M 312
D
T S
P
x2 z2 dS where S is the surface of the sphere x2 + y2 + z2 = a2 .
1. Verify the divergence theorem for
2. Calculate the surface integral
S
v · n dS where v = x − z2 , 0, xz + 1 and S is the surface that
S
encloses the solid region x2 + y2 + z 2 ≤ 4, z ≥ 0.
(a) directly,
(b) by the divergence theorem.
3. A closed surface is made up of S1 the part of the paraboloid z = 1 − x2 − y2 with z ≥ 0, together with
S2 , the circular disc x2 + y2 ≤ 1, z = 0.
Verify the divergence theorem for this closed surface for F = 2x, 2y, 1 .
3 2 y2
4. Verify the divergence theorem for 0 0 0 div F dz dy dx where F = x2 , 2y, 0 .
5. Verify the divergence theorem for F = x, y, 0 and for the closed surface S = S1 ∪ S2 ∪ S3 consisting of
• S1 : half of a circular disc x2 + y2 ≤ 1, x ≤ 0, z = 0,
√
• S2 : the part of the plane z = − 3x with x2 + y2 ≤ 1 and z ≥ 0, and
√
• S3 : the part of the cylinder x2 + y2 = 1 with x ≤ 0 and 0 ≤ z ≤ − 3x.
6. Calculate the surface integral
F · n dS where F = xyz, 0, 1 and S is the surface that encloses the
S
solid region x2 + y2 ≤ z, y ≥ 0, 1 ≤ z ≤ 4
(a) directly,
(b) by the divergence theorem.
2
+ y2 = z2 , above the hemisphere
7. Let F
below the cone x
= xz, yz, 0 and let V be the solid region
2
2
2
2
z = 4 − x − y , and inside the cylinder x + y = 4. Calculate
div F dV
V
(a) directly,
(b) by the divergence theorem.
8. Let V be the pyramid with the rectangular base z = 0, 0 ≤
x ≤ 2, 0 ≤ y ≤ 1 and the vertex above the
div F dV where F = 3, y3 , 1 .
base (apex) at (0, 0, 2). Verify the divergence theorem for
V
9. Calculate the surface integral
F · n dS where F = 1, 0, z2 and S is the surface that encloses the
S
solid region x2 + y2 + z2 ≤ 9, x, y, z ≤ 0
(a) directly,
(b) by the divergence theorem.
10. Let F = x, y, z and let D be the circular disc x2 + y2 ≤ 4, z = 0. Verify the divergence theorem for
4−x−y
div F dz dy dx.
D 0
Divergence Theorem Supplementary Problems - SOLUTION KEY
x2 z2 dS where S is the surface of the sphere x2 + y2 + z2 = a2 .
1. Verify the divergence theorem for
S
z
a
V
x
S
y
Using spherical
ρ = a, dS = a2 sin φdφdθ;
coordinateswith
2π π
LHS =
F · n dS =
x2 z 2 dS = 0 0 (a sin φ cos θ)2 (a cos φ)2 a2 sin φdφdθ
S
π
2π S
= a6 0 cos2 θdθ 0 sin3 φ cos2 φdφ
2θ
The first integral is evaluated by the power-reducing formula: cos2 θ = 1+cos
2
In
second integral, we substitute u = cos φ, du = − sin φ dφ so that
the
sin3 φ cos2 φdφ = (1 − cos2 φ) cos2 φ sin φdφ = − (1 − u2 )u2 du = u4 − u2 du =
π
2π 5
3
1
1
1
4
6
LHS = a6 12 θ + 14 sin 2θ 0 cos5 φ − cos3 φ = a6 (π) −1
5 + 3 − 5 + 3 = 15 πa
u5
5
−
u3
3
+C
0
The surface S is a level surface of g(x, y, z) = x2 + y2 + z2 .
ñ2x,2y,2z
Its unit normal vector is ±∇g
= ± xa , ya , za
|∇g| =
2
2
2
4x +4y +4z
To ensure the outward normal orientation, we pick the “+” sign: n = xa , ya , za .
An example of a vector field F such that F · n = x2 z2 is F = axz2 , 0, 0 .
∂
∂
∂
axz2 + ∂y
divF = ∂x
(0) + ∂z
(0) = az 2
Using spherical
coordinates with
dV = ρ2 sin φ dρ dφ dθ we have π
a 4
2π π a
2π
2 2
2
RHS =
0 0 (ρ cos φ) ρ sin φ dρ dφ dθ = a 0 dθ 0 cos φ sin φdφ 0 ρ dρ
V divF dV = a 0
Using substitution in the middle integral u = cos φ, we obtain
π 5 a
5
3
2π
φ
ρ
1
1
a
4
= 15
=
a(2π)(
+
)
πa6
RHS = a [θ]0 − cos
3
5
3
3
5
0
0
2. Calculate the surface integral
v · ndS where v = x − z2 , 0, xz + 1 and S is the surface that
S
encloses the volume x2 + y2 + z2 ≤ 4, z ≥ 0
(a) directly,
(b) by the divergence theorem.
z
z
2
2
z
S1
2
V
x
y
2
2
x
x
y
2
2
2
2
y
S2
v · ndS =
Part (a):
S
v · ndS +
S1
v · ndS
S2
S1 g(x, y, z) = x2 + y2 + z2 ;
±∇g
|∇g|
=
±2x,2y,2z
2
2
4x + 4y + 4z 2
2
2
16
2
=±
x
y z
2, 2, 2
; choose “+” sign: n =
2
v · n = (x − z2 ) x2 + (xz + 1) z2 = x2 − xz2 + xz2 + z2 = x2 + z2
Using spherical coordinates with ρ = 2, we have dS = 22 sin φ dφ dθ :
2π π/2 (2 sin φ cos θ)2 +2 cos φ
(4) sin φ dφ dθ
S1 v · ndS = 0
2
2π
π/20 3
2π π/2
2
= 8 0 cos θdθ 0 sin φdφ + 4 0 dθ 0 cos φ sin φ dφ
The integral sin3 φdφ is evaluated by substitution u = cos φ so that it becomes
3
− (1 − u2 )du = −u + u3 + C
2 π/2
θ sin 2θ 2π π/2
sin φ
− cos φ + 13 cos3 φ 0 + 4 [θ]2π
0
S1 v · ndS = 8 2 +
4
2
0
0
1
4 3
1
28
= 8π 1 − 3 + 8π 2 = 8π 6 + 6 = 3 π
z = 0; n = −k; z = 0; v · n = −1;
(−1)dS = −(area of S2 ) = −4π
S2
LHS =
v · ndS =
v · ndS +
v · ndS =
S2
S
S1
28
π
3
− 4π =
16π
3
S2
∂
∂
∂
Part (b): divv = ∂x
(0) + ∂z
(xz + 1) = 1 + x
x − z2 + ∂y
Using spherical coordinates with dV = ρ2 sin φ dρ dφ dθ
2π π/2 2
RHS =
divv dV = 0 0
(1 + ρ sin φ cos θ) ρ2 sin φ dρ dφ dθ
0
V
2π
2π π/2
2
2
π/2
= 0 dθ 0 sin φdφ 0 ρ2 dρ +
cos θdθ 0 sin2 φdφ 0 ρ3 dρ
0 π/2
= [θ]2π
0 [− cos φ]0
3 2
ρ
3
0
[sin θ]2π
0 =0
= (2π) (0 + 1)( 83 ) =
16
3 π
x
y z
2, 2, 2
;
3. A closed surface is made up of S1 the part of the paraboloid z = 1 − x2 − y2 with z ≥ 0, together with
S2 , the circular disc x2 + y2 ≤ 1, z = 0.
Verify the divergence theorem for this closed surface for F = 2x, 2y, 1 .
z
z
1
z
1
S1
1
V
x
1
1
x
y
1
1
y
x
1
1
y
S2
LHS =
F · ndS =
F · ndS +
S
S1
2
2
S1 g(x, y, z) = x + y + z;
F · ndS
S2
4x +4y +1
4x +4y +1
Projecting onto the xy-plane, then switching to polar coordinates we have
F · ndS =
S1
=
S2
[θ]2π
0
r4 +
r2
2
S2
1
0
2
√4x
2
+4y +1
4x2 +4y 2 +1
2
+4y 2 +1
4x2 +4y 2 +1
= ñ2x,2y,1
; choose “+”: n = √2x,2y,1
; F · n = √4x
2
2
2
2
∇g
|∇g|
= 3π
1/|n3 |
2π 1
4x2 + 4y2 + 1dx dy = 0 0 (4r2 + 1)r dr dθ
z = 0; n = −k; F · n = −1;
F · ndS =
S2
(−1)dS = −1(area of S2 ) = −π
S2
LHS = 3π − π = 2π
∂
∂
∂
divF = ∂x
(2x) + ∂y
(2y) + ∂z
(1) = 4
Using cylindrical coordinates with dV = r dz dr dθ, we obtain
2π 1 1−r2
2π 1
2
RHS =
divF dV = 4 0 0 0
r dz dr dθ = 4 0 0 r [z]z=1−r
dr dθ
z=0
V
1
1
2π 1
2π r 2
r4
2
= 4 0 0 r(1 − r )dr dθ = 4 [θ]0 2 − 4 = 4(2π) 4 = 2π
0
3 2 y2
4. Verify the divergence theorem for
0
0
0
div F dz dy dx where F = x2 , 2y, 0 .
z
z
z
z
z
4
4
4
4
4
S4
S1
S3
V
x 3
y
2
LHS =
x 3
2
F · ndS =
S
F · ndS +
S1
S1 g(x, y, z) = y − z = 0;
F · n = √−4y2
2
4y +1
F · ndS =
y
S2
x 3
F · ndS +
4y 2 +1
x 3
2
F · ndS +
S3
±0,2y,−1
= √
y
2
S2
∇g
|∇g|
2
F · ndS +
x 3
F · ndS
S4
S5
−0,2y,−1
. Choose “−”: n = √
y
4y 2 +1
0,−2y,1
= √
4y 2 +1
;
; project onto the xy-plane:
3 2
0
0
1/|n3 |
√−4y2
2
4y +1
2
3
2
3 −4y 3
2
4y2 + 1dy dx = 0 dx
−4y
dy
=
[x]
= −32
0
0
3
0
S1
S2 z = 0; n = −k; F · n = 0;
F · ndS = 0
S3 x = 3; n = i; . F · n = 9;
S2
F · ndS =
2 y2
0
0
9 dz dy =
2
0
2
[9z]z=y
z=0 dy =
2
0
2
9y2 dy = 3y3 0 = 24
S3
S4 x = 0; n = −i; F · n = 0;
F · ndS = 0
S4
S5 y = 2; n = j; F · n = 4;
F · ndS = 4 (Area of the rectangle S5 ) = 4(3)(4) = 48
S5
LHS = −32 + 0 + 24 + 0 + 48 = 40
2
∂
∂
x + ∂y
(2y) + ∂z
(0) = 2x + 2
2
32 y
3 2 y2
3
2
2
RHS = 0 0 0 div F dz dy dx = 0 0 0 (2x + 2) dz dy dx = 0 (2x + 2) 0 [z]z=y
z=0 dy dx
3 3 y=2
3
2 2
y dy = x2 + 2x 0 y3
= (15)( 83 ) = 40
= 0 (2x + 2) dx
0
div F =
S5
∂
∂x
y=0
2
y
5. Verify the divergence theorem for F = x, y, 0 and for the closed surface S = S1 ∪ S2 ∪ S3 consisting of
• S1 : half of a circular disc x2 + y2 ≤ 1, x ≤ 0, z = 0,
√
• S2 : the part of the plane z = − 3x with x2 + y2 ≤ 1 and z ≥ 0, and
√
• S3 : the part of the cylinder x2 + y2 = 1 with x ≤ 0 and 0 ≤ z ≤ − 3x.
z
z
z
z
√⎯3
√⎯3
√⎯3
√⎯3
S2
S3
V
x
−1
−1
1
−1
−1
y
x
y
1
x
−1
−1
1
y
x
−1
−1
1
S1
LHS =
F · ndS +
S
S1
z = 0; n = −k; F · n = 0;
S1
S2
F · ndS =
S2
3
, 0, 12
2
F · ndS
S3
F · ndS = 0 ;
S1
√
√
F · ndS +
√
; F · n = 23 x;
3x + z = 0; n =
Projecting onto the xy-plane, then switching to polar coordinates yields
3|
√ 1/|n
√ 3π/2 1
3
F · ndS =
x (2) dy dx = 3 π/2 0 r cos θ r dr dθ
2
S2
=
√
S1
3π/2
3 [sin θ]π/2
2
r3
3
1
0
2
S3 g(x, y, z) = x + y ;
=
√
3 (−1 − 1) ( 13 ) =
∇g
|∇g|
√
−2 3
3
±2x,2y,0
=√
= ± x, y, 0 . Choose “+” : n = x, y, 0 ; F · n = x2 + y2 = 1.
2
2
4x +4y
Integrating
directly in cylindrical
coordinates with dS = r dz dθ we have
√ 3π/2
√
√
√
3π/2 −√3 cos θ
F · ndS = π/2 0
1 dz dθ = − 3 π/2 cos θ dθ = − 3 [sin θ]3π/2
= − 3(−1 − 1) = 2 3
π/2
S3
LHS = 0 −
√
2 3
3
√
+2 3 =
√
4 3
3
∂
∂
∂
div F = ∂x
(x) + ∂y
(y) + ∂z
(0) = 2
Using cylindrical coordinates with dV √
= r dz dr dθ, we obtain
√ 3π/2 1
3π/2 1 − 3r cos θ
RHS =
divF dV = 2 π/2 0 0
r dz dr dθ = −2 3 π/2 0 r2 cos θ dr dθ
V
√
√ 3 1
√
3π/2
= −2 3 r3 [sin θ]π/2 = −2 3( 13 )(−1 − 1) = 4 3 3
0
y
F · ndS where F = xyz, 0, 1 and S is the surface that encloses the
6. Calculate the surface integral
S
solid region x2 + y2 ≤ z, y ≥ 0, 1 ≤ z ≤ 4
(a) directly,
(b) by the divergence theorem.
z
z
S2
4
z
z
z
4
4
4
S4
1
x
V
1
y
1
2
x
y
1
S
x
F · ndS +
S1
F · ndS +
2
1
y
1
x
2
1
F · ndS +
S3
F · ndS
S4
F · n dS = −1(Area of S1 ) = − π2
S1 z = 1; n = −k; F · n = −1;
S2 z = 4; n = k; F · n = 1;
x
S2
y
2
F · ndS =
Part (a):
1
S1
S3
S1
F · n dS = (Area of S2 ) = 2π
S2
±2x,2y,−1
∇g
S3 satisfies x2 + y2 = z, i.e., g(x, y, z) = x2 + y2 − z = 0; |∇g|
=√
;
2
2
2
choose “+” : n = √2x,2y,−1
; F · n = √ 2x2 yz−12
2
2
4x +4y +1
4x +4y +1
4x +4y +1
;
Denote D to be the projection of S3 onto the xy-plane. Switching to polar coordinates yields
F · ndS =
1/|n 3 |
2
D
2
2x y(x
√
+y )−1
2 + 4y 2 + 1dy dx = π 2 2r 2 cos2 θ r sin θ r 2 − 1 r dr dθ
4x
0 1
2
2
4x2 +4y +1
S3
2 6 π 2
2
cos
θ
sin
θ
dθ
r
dr
−
dθ
r
dr
=
0
1
0
1
128
1
3π
508
3π
= −2
3 (−1 − 1)
7 − 7 − 2 = 21 − 2
S4 y = 0; n = −j; F · n = 0;
F · ndS = 0
2
π
−2
3
2 2
3 π r 7 2
π
cos θ 0 7 − [θ]0 r2
1
1
S4
F · ndS = − π2 + 2π +
508
21
−
3π
2
+0=
508
21
S
∂
∂
∂
Part (b): div F = ∂x
(xyz) + ∂y
(0) + ∂z
(1) = yz
Using cylindrical coordinates with dV = r dr dz dθ, we obtain
√
2 7/2 4
π 4 √z
π
4 r3 r= z
divF
dV
=
r
sin
θ
zr
dr
dz
dθ
=
sin
θ
z
dz dθ = [− cos θ]π0 21
z
V
3
0 1 0
0
1
1
r=0
= (−(−1) −
(−1))( 256
21
−
2
)
21
=
508
21
y
2
7. Let F
below the cone x
+ y2 = z2 , above the hemisphere
= xz, yz, 0 and let V be the solid region
2
2
2
2
div F dV
z = 4 − x − y , and inside the cylinder x + y = 4. Calculate
V
(a) directly,
(b) by the divergence theorem.
z
z
z
z
2
2
√⎯2
2
S2
S1
S3
V
x
y
2
2
x
2
2
y
x
y
2 √⎯2
x
2
2
y
∂
∂
∂
Part (a): div F = ∂x
(xz) + ∂y
(yz) + ∂z
(0) = 2z
Using
cylindrical
coordinates
with
dV
= r dr dz dθ, we obtain
2π 2 r
2π 2 z=r
√
√
divF
dV
=
2zr
dz
dr
dθ = 0 √2 r z2 z=√4−r2 dr dθ
2
0
V
2
4−r
4
2
2π 2
2π
= 0 √2 r(r2 − (4 − r2 )) dr dθ = [θ]0 r2 − 2r2 √ = (2π)(8 − 8 − 2 + 4) = 4π
2
(Note that
divF dV can also be evaluated using spherical coordinates with dV = ρ2 sin φ dρ dφ dθ as
V
2π π/2 2/ sin φ
2ρ cos φ ρ2 sin φ dρ dφ dθ)
0
π/4 2
F · ndS =
Part (b):
S
F · ndS +
S1
F · ndS +
S2
F · ndS
S3
ñ2x,2y,2z
S1 g(x, y, z) = x2 + y2 + z2 = 4; ±∇g
= ± x2 , y2 , z2 ;
|∇g| =
4x2 +4y 2 +4z 2
2
−y −z
−x2 z−y 2 z
choose “−” sign: n = −x
= −r2 z
2 , 2 , 2 ; F·n=
2
Using spherical coordinates with ρ = 2, we have dS = 22 sin φ dφ dθ :
π/2
2π π/2 −(2 sin φ)2 (2 cos φ)
2π sin4 φ
1
F·ndS
=
(4)
sin
φ
dφ
dθ
=
−16
[θ]
= −16(2π)( 14 − 16
) = −6π
0
0
π/4
S1
2
4
π/4
2
2
2
S2 g(x, y, z) = x + y − z = 0;
±∇g
|∇g|
±2x,2y,−2z
=√
; choose “−” sign: n = √−2x,−2y,2z
.
2
2
2
4x2 +4y2 +4z 2
2
2
2
2
2
−(ρ sin π
)
(ρ
cos π
−2x
z−2y
z
−r
z
4
4)
√ .
F·n = √ 2 2 2 = ρ =
= −ρ
ρ
2 2
4x +4y +4z
Using spherical coordinates with φ = π4 , we have dS = ρ sin π4
2√2
2π 2√2 −ρ2 ρ 2π ρ4
−1
√
√
dρ
dθ
=
F · ndS = 0 2
[θ]
0
4
4
2 2
2
2
S2
4x +4y +4z
dρ dθ :
=
−1
4 (2π)(16
√ ±2x,2y,0 = ± x2 , y2 , 0 ;
S3 g(x, y, z) = x2 + y2 = 4; ±∇g
|∇g| =
4x2 +4y2 +4z 2
2
2
z
choose “+” sign: n = x2 , y2 , 0 ; F · n = x z+y
= 2z
2
Using
cylindrical coordinates with r = 2, we have dS = 2 dz dθ :
2π 2
2 2
F · ndS = 0 0 (2z) (2) dz dθ = [θ]2π
2z 0 = (2π)(8) = 16π
0
S3
F · ndS = −6π − 6π + 16π = 4π
S
− 4) = −6π
8. Let V be the pyramid with the rectangular base z = 0, 0 ≤
x ≤ 2, 0 ≤ y ≤ 1 and the vertex above the
base (apex) at (0, 0, 2). Verify the divergence theorem for
div F dV where F = 3, y3 , 1 .
V
z
z
z
z
z
2
2
2
2
2
S2
S4
S3
S5
V
x
1
2
LHS =
S
y x
F · ndS =
S1
2
1
S1
F · ndS +
S2
S1 z = 0; n = −k; F · n = −1;
S2 y = 0; n = −j; F · n = 0;
S1
S2
y x
F · ndS +
1
2
S3
F · ndS +
S4
y x
1
2
F · ndS +
S5
y x
2
1
F · ndS
F · ndS = −(Area of the rectangle S1 ) = −2
F · ndS = 0
−→ −→
S3 part of the plane containing A(2, 0, 0), B(2, 1, 0), D(0, 0, 2); normal vector AB × AD = 2, 0, 2 .
The plane equation is 2(x − 2) + 2z = 0,which simplifies
to x + z = 2.
√
1
1
√ = 2 2.
The outward unit normal vector is n = √2 , 0, √2 ; F · n = 3+1
2
√
√
√
F · ndS = 2 2(Area of triangle S3 ) = 2 2( 12 ) (1) (2 2) = 4
S3
(same result could also be calculated by projecting onto the yz-plane
1/|n 1 |
S3
√ √ 2 2
F · ndS =
2 dy dz = 4(Area of triangle S4 ))
S4
S4 x = 0; n = −i; F · n = −3;
S4
F · ndS = −3(Area of triangle S4 ) = −3
−→ −→
S5 part of the plane containing B(2, 1, 0), C(0, 1, 0), D(0, 0, 2); normal vector CB × CD = 0, −4, −2 .
The plane equation is −4(y − 1) − 2z =0, which simplifies
to 2y + z = 2.
3
2
1
The outward unit normal vector is n = 0, √5 , √5 ; F · n = 2y√+1
.
5
Projecting onto the xz-plane we obtain
1/|n 2 |
√
z=2−x
2 2−x 2 −2 2(1− z2 )3 +1
5
z 3
1
z 4
z
√
1
−
dz
dx
=
1
−
dz
dx
=
F·ndS =
+
+
dx
2
2
4
2
2
0 0
0
5
2
z=0
S5
S2
5
2
2 4
x
1
−x
x2
3x
= 95
= 0 −x
32 + 1 − 2 + 2 dx = 160 − 4 + 2
0
S
F · ndS = −2 + 0 + 4 − 3 +
9
5
=
4
5
3
∂
∂
y + ∂z
(3) + ∂y
(1) = 3y2
1 2−2y 2−z 2
1 2−2y x=2−z
RHS =
divF dV = 0 0
3y dx dz dy = 3 0 y2 0
[x]x=0 dz dy
0
V
1 2−2y
1 1
2 z=2−2y
= 3 0 y2 0
(2 − z) dz dy = 3 0 y2 2z − z2
dy = 3 0 y2 (4 − 4y − 2 + 4y − 2y2 )dy
z=0
1
= 2y3 − 65 y5 0 = 45
div F =
∂
∂x
y
9. Calculate the surface integral
F · n dS where F = 1, 0, z2 and S is the surface that encloses the
S
solid region x2 + y2 + z2 ≤ 9, x, y, z ≤ 0.
(a) directly,
(b) by the divergence theorem.
z
z
−3
−3
−3
−3
y
x
z
−3
−3
y
x
−3
S
−3
F · n dS =
S1
z
−3
y
x
F · n dS +
S1 x = 0; n = i; F · n = 1;
S1
S2 y = 0; n = j; F · n = 0;
S2
S3 z = 0; n = k; F · n = 0;
−3
S2
F · n dS +
y
x
S4
S3
−3
F · n dS +
F · n dS = (Area of S1 ) =
S4
−3
F · n dS
9π
4
F · n dS = 0
S3
F · n dS = 0
S4 g(x, y, z) = x2 + y2 + z2 = 9;
±∇g
|∇g|
= ñ2x,2y,2z
2
2
4x +4y
3
+4z 2
=±
x
y z
3, 3, 3
; choose “+”: n =
x
y z
3, 3, 3
F · n = x+z
3 ;
Using spherical coordinates with ρ = 3, we have dS = 32 sin φ dφ dθ
3π/2 π 3 cos θ sin φ+(3 cos φ)3 2
F · n dS = π
(3 ) sin φ dφ dθ
3
π/2
S4
π
3π/2
π
3π/2
2
3
=9 π
cos θ dθ ( π/2 sin φ dφ) + 81 π
dθ
cos φ sin φ dφ
π/2
1−cos(2φ)
2
sub stitute u=cos φ
π
π
3π/2 φ
3π/2 cos 4 φ
= 9 [sin θ]π
− sin(2φ)
− 81 [θ]π
= 9(−1)( π2 − π4 ) − 81( π2 )( 14 ) = − 99
π
2
4
4
8
π/2
S
F · n dS =
9π
4
+0+0−
99
π
8
π/2
= − 81
π
8
2
∂
∂
∂
Part (b): div F = ∂x
(1) + ∂y
(0) + ∂z
z = 2z = 2ρ cos φ
3π/2 π 3
divF dV = π
2ρ cos φ ρ2 sin φ dρ dφ dθ
π/2 0
V
π
3π/2
3 3
dθ
cos φ sin φ dφ
ρ
dρ
=2 π
0
=
π/2
−3
−3
S2
S1
z
−3
y
x
V
Part (a):
S3
su bstitu te u=sin φ
2 π 4 3
3π/2 sin φ
ρ
81
= 2( π2 )( −1
2 [θ]π
2
4
2 )( 4 )
π/2
0
= − 81
8 π
10. Let F = x, y, z and let D be the circular disc x2 + y2 ≤ 4, z = 0. Verify the divergence theorem for
4−x−y
div F dz dy dx.
D 0
z
z
z
z
S3
S2
V
2
x
LHS =
S
y
y
2
D=S1
F · n dS =
S1
2
x
F · n dS +
S2
S1 z = 0; n = −k; F · n = −z = 0;
S2 g(x, y, z) = x2 + y2 = 4;
2
2
±∇g
|∇g|
F · n dS +
S1
y
2
x
S3
y
2
2
x
2
2
F · n dS
F · n dS = 0
= √ ±2x,2y,0
2
2
4x +4y +4z 2
=±
x
y
2, 2,0
; choose “+”: n =
x
y
2, 2,0
;
2
= r2 = 2;
F · n = x +y
2
using cylindrical coordinates with r = 2, we have dS = 2 dz dθ :
2π 4−2 cos θ−2 sin θ
2π
F · n dS = 0 0
(2)(2) dz dθ = 4 0 (4 − 2 cos θ − 2 sin θ) dθ
S2
= 4 [4θ − 2 sin θ + 2 cos θ]2π
0 = 32π
S3 part of the plane z = 4 − x − y, i.e., x + y + z = 4;
a normal vector is ± 1, 1, 1 ; the outward unit normal is n =
we project onto the xy-plane
√1 , √1 , √1
3
3
3
1/|n3 |
S
S3
F · n dS =
S1
√
√4
3 dy dx = 4(Area of S1 ) = 16π
3
F · n dS = 0 + 32π + 16π = 48π
∂
∂
(x) + ∂y
(y) + ∂z
(z) = 3
2π 2 4−r cos θ−r sin θ
RHS =
divF dV = 0 0 0
3r dz dr dθ
2π 2
r=2
2π 2 V
= 0 0 3r (4 − r cos θ − r sin θ) dr dθ = 0 6r − r3 cos θ − r3 sin θ r=0
2π
2π
= 0 (24 − 8 cos θ − 8 sin θ) dθ = [24θ − 8 sin θ + 8 cos θ]0 = 48π
div F =
∂
∂x
; F·n=
x+y+z
√
3
=
√4
3