AB2.5: Surfaces and Surface Integrals. Divergence Theorem of Gauss

AB2.5: Surfaces and Surface Integrals.

Divergence Theorem of Gauss

Representations of surfaces

Representation of a surface S as projections on the xy- and xz-planes, etc. are z = f (x, y), x = g(x, z)

or

g(x, y, z) = 0.

For example,

z = +^qa²− x²− y² or x²+ y² + z² = a², z ≥ 0 represents a hemisphere of radius a and center O.

A surface S can be represented by a vector function

r(u, v) = [x(u, v), y(u, v), z(u, v)] = x(u, v)i + y(u, v)j + z(u, v)k, u, v ∈ R

This is called a parametric representation of a surface, u, v varying in a two-dimensional region R are the parameters of the representation.

EXAMPLE 1

Parametric representation of a cylinder

A circular cylinder x²+ y² = a², −1 ≤ z ≤ 1 has radius a, height 2, and the z-axis as the axis.

A parametric representation is

r(u, v) = [a cos u, a sin u, v] = a cos ui + a sin uj + vk, u, v in rectangle R : 0 ≤ u ≤ 2π, −1 ≤ v ≤ 1.

The components of r(u, v) are

x = a cos u, y = a sin u, z = v.

EXAMPLE 2

Parametric representation of a sphere A sphere x²+ y²+ z² = a² has the parametric representation

r(u, v) = a cos v cos ui + a cos v sin uj + a sin vk, u, v in rectangle R : 0 ≤ u ≤ 2π, −π/2 ≤ v ≤ π/2.

The components of r(u, v) are

x = a cos v cos u, y = a cos v sin u, z = a sin v.

EXAMPLE 3

Parametric representation of a cone A circular cone z = +√

x²+ y², 0 ≤ z ≤ H has the parametric representation

r(u, v) = u cos vi + u sin vj + uk, u, v in rectangleR : 0 ≤ v ≤ 2π, 0 ≤ u ≤ H.

The components of r(u, v) are

x = u cos v, y = u sin v, z = u.

Indeed, this yields x²+ y² = z². Tangent to a surface

Get a curve C on S by a pair of continuous functions u = u(t), v = v(t)

so that C has the position vector ˜r(u(t), v(t)). By the chain rule, we get a tangent vector of a curve C

˜r⁰(t) = d˜r dt = ∂r

∂uu⁰+ ∂r

∂vv⁰.

Hence the partial derivatives ru and rv at a point P are tangential to S at P and we assume that are linearly indepedent. Then their vector product gives a normal vector N of S at P ,

N = r_u× r^v 6= 0.

The corresponding unit normal vector n n = 1

|N|N = 1

|r^u× r^v|r_u× r^v. If S is represented by

g(x, y, z) = 0, then

n = 1

|grad g|grad g.

EXAMPLE 4

Unit normal vector of a sphere g(x, y, z) = x²+ y²+ z²− a² = 0:

n = 1

|grad g|grad g = 1

agrad g =

x a,y

a,z a



= x ai + y

aj + z ak.

EXAMPLE 5

Unit normal vector of a cone g(x, y, z) = −z +

q

x²+ y² = 0:

n = 1

|grad g|grad g = 1

√2

"

√ x

x²+ y², y

√x²+ y², −1

#

=

√ x

x² + y²i + y

√x²+ y²j − k.

Definition and evaluation of surface integrals

A surface integral of a vector function F(r) over a surface S is defined as

Z

S

Z

F · ndA =

Z

R

Z

F(r(u, v)) · N(u, v)dudv, where

r(u, v) = [x(u, v), y(u, v), z(u, v)] = x(u, v)i + y(u, v)j + z(u, v)k, u, v ∈ R is a parametric representation of S with a normal vector

N = r_u× r^v 6= 0 and the corresponding unit normal vector

n = 1

|N|N.

Note that

ndA = n|N|dudv = |N|dudv,

and it is assumed that the parameters u, v vary in a domain R on the u, v-plane.

In terms of components

F = [F₁, F₂, F₃) = F₁i + F₂j + F₃k,

n = [cos α, cos β, cos γ] = cos αi + cos βj + cos γk, N = [N₁, N₂, N₃) = N₁i + N₂j + N₃k,

and _Z

S

Z

F · ndA =

Z

S

Z

(F₁cos α + F₂cos β + F₃cos γ)dA =

Z

S

Z

(F₁N₁+ F₂N₂+ F₃N₃)dudv.

EXAMPLE 1

Flux through a surface

Compute the flux of water through the parabolic cylinder S : y = x², 0 ≤ x ≤ 2, 0 ≤ z ≤ 3 if the velocity vector is v = F = [3z², 6, 6xz].

Solution. S is represented by

r(u, v) = [u, u², v] = ui + u²j + vk, 0 ≤ u ≤ 2, 0 ≤ v ≤ 3 because one can set x = u, z = v, and y = x² = u².

From this

r_u = [1, 2u, 0], r_v = [0, 0, 1];

the vector product of r_u and r_v gives a normal vector N 6= 0 of the parabolic cylinder

N = r_u × r^v =

   

i j k

1 2u 0

0 0 1

  

= 2ui − j = [2u, −1, 0].

The corresponding unit normal vector n = 1

|N|N = 1

√1 + 4u²(2ui − j).

On S

F(r(u, v)) = F(S) = [3v², 6, 6uv] = 3(v²i + 2j + 2uvk).

Hence

F(r(u, v)) · N(u, v) = 3[v², 2, 2uv] · [2u, −1, 0] = 3(2uv²− 2) = 6(uv²− 1).

The parameters u, v vary in the rectangle R : 0 ≤ u ≤ 2, 0 ≤ v ≤ 3. Now we can write and calculate the flux integral:

Z

S

Z

F · ndA =

Z

R

Z

F(r(u, v)) · N(u, v)dudv =

Z 3 0

Z 2 0

6(uv²− 1)dudv = 6(

Z 3 0

v²dv

Z 2

0 udu −

Z 3 0

Z 2 0

dudv) = 6(3²· 2 − 6) = 72.

EXAMPLE 2

Surface integral

Compute the surface integral for S being a portion of the plane S : x + y + z = 1, 0 ≤ x, y, z ≤ 1.

for F = [x², 0, 3y²].

Solution. Setting x = u and y = v, we have z = 1 − u − v, so that S can be represented by r(u, v) = [u, v, 1 − u − v], 0 ≤ v ≤ 1, 0 ≤ u ≤ 1 − v.

From this

r_u = [1, 0, −1], r_v = [0, 1, −1];

a normal vector

N = r_u× r^v =

   

i j k

1 0 −1 0 1 −1

   

= i + j + k = [1, 1, 1].

The corresponding unit normal vector n = 1

|N|N = 1

√3(i + j + k).

On S

F(r(u, v)) = F(S) = [u², 0, 3v²] = u²i + 3v²k).

Hence

F(r(u, v)) · N(u, v) = [u², 0, 3v²] · [1, 1, 1] = u²+ 3v².dudv.

The parameters u, v vary in the triangle R : 0 ≤ v ≤ 1, 0 ≤ u ≤ 1 − v. Now we can write and calculate the surface integral:

Z

S

Z

F · ndA =

Z

R

Z

F(r(u, v)) · N(u, v)dudv =

Z

R

Z

(u²+ 3v²)dudv =

Z ₁

0

Z _1−v

0

(u²+ 3v²)dudv =

Z ₁

0

dv

Z _1−v

0

u²du + 3

Z ₁

0

v²dv

Z _1−v

0

du =

= (1/3)

Z ₁

0 (1 − v)³dv + 3

Z ₁

0

v²(1 − v)dv = (1/3)

Z ₁

0

t³dt + 3

Z ₁

0

(v² − v³)dv = (1/3) · (1/4) + 3(1/3 − 1/4) = 1/3.

Divergence theorem of Gauss

Recall that if v(x, y, z) is a differentiable vector function,

v(x, y, z) = v₁(x, y, z)i + v₂(x, y, z)j + v₃(x, y, z)k, then the function

div v = ∂v₁

∂x + ∂v₂

∂y + ∂v₃

∂z is called the divergence of v.

Formulate the divergence theorem of Gauss.

Let T be a closed bounded region in space whose boundary is a piecewise smooth orientable surface S (consists of finitely many smooth surfaces). Let F(x, y, z) be a vector function that is continuous and have continuous first partial derivatives everywhere in some domain containing

T . Then _{Z Z}

T

Z

div FdV =

Z

S

Z

F · ndA.

Here n is an outer unit normal vector of S pointing to the outside of S.

In components,

Z Z

T

Z ∂F₁

∂x +∂F₂

∂y +∂F₃

∂z

!

dxdydz =

Z

S

Z

(F₁cos α + F₂cos β + F₃cos γ)dA.

or _{Z Z}

T

Z ∂F₁

∂x +∂F₂

∂y +∂F₃

∂z

!

dxdydz =

Z

S

Z

(F₁dydz + F₂dzdx + F₃dxdy).

EXAMPLE 1

Evaluation of a surface integral by the divergence theorem Evaluate

I =

Z

S

Z

(x³dydz + x²ydzdx + x²zdxdy),

where S is a piecewise smooth surface consisting of the cylinder x²+ y² = a² (0 ≤ z ≤ b) and the circular disks z = 0 and z = b (x²+ y² ≤ a²) (S consists of three parts of smooth surfaces).

Solution. We have

F₁ = x³, F₂ = x²y, F₃ = x²z.

Hence the divergence of F = [F₁, F₂, F₃] is div F = ∂F₁

∂x + ∂F₂

∂y +∂F₃

∂z = 3x²+ x²+ x² = 5x². Introducing the polar coordinates

x = r cos θ, y = r sin θ (cylindrical coordinates r, θ, z)

we have

dxdydz = rdrdθdz,

and by the divergence theorem, the surface integral is transformed to a triple integral over the closed region T in space whose boundary is the surface S of the cylinder,

Z

S

Z

(x³dydz + x²ydzdx + x²zdxdy) =

Z Z

T

Z

div FdV =

Z Z

T

Z

5x²dxdydz =

5

Z b z=0

Z a r=0

Z 2π θ=0

r²cos²θrdrdθdz = 5b

Z _a

0

Z _2π

0

r³cos²θdrdθ = 5ba⁴ 4

Z _2π

0

cos²θdθ = 5ba⁴

8

Z 2π 0

(1 + 2 cos θ)dθ = 5 4πba⁴.

EXAMPLE 2

Verification of the divergence theorem Evaluate

I =

Z

S

Z

F · ndA), F = 7xi − zk

over the sphere S : x²+ y²+ z² = 4 by the divergence theorem and directly.

Solution. We have

F = [F₁, 0, F₃], F₁ = 7x, F₃ = −z.

The divergence of F is

div F = ∂F1

∂x + ∂F2

∂y + ∂F3

∂z = 7 + 0 − 1 = 6, and, by the divergence theorem,

I =

Z Z

T,ball

Z

div FdV = 6

Z Z

T,ball

Z

dxdydz = 6 ·4

3π2³ = 64π.

Now we will calculate the surface integral over S directly. Use the parametric representation of the sphere of radius 2

S : r(u, v) = 2 cos v cos ui + 2 cos v sin uj + 2 sin vk, u, v in rectangle R : 0 ≤ u ≤ 2π, −π/2 ≤ v ≤ π/2.

Then

r_u = [−2 sin u cos v, 2 cos v cos u, 0], r_v = [−2 sin v cos u, −2 sin v sin u, 2 cos v],

N = r_u×r^v =

   

i j k

−2 sin u cos v 2 cos v cos u 0

−2 sin v cos u −2 sin v sin u 2 cos v

   

= [4 cos²v cos u, 4 cos²v sin u, 4 cos v sin v].

On S we have

x = 2 cos v cos u, z = 2 sin v, and, correspondingly,

F(r(u, v)) = F(S) = [7x, 0, −z] = [14 cos v cos u, 0, −2 sin v].

Hence

F(r(u, v)) · N(u, v) =

(14 cos v cos u)4 cos²v cos u + (−2 sin v)(4 cos v sin v) = 56 cos³v cos²u − 8 cos v sin²u.

The parameters u, v vary in the rectangle R : 0 ≤ u ≤ 2π, −π/2 ≤ v ≤ π/2. Now we can write and calculate the surface integral:

Z

S

Z

F · ndA =

Z

R

Z

F(r(u, v)) · N(u, v)dudv =

8

Z _2π

0

Z _−π/2

−π/2

(7 cos³v cos²u − cos v sin²v)dudv =

8

(7 2

Z _2π

0

(1 + cos 2u)du

Z _π/2

−π/2

cos³vdv − 2π

Z _π/2

−π/2

cos v sin²vdv

)

=

56π

Z π/2

−π/2

cos³vdv − 16π

Z π/2

−π/2

cos vdv sin²vdv =

8π

(

7

Z π/2

−π/2(1 − sin²v)d sin v − 2

Z π/2

−π/2

dv sin²vd sin v

)

=

8π



7

Z 1

−1(1 − t²)dt − 2

Z 1

−1

t²dt



= 8π[7 · (2 − 2/3) − 4/3] = 8π · 4/3 · 6 = 64π.

The values obtained by both methods coincide.

EXAMPLE 2

Applications of the divergence theorem By the mean value theorem for triple integrals,

Z Z

T

Z

f (x, y, z)dV = f (x₀, y₀, z₀)V (T ) where (x₀, y₀, z₀) is a certain point in T and V (T ) is the volume of T .

By the divergence theorem, div F(x₀, y₀, z₀) = 1

V (T )

Z Z

T

Z

div FdV = 1 V (T )

Z

S(T )

Z

F · ndA.

Choosing a fixed point P : (x₁, y₁, z₁) in T and shrinking T down to onto P so that the maximum distance d(T ) of the points of T from P tends to zero, we obtain

div F(x₁, y₁, z₁) = lim

d(T )→0

1 V (T )

Z

S(T )

Z

F · ndA,

which is sometimes used as the definition of the divergence. From this expression, it follows that the divergence is independent of a particular choice of Cartesian coordinates.

EXAMPLE 4

A basic property of solutions of Laplaces equation

Recall that we can transform the double integral of the Laplacian of a function into a line integral of its normal derivative. In the same manner, by the divergence theorem, we can transform the triple integral of the the Laplacian of a function into a surface integral of its normal derivative. Indeed, setting

F = grad f we have

div F = div grad f = ∂²f

∂x² + ∂²f

∂y² + ∂²f

∂z² = ∇²f.

On the other hand,

F · n = grad f · n

is, by definition, the normal derivative of f (the directional derivative in the direction of the outer normal vector to S, the boundary of T ), ∂f

∂n. Thus, by the divergence theorem, the desired formula for the integral of the Laplacian of f becomes

Z Z

T

Z

∇²f dxdydz =

Z

S

Z ∂f

∂ndA.

Thus, if f (x, y, z) is a harmonic function in T (∇²f = 0 in T ), then the integral of the normal derivative of this function over any piecewise smooth orientable surface S in T whose entire interior belongs to T is zero.

PROBLEM 9.5.1

Find the normal vector to the xy-plane

r(u, v) = [u, v] = ui + vj and the parameter curves u = const and v = const.

Solution. Recall that the vector product a × b of two vectors a = [a¹, a₂, a₃] and b = [b1, b2, b3] is a vector v = a×b perpendicular to both a and b so that a, b, v form a right-handed triple:

v = [v₁, v₂, v₃] = a × b =

   

i j k

a₁ a₂ a₃ b₁ b₂ b₃

   

= v₁i + v₂j + v₃k, or

v₁ =

  

a₂ a₃ b₂ b₃

 

, v₂ =

  

a₃ a₁ b₃ b₁

 

, v₃ =

  

a₁ a₂ b₁ b₂

  

For the xy-plane

r(u, v) = [u, v, 0] = ui + vj;

ru = [1, 0, 0] = i, rv = [0, 1, 0] = j,

and the vector product of r_u and r_v gives a normal vector N 6= 0 of the xy-plane

N = r_u× r^v =

   

i j k 1 0 0 0 1 0

   

= k.

The corresponding unit normal vector n = 1

|N|N = 1

1k = k.

The parameter curves u = const and v = const are straight lines.

PROBLEM 9.5.3

Find the normal vector to the cone

r(u, v) = u cos vi + u sin vj + cuk = [u cos v, u sin v, cu]

and the parameter curves u = const and v = const.

Solution. The cone is given by the representation z = c√

x²+ y². We have r_u = [cos v, sin v, c], r_v = [−u sin v, u cos v, 0],

and a normal vector N 6= 0 of the cone

N = r_u× r^v =

   

i j k

cos v sin v c

−u sin v u cos v 0

  

= −cu cos vi − cu sin vj + uk = −u[c cos v, c sin v, −1].

The parameter curves u = const are circles x² + y² = u², z = cu and v = const are straight lines y = x tan v.

PROBLEM 9.5.13

Find a parametric representation of the plane 3x + 4y + 6z = 24.

Solution. We have z = 4 − (1/2)x − (2/3)y. Therefore, setting x = 8u and y = 6v, we obtain a parametric representation

r(u, v) = [8u, 6v, 4(1 − u − v)] = 8ui + 6vj + 4(1 − u − v)k.

Another parametric representation can be obtained by setting x = u and y = v

˜r(u, v) = [u, v, 4 − (1/2)u − (2/3)v] = ui + vj + (4 − (1/2)u − (2/3)v)k.

Take a parametric representation

r(u, v) = [8u, 6v, 4(1 − u − v)]

of the plane 3x + 4y + 6z = 24. Then

r_u = [8, 0, −4], r_v = [0, 6, −4], and a normal vector N 6= 0

N = r_u× r^v =

   

i j k 8 0 −4 0 6 −4

   

=

24i + 32j + 48k = 8(3i + 4j + 6k) = 8[3, 4, 6].

The corresponding unit normal vector n = 1

|N|N = 1

√61(3i + 4j + 6k).

PROBLEM 9.5.15

Find a parametric representation of the ellipsoid x²+ y²+ (1/4)z² = 1.

Solution. Setting

x = cos v cos u, y = cos v sin u, z = 2 sin v.

we see that x²+ y²+ (1/4)z² = 1, which yields the parametric representation of the ellipsoid r(u, v) = cos v cos ui + cos v sin uj + 2 sin vk,

Then

r_u = − cos v sin ui + cos v cos uj, r_v = − sin v sin ui − sin v cos uj + 2 cos vk.

The normal vector N 6= 0

N = r_u×r^v =

   

i j k

− cos v sin u cos v cos u 0

− sin v sin u − sin v cos u 2 cos v

   

= 2 cos²v cos ui+2 cos²v sin uj+sin v cos vk.

PROBLEM 9.5.24

Find the unit normal vector to the ellipsoid 4x²+ y² + 9z² = 36.

Solution. We have g(x, y, z) = 4x²+ y²+ 9z²− 36 = 0. Find the partial derivatives

∂g

∂x = 8x, ∂g

∂y = 2y, ∂g

∂z = 18z.

Then

grad g = 2[4x, y, 9z], |grad g| = 2

q

16x² + y²+ 81z², and the unit normal vector is given by

n = 1

|grad g|grad g = 1 2√

16x²+ y²+ 81z²grad g =

√ 1

16x²+ y²+ 81z²[4x, y, 9z] = 1

√16x² + y²+ 81z²(4xi + yj + 9zk).

PROBLEM 9.5.25

Find the unit normal vector of the plane 4x − 4y + 7z = −3.

Solution. We have z = 1/7(−3 − 4x + 4y). Therefore, setting x = u and y = v, we obtain a parametric representation

r(u, v) = [u, v, 1/7(−3 − 4u + 4v)] = ui + vj + 1/7(−3 − 4u + 4v)k.

Then

ru = [1, 0, −4/7], rv = [0, 1, 4/7],

and a normal vector N 6= 0

N = r_u × r^v =

   

i j k

1 0 −4/7 0 1 4/7

   

=

(4/7)i − (4/7)j + k = (1/7)(4i − 4j + 7k) = (1/7)[4, −4, 7].

The corresponding unit normal vector n = 1

|N|N = 1

9(4i − 4j + 7k).

On the other hand, we have the representation of the plane in the form g(x, y, z) = 4x − 4y + 7z + 3 = 0. Find the partial derivatives

∂g

∂x = 4, ∂g

∂y = −4, ∂g

∂z = 7.

Thus

grad g = [4, −4, 7], |grad g| =√

16²+ 16²+ 49 = 9, and the unit normal vector is given by

n = 1

|grad g|grad g = 1

9(4i − 4j + 7k) which coincides with the previous result.

PROBLEM 9.6.1

Compute the surface integral for F = [3x², y², 0] and S being a portion of the plane r(u, v) = [u, v, 2u + 3v], 0 ≤ u ≤ 2, −1 ≤ v ≤ 1.

Solution. We have

r_u = [1, 0, 2], r_v = [0, 1, 3];

a normal vector

N = r_u× r^v =

   

i j k 1 0 2 0 1 3

  

= −2i − 3j + k = [−2, −3, 1].

The corresponding unit normal vector n = 1

|N|N = 1

√14(−2i − 3j + k).

On S

F(r(u, v)) = F(S) = [3u², v², 0] = 3u²i + v²j).

Hence

F(r(u, v)) · N(u, v) = [3u², v², 0] · [−2, −3, 1] =

−6u²− 3v² = −3(2u²+ v²).

The parameters u, v vary in the rectangle R : 0 ≤ u ≤ 2, −1 ≤ v ≤ 1. Now we can write and calculate the surface integral: _Z

S

Z

F · ndA =

Z

R

Z

F(r(u, v)) · N(u, v)dudv = −3

Z

R

Z

(2u²+ v²)dudv =

= −6

Z 1

−1

dv

Z 2 0

u²du − 3

Z 1

−1

v²dv

Z 2

0 du == −12

Z 2 0

u²du − 6

Z 1

−1

v²dv =

−6[2 · (8/3) + 2/3] = −32 − 4 = −36.

PROBLEM 9.6.5

Compute the surface integral for F = [x − z, y − x, z − y] and S being a portion of the cone r(u, v) = u cos vi + u sin vj + uk, u, v in rectangleR : 0 ≤ v ≤ 2π, 0 ≤ u ≤ 3.

Solution. We have

ru = [cos v, sin v, 1], rv = [−u sin v, u cos v, 0], and a normal vector N 6= 0 of the cone

N = r_u× r^v =

   

i j k

cos v sin v 1

−u sin v u cos v 0

  

= −u cos vi − u sin vj + uk = −u[cos v, sin v, −1].

On S

F(r(u, v)) = F(S) = [u cos v − u, u sin v − u cos v, u − u sin v] = u[(cos v − 1)i + (sin v − cos v)j + (1 − sin v)k].

Hence

F(r(u, v)) · N(u, v) = u[cos v − 1, sin v − cos v, 1 − sin v] · (−u)[cos v, sin v, −1] =

−u²[cos v(cos v − 1) + sin v(sin v − cos v) + sin v − 1] = −u²(1 − cos v − sin v cos v + sin v − 1) =

−u²(sin v − cos v − sin v cos v).

The parameters u, v vary in the rectangle R : 0 ≤ v ≤ 2π, 0 ≤ u ≤ 3. Now we can write and calculate the surface integral:

Z

S

Z

F · ndA =

Z

R

Z

F(r(u, v)) · N(u, v)dudv =

−

Z

R

Z

u²(sin v − cos v − sin v cos v)dudv =

= −

Z 2π

0 (sin v − cos v − sin v cos v)dv

Z 3 0

u²du =

(−1/3)(

Z _2π

0 sin vdv −

Z _2π

0 cos vdv −

Z _2π

0 sin v cos vdv) = (−1/3)(0 + 0 + 0) = 0.

PROBLEM 9.7.1

Find the total mass of the mass distribution of the density σ = x²+ y²+ z² in the box T : |x| ≤ 1, |y| ≤ 3, |z| ≤ 2.

Solution. The required total mass M is given by the triple integral M =

Z Z

T

Z

σ(x, y, z)dxdydz =

Z ₂

z=−2

dz

Z ₃

y=−3

dy

Z ₁

x=−1

(x²+ y²+ z²)dx =

Z 2

−2

dz

Z 3

−3

dy

Z 1

−1

x²dx +

Z 2

−2

dz

Z 3

−3

y²dy

Z 1

−1

dx +

Z 2

−2

z²dz

Z 3

−3

dy

Z 1

−1

dx =

= 4 · 6 · 2

3 + 4 · 2 · 2 · 3²+ 6 · 2 · 16

3 = 16 + 144 + 64 = 224.

PROBLEM 9.7.13

Evaluate the surface integral for F = [x², 0, z²] over the surface of the box T : |x| ≤ 1, |y| ≤ 3, |z| ≤ 2.

Solution. We have

F₁ = x², F₂ = 0, F₃ = z². Hence the divergence of F = [F₁, F₂, F₃] is

div F = ∂F₁

∂x + ∂F₂

∂y + ∂F₃

∂z = 2x + 2z.

By the divergence theorem, the desired surface integral equals a triple integral over the box

T _Z

S

Z

F · ndA =

Z Z

T

Z

div FdV = 2

Z Z

T

Z

(x + z)dxdydz = 2

Z ₂

z=−2

dz

Z ₃

y=−3

dy

Z ₁

x=−1

(x + z)dxdydz =

Z 2

−2

dz

Z 3

−3

dy

Z 1

−1

xdx +

Z 2

−2

zdz

Z 3

−3

dy

Z 1

−1

dx = 0.

PROBLEM 9.7.15

Evaluate the surface integral for F = [cos y, sin x, cos z] over the surface of the cylinder S consisting of the cylinder x² + y² = 4 (|z| ≤ 2) and the circular disks z = −2 and z = 2 (x²+ y² ≤ 4)

Solution. We have

F₁ = cos y, F₂ = sin x, F₃ = cos z.

Hence the divergence of F = [F₁, F₂, F₃] is div F = ∂F₁

∂x + ∂F₂

∂y + ∂F₃

∂z = − sin z.

Introducing the polar coordinates

x = r cos θ, y = r sin θ (cylindrical coordinates r, θ, z) we have

dxdydz = rdrdθdz,

and by the divergence theorem, the surface integral is transformed to a triple integral over the closed region T in space whose boundary is the surface S of the cylinder of radius 2 and height 4,

Z

S

Z

F · ndA =

Z Z

T

Z

div FdV = −

Z Z

T

Z

sin zdxdydz = −

Z ₂

z=−2

sin zdz

Z ₂

r=0

Z _2π

θ=0

rdrdθ = 0.

PROBLEM 9.8.1

Verify basic property of solutions of Laplaces equation for f (x, y, z) = 2z²− x²− y² and S the surface of the box T : 0 ≤ x ≤ 1, 0 ≤ y ≤ 2, 0 ≤ z ≤ 4.

Solution. Set

F = grad f = [−2x, −2y, 4z].

Then

div F = div grad f = ∂²f

∂x² +∂²f

∂y² +∂²f

∂z² = ∇²f = −2 − 2 + 4 = 0, so that f (x, y, z) = 2z²− x²− y² is a harmonic function.

Now calculate directly the surface integral over the sixe successive sides of the surface of the box T beginning from its upper side parallel to the x, y-plane and situated in the plane z = 4, then taking the opposite side in the plane z = 0 etc.:

Z

S

Z ∂f

∂ndA =

Z

S

Z ∂f

∂z

 

_z=4dxdy −

Z

S

Z ∂f

∂z

 

_z=0dxdy+

Z

S

Z ∂f

∂x

  x=1

dydz −

Z

S

Z ∂f

∂x

  x=0

dydz+

Z

S

Z ∂f

∂y

  y=2

dxdz −

Z

S

Z ∂f

∂y

  y=0

dxdz = 4 · 4 · 2 − 0 + (−2) · 8 − 0 + (−4) · 4 − 0 = 0.

PROBLEM 9.8.3

Evaluate

I =

Z

S

Z

F · ndA, F = [x, z, y]

over the hemisphere S : x²+ y²+ z² = 4, z ≥ 0 by the divergence theorem.

Solution. We have

F = [x, z, y], F₁ = x, F₂ = z, F₃ = y.

The divergence of F is

div F = ∂F₁

∂x + ∂F₂

∂y + ∂F₃

∂z = 1 + 0 + 0 = 1, and, by the divergence theorem,

I =

Z Z

T,one half of ball

Z

div FdV =

Z Z

T

Z

dxdydz = 1 2 · 4

3π2³ = 16π 3 .