPLICATIONS—ELECTRIC FIELD

  • The point of this calculation is that we can compute the surface integral over S1 because S1 is a sphere.


Applications electric field3

APPLICATIONS—ELECTRIC FIELD

  • The normal vector at x is x/|x|.

  • Therefore.

  • since the equation of S1 is |x| = a.


Applications electric field4

APPLICATIONS—ELECTRIC FIELD

  • Thus, we have:


Applications electric field5

APPLICATIONS—ELECTRIC FIELD

  • This shows that the electric flux of E is 4πεQthrough anyclosed surface S2 that containsthe origin.

    • This is a special case of Gauss’s Law (Equation 11 in Section 12.7) for a single charge.

    • The relationship between ε and ε0 is ε= 1/4πε0.


Applications fluid flow

APPLICATIONS—FLUID FLOW

  • Another application of the Divergence Theorem occurs in fluid flow.

    • Let v(x, y, z) be the velocity field of a fluid with constant density ρ.

    • Then, F = ρv is the rate of flow per unit area.


Applications fluid flow1

APPLICATIONS—FLUID FLOW

  • Suppose:

  • P0(x0, y0, z0) is a point in the fluid.

  • Ba is a ball with center P0 and very small radius a.

    • Then,div F(P) ≈ div F(P0) for all points in Ba since div F is continuous.


Applications fluid flow2

APPLICATIONS—FLUID FLOW

  • We approximate the flux over the boundary sphere Sa as follows:


Applications fluid flow3

APPLICATIONS—FLUID FLOW

Equation 8

  • This approximation becomes better as a → 0 and suggests that:


Source and sink

SOURCE AND SINK

  • Equation 8 says that div F(P0) is the net rate of outward flux per unit volume at P0.

  • (This is the reason for the name divergence.)

    • If div F(P) > 0, the net flow is outward near Pand P is called a source.

    • If div F(P) < 0, the net flow is inward near Pand P is called a sink.


Source

SOURCE

  • For this vector field, it appears that the vectors that end near P1 are shorter than the vectors that start near P1.

    • Thus, the net flow is outward near P1.

    • So, div F(P1) > 0 and P1 is a source.


Vector calculus

SINK

  • Near P2, the incoming arrows are longer than the outgoing arrows.

    • The net flow is inward.

    • So, div F(P2) < 0 and P2 is a sink.


Source and sink1

SOURCE AND SINK

  • We can use the formula for F to confirm this impression.

    • Since F = x2i + y2j, we have div F = 2x + 2y, which is positive when y > –x.

    • So, the points above the line y = –xare sources and those below are sinks.


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