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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)
org(x, y, z) = 0.
For example,z = +
√a2 − 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-dimensionalregion 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 cosu, a sinu, v] = a cosui + a sinuj + vk,
u, v in rectangle R : 0 ≤ u ≤ 2π, −1 ≤ v ≤ 1.
The components of r(u, v) are
x = a cosu, y = a sinu, 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 cosui + a cos v sinuj + 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 cosu, y = a cos v sinu, 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 r̃(u(t), v(t)). By the chain rule, we get a tangent vector of acurve C
r̃′(t) =dr̃dt
=∂r∂uu′ +
∂r∂vv′.
Hence the partial derivatives ru and rv at a point P are tangential to S at P and we assumethat are linearly indepedent. Then their vector product gives a normal vector N of S at P ,
N = ru × rv 6= 0.
The corresponding unit normal vector n
n =1|N|
N =1
|ru × rv|ru × rv.
If S is represented byg(x, y, z) = 0,
thenn =
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 =
1a
grad 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 +√x2 + y2 = 0:
n =1
|grad g|grad g =
1√2
[x√
x2 + y2,
y√x2 + y2
,−1]
=
x√x2 + y2
i +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∫S
∫F · ndA =
∫R
∫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 = ru × rv 6= 0
and the corresponding unit normal vector
n =1|N|
N.
Note thatndA = 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 = [cosα, cosβ, cos γ] = cosαi + cos βj + cos γk,
N = [N1, N2, N3) = N1i +N2j +N3k,
and ∫S
∫F · ndA =
∫S
∫(F1 cosα + F2 cos β + F3 cos γ)dA =
∫S
∫(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
ru = [1, 2u, 0], rv = [0, 0, 1];
the vector product of ru and rv gives a normal vector N 6= 0 of the parabolic cylinder
N = ru × rv =
∣∣∣∣∣∣∣i j k1 2u 00 0 1
∣∣∣∣∣∣∣ = 2ui− j = [2u,−1, 0].
The corresponding unit normal vector
n =1|N|
N =1√
1 + 4u2(2ui− j).
On SF(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 andcalculate the flux integral:∫
S
∫F · ndA =
∫R
∫F(r(u, v)) ·N(u, v)dudv =
∫ 3
0
∫ 2
06(uv2 − 1)dudv = 6(
∫ 3
0v2dv
∫ 2
0udu−
∫ 3
0
∫ 2
0dudv) = 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 thisru = [1, 0,−1], rv = [0, 1,−1];
a normal vector
N = ru × rv =
∣∣∣∣∣∣∣i j k1 0 −10 1 −1
∣∣∣∣∣∣∣ = i + j + k = [1, 1, 1].
The corresponding unit normal vector
n =1|N|
N =1√3
(i + j + k).
On SF(r(u, v)) = F(S) = [u2, 0, 3v2] = u2i + 3v2k).
HenceF(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 andcalculate the surface integral:∫
S
∫F · ndA =
∫R
∫F(r(u, v)) ·N(u, v)dudv =
∫R
∫(u2 + 3v2)dudv =
∫ 1
0
∫ 1−v
0(u2 + 3v2)dudv =
∫ 1
0dv∫ 1−v
0u2du+ 3
∫ 1
0v2dv
∫ 1−v
0du =
= (1/3)∫ 1
0(1− v)3dv + 3
∫ 1
0v2(1− v)dv = (1/3)
∫ 1
0t3dt+ 3
∫ 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 functiondiv 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 iscontinuous and have continuous first partial derivatives everywhere in some domain containingT . Then ∫ ∫
T
∫div FdV =
∫S
∫F · ndA.
Here n is an outer unit normal vector of S pointing to the outside of S.In components,∫ ∫
T
∫ (∂F1
∂x+∂F2
∂y+∂F3
∂z
)dxdydz =
∫S
∫(F1 cosα + F2 cosβ + F3 cos γ)dA.
or ∫ ∫T
∫ (∂F1
∂x+∂F2
∂y+∂F3
∂z
)dxdydz =
∫S
∫(F1dydz + F2dzdx+ F3dxdy).
EXAMPLE 1 Evaluation of a surface integral by the divergence theorem
EvaluateI =
∫S
∫(x3dydz + x2ydzdx+ x2zdxdy),
where S is a piecewise smooth surface consisting of the cylinder x2 + y2 = a2 (0 ≤ z ≤ b) andthe circular disks z = 0 and z = b (x2 + y2 ≤ a2) (S consists of three parts of smooth surfaces).
Solution. We haveF1 = 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 havedxdydz = rdrdθdz,
and by the divergence theorem, the surface integral is transformed to a triple integral over theclosed region T in space whose boundary is the surface S of the cylinder,∫
S
∫(x3dydz + x2ydzdx+ x2zdxdy) =
∫ ∫T
∫div FdV =
∫ ∫T
∫5x2dxdydz =
5∫ b
z=0
∫ a
r=0
∫ 2π
θ=0r2 cos2 θrdrdθdz =
5b∫ a
0
∫ 2π
0r3 cos2 θdrdθ = 5b
a4
4
∫ 2π
0cos2 θdθ =
5ba4
8
∫ 2π
0(1 + 2 cos θ)dθ =
54πba4.
EXAMPLE 2 Verification of the divergence theorem
EvaluateI =
∫S
∫F · ndA), F = 7xi− zk
over the sphere S : x2 + y2 + z2 = 4 by the divergence theorem and directly.
Solution. We haveF = [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 =∫ ∫
T,ball
∫div FdV = 6
∫ ∫T,ball
∫dxdydz = 6 · 4
3π23 = 64π.
Now we will calculate the surface integral over S directly. Use the parametric representationof the sphere of radius 2
S : r(u, v) = 2 cos v cosui + 2 cos v sinuj + 2 sin vk,
u, v in rectangle R : 0 ≤ u ≤ 2π, −π/2 ≤ v ≤ π/2.
Thenru = [−2 sin u cos v, 2 cos v cosu, 0],
rv = [−2 sin v cosu,−2 sin v sinu, 2 cos v],
N = ru×rv =
∣∣∣∣∣∣∣i j k
−2 sinu cos v 2 cos v cosu 0−2 sin v cosu −2 sin v sinu 2 cos v
∣∣∣∣∣∣∣ = [4 cos2 v cosu, 4 cos2 v sinu, 4 cos v sin v].
On S we havex = 2 cos v cosu, z = 2 sin v,
and, correspondingly,
F(r(u, v)) = F(S) = [7x, 0,−z] = [14 cos v cosu, 0,−2 sin v].
HenceF(r(u, v)) ·N(u, v) =
(14 cos v cosu)4 cos2 v cosu+ (−2 sin v)(4 cos v sin v) = 56 cos3 v cos2 u− 8 cos v sin2 u.
The parameters u, v vary in the rectangle R : 0 ≤ u ≤ 2π, −π/2 ≤ v ≤ π/2. Now we canwrite and calculate the surface integral:∫
S
∫F · ndA =
∫R
∫F(r(u, v)) ·N(u, v)dudv =
8∫ 2π
0
∫ −π/2−π/2
(7 cos3 v cos2 u− cos v sin2 v)dudv =
8{
72
∫ 2π
0(1 + cos 2u)du
∫ π/2
−π/2cos3 vdv − 2π
∫ π/2
−π/2cos v sin2 vdv
}=
56π∫ π/2
−π/2cos3 vdv − 16π
∫ π/2
−π/2cos vdv sin2 vdv =
8π{
7∫ π/2
−π/2(1− sin2 v)d sin v − 2
∫ π/2
−π/2dv sin2 vd sin v
}=
8π{
7∫ 1
−1(1− t2)dt− 2
∫ 1
−1t2dt
}=
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,∫ ∫T
∫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 )
∫ ∫T
∫div FdV =
1V (T )
∫S(T )
∫F · ndA.
Choosing a fixed point P : (x1, y1, z1) in T and shrinking T down to onto P so that themaximum distance d(T ) of the points of T from P tends to zero, we obtain
div F(x1, y1, z1) = limd(T )→0
1V (T )
∫S(T )
∫F · ndA,
which is sometimes used as the definition of the divergence. From this expression, it followsthat 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 lineintegral of its normal derivative. In the same manner, by the divergence theorem, we cantransform the triple integral of the the Laplacian of a function into a surface integral of itsnormal 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∫ ∫T
∫∇2fdxdydz =
∫S
∫ ∂f
∂ndA.
Thus, if f(x, y, z) is a harmonic function in T (∇2f = 0 in T ), then the integral of thenormal derivative of this function over any piecewise smooth orientable surface S in T whoseentire 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-handedtriple:
v = [v1, v2, v3] = a× b =
∣∣∣∣∣∣∣i j ka1 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 ru and rv gives a normal vector N 6= 0 of the xy-plane
N = ru × rv =
∣∣∣∣∣∣∣i j k1 0 00 1 0
∣∣∣∣∣∣∣ = k.
The corresponding unit normal vector
n =1|N|
N =11k = 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
ru = [cos v, sin v, c], rv = [−u sin v, u cos v, 0],
and a normal vector N 6= 0 of the cone
N = ru × rv =
∣∣∣∣∣∣∣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 straightlines 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, weobtain 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
r̃(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
ru = [8, 0,−4], rv = [0, 6,−4],
and a normal vector N 6= 0
N = ru × rv =
∣∣∣∣∣∣∣i j k8 0 −40 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 cosu, y = cos v sinu, 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 cosui + cos v sinuj + 2 sin vk,
Then
ru = − cos v sinui + cos v cosuj, rv = − sin v sinui− sin v cosuj + 2 cos vk.
The normal vector N 6= 0
N = ru×rv =
∣∣∣∣∣∣∣i j k
− cos v sinu cos v cosu 0− sin v sinu − sin v cosu 2 cos v
∣∣∣∣∣∣∣ = 2 cos2 v cosui+2 cos2 v sinuj+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.
Thengrad g = 2[4x, y, 9z], |grad g| = 2
√16x2 + y2 + 81z2,
and the unit normal vector is given by
n =1
|grad g|grad g =
12√
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 obtaina parametric representation
r(u, v) = [u, v, 1/7(−3− 4u+ 4v)] = ui + vj + 1/7(−3− 4u+ 4v)k.
Thenru = [1, 0,−4/7], rv = [0, 1, 4/7],
and a normal vector N 6= 0
N = ru × rv =
∣∣∣∣∣∣∣i j k1 0 −4/70 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 =19
(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.
Thusgrad g = [4,−4, 7], |grad g| =
√162 + 162 + 49 = 9,
and the unit normal vector is given by
n =1
|grad g|grad g =
19
(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 haveru = [1, 0, 2], rv = [0, 1, 3];
a normal vector
N = ru × rv =
∣∣∣∣∣∣∣i j k1 0 20 1 3
∣∣∣∣∣∣∣ = −2i− 3j + k = [−2,−3, 1].
The corresponding unit normal vector
n =1|N|
N =1√14
(−2i− 3j + k).
On SF(r(u, v)) = F(S) = [3u2, v2, 0] = 3u2i + v2j).
HenceF(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 andcalculate the surface integral: ∫
S
∫F · ndA =∫
R
∫F(r(u, v)) ·N(u, v)dudv = −3
∫R
∫(2u2 + v2)dudv =
= −6∫ 1
−1dv∫ 2
0u2du− 3
∫ 1
−1v2dv
∫ 2
0du == −12
∫ 2
0u2du− 6
∫ 1
−1v2dv =
−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 = ru × rv =
∣∣∣∣∣∣∣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 SF(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 andcalculate the surface integral:∫
S
∫F · ndA =
∫R
∫F(r(u, v)) ·N(u, v)dudv =
−∫R
∫u2(sin v − cos v − sin v cos v)dudv =
= −∫ 2π
0(sin v − cos v − sin v cos v)dv
∫ 3
0u2du =
(−1/3)(∫ 2π
0sin vdv −
∫ 2π
0cos vdv −
∫ 2π
0sin 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 =∫ ∫
T
∫σ(x, y, z)dxdydz =
∫ 2
z=−2dz∫ 3
y=−3dy∫ 1
x=−1(x2 + y2 + z2)dx =
∫ 2
−2dz∫ 3
−3dy∫ 1
−1x2dx+
∫ 2
−2dz∫ 3
−3y2dy
∫ 1
−1dx+
∫ 2
−2z2dz
∫ 3
−3dy∫ 1
−1dx =
= 4 · 6 · 23
+ 4 · 2 · 2 · 32 + 6 · 2 · 163
= 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 haveF1 = 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 boxT ∫
S
∫F · ndA =
∫ ∫T
∫div FdV =
2∫ ∫
T
∫(x+ z)dxdydz = 2
∫ 2
z=−2dz∫ 3
y=−3dy∫ 1
x=−1(x+ z)dxdydz =
∫ 2
−2dz∫ 3
−3dy∫ 1
−1xdx+
∫ 2
−2zdz
∫ 3
−3dy∫ 1
−1dx = 0.
PROBLEM 9.7.15
Evaluate the surface integral for F = [cos y, sinx, cos z] over the surface of the cylinder Sconsisting of the cylinder x2 + y2 = 4 (|z| ≤ 2) and the circular disks z = −2 and z = 2(x2 + y2 ≤ 4)
Solution. We haveF1 = cos y, F2 = sinx, 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 havedxdydz = rdrdθdz,
and by the divergence theorem, the surface integral is transformed to a triple integral over theclosed region T in space whose boundary is the surface S of the cylinder of radius 2 and height4,∫S
∫F · ndA =
∫ ∫T
∫div FdV = −
∫ ∫T
∫sin zdxdydz = −
∫ 2
z=−2sin zdz
∫ 2
r=0
∫ 2π
θ=0rdrdθ = 0.
PROBLEM 9.8.1
Verify basic property of solutions of Laplaces equation for f(x, y, z) = 2z2 − x2 − y2 and S thesurface of the box T : 0 ≤ x ≤ 1, 0 ≤ y ≤ 2, 0 ≤ z ≤ 4.
Solution. SetF = 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.:∫
S
∫ ∂f
∂ndA =
∫S
∫ ∂f
∂z
∣∣∣∣∣z=4
dxdy −∫S
∫ ∂f
∂z
∣∣∣∣∣z=0
dxdy+
∫S
∫ ∂f
∂x
∣∣∣∣∣x=1
dydz −∫S
∫ ∂f
∂x
∣∣∣∣∣x=0
dydz+
∫S
∫ ∂f
∂y
∣∣∣∣∣y=2
dxdz −∫S
∫ ∂f
∂y
∣∣∣∣∣y=0
dxdz =
4 · 4 · 2− 0 + (−2) · 8− 0 + (−4) · 4− 0 = 0.
PROBLEM 9.8.3
EvaluateI =
∫S
∫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 =∫ ∫
T,one half of ball
∫div FdV =
∫ ∫T
∫dxdydz =
12· 4
3π23 =
16π3.
