Gaus Div Theorm

7/30/2019 Gaus Div Theorm

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The Gauss Divergence Theorem

Theorem 1 Let F(X) be a continuously differentiable vector field ina domain D Rn. Let R D be a closed, bounded region whoseboundary is a smooth surface, D. For each point x let(X) bethe unit outward, or exterior, normal to with respect to the region

R. Then, with dX

dx1dx2...dxn and with d indicating integration

with respect to surface area on ,R

F(X) dX =

F(X) (X) d.

Proof We will give the proof for the case n = 3; much the sameproof can be given in the general n-dimensional situation. Supposingthat

F(X) = F(x,y,z) =

f(x, y, z )g(x, y, z )h(x, y, z )

,

we have

F(X) =

f

x+

g

y+

h

z

(x,y,z).

ThusR

F(X) dX =R

f

xdX +

R

g

ydX +

R

h

zdX.

Initially we will concentrate our attention on the last integral appearinghere.

We will also initially suppose (see figure on page 5) that the boundingsurface for R consists of an upper surface and a lower surface: : (x, y, z ) = z (x, y) = 0; : (x, y, z ) = z (x, y) = 0,together with some possible surface components parallel to the z-axiswhich we will denote collectively by z.

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If we take a point on the upper surface, namely (x, y, (x, y)), the unitnormal vector there is

(X) =i

x j

y+ k

x

2+

y

2+ 1

.

On the lower surface the unit outward normal vector is

(X) =ix + jy kx

2+

y

2+ 1

.

On the upper surface, z = (x, y), the element of surface area is

d =

x

2+

y

2+ 1 dxdy,

while on the lower surface the corresponding surface area element is

d =

x

2+

y

2+ 1 dxdy.

Therefore, with Rxy denoting the projection ofR onto the x, y plane,R

h

z(x,y,z) dx dy dz =

Rxy

z=(x,y)z=(x,y)

h

z(x, y, z ) dz

dx dy

=Rxy

(h(x, y, (x, y)) h(x, y, (x, y))) dxdy

=

h(x,y, (x, y))d

x

2+

y

2+ 1

h(x, y, (x, y))d

x

2+

y

2+ 1

.

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But we clearly have

1x

2+

y

2+ 1

= k (X),1

x

2+

y

2+ 1

= k (X).

Consequently R

h

z(x, y, z ) dx dy dz

=

h(x, y, z ) k (X) d +

h(x, y, z ) k (X) d.On z we have k (X) 0, so the above can be extended to

R

h

z(x, y, z ) dx dy dz =

h(x, y, z ) k (X) d.

Complicated regions R can be reduced, by means of cuts parallel to thez-axis to regions satisfying the hypothesis above. So we may assume

the result obtained so far is not limited with respect to the geometryofR. (See figure on p. 5.)

In the same way we can see that

R

f

x(x, y, z ) dx dy dz =

f(x,y,z) i (X) d;R

g

y(x, y, z ) dx dy dz =

g(x, y, z )j (X) d.

Adding the three results separately obtained for f, g and h we have

R

F(X) dX =R

f

x(x, y, z ) +

g

y(x, y, z ) +

h

z(x, y, z )

dx dy dz

=

(f(x,y,z) i + g(x, y, z )j + h(x, y, z ) k) (X) d=

F(X) (X) dand the theorem is proved.

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Example 1 Let F(x, y, z ) = x i + yj + zk and let be the sphereof radius 1 in R3. We first compute =

F(X) (X) d. Here we

have

(X) =1

x2 + y2 + z2

x

y

z

F(X) (X) = x

2 + y2 + z2x2 + y2 + z2

= x2 + y2 + z2 1 on .

So we have

F(X) (X) d = 1

d = 4.

On the other hand, with R denoting the unit ball inside the unit sphere as described above, we have

R F(X) dX = R

x

y

z

dx dy dz = R 3 dx dy dz

= 3R

dx dy dz = 3 43

= 4.

The Divergence Theorem equates two different integrals; one overthe region R and one over , the boundary of R. It can happen thatone or the other of these integrals is substantially easier to calculatethan the other, so it may be advantageous to switch from one integral

to the other to facilitate computation.

Example 2 Let R and be as in the preceding example, but now

F(X) = F(x, y, z ) =

x3

y3

z3

.

Then

F(X) (X) = x4 + y4 + z4x2 + y2 + z2

.

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In this case the surface integral,

x4 + y4 + z4x2 + y2 + z2

d

is not easy to compute. But F(x, y, z ) = 3

x2 + y2 + z2

andthus, using spherical coordinates,

R

F(x, y, z ) dx dy dz = 2

0

01

0

3r2 r2 sin drdd

= 20

sin d10

3r4 dr = 1210

r4 dr = 12r5

5

1

0

=12

5.

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