M 312
S T
S
P
F · dr where F = x + y + z, y + z, z and C is the intersection of the plane
1. Calculate the integral
C
x = y and the cylinder y2 + z2 = 1
(a) directly,
(b) by Stokes’ theorem.
2. Verify
Stokes’ theorem on the triangle with vertices (2, 0, 0) , (0, 2, 0) , (0, 0, 2) and
F = x2 + y2 , y2 , x2 + z2 .
3. Verify Stokes’ theorem for v = y2 − xyz, 3x + 2x2 , x3 + y and C is the curve of intersection of the
sphere x2 + y2 + z2 = 25 and the plane z = −4.
4. Verify Stokes’ theorem for F = y, −x, yz over the part of the paraboloid z = 2 x2 + y2 for which
z ≤ 12 .
5. Calculate the integral 2ydx − zdy + ydz with C the intersection of the plane 2x + y + z = 0 and the
cylinder x2 + y2 = 2y
C
(a) directly,
(b) by Stokes’ theorem.
6. Let S be the part of the surface z = y1 with 0 ≤ x ≤ 3 and 1 ≤ y ≤ 2 oriented upwards and let
F = z, x2 , x . Calculate S curl F · n dS
(a) directly,
(b) by Stokes’ theorem.
7. Verify Stokes’ theorem for F = y, −x, xy over the part of the sphere x2 + y2 + z2 = 1 for which x ≥ 0,
z ≥ 0.
8. Verify Stokes’ theorem for F = 0, 0, x2 − y2 over the part of the surface x2 + y2 = z2 inside the first
octant with 1 ≤ z ≤ 2 .
9. Verify Stokes’ theorem for F = x − y, y − z, z − x over the part of the surface x2 + y2 = 9 inside the
first octant with z ≤ 4 and y ≤ x.
10. Consider the points K(2, 0, 0), L(0, 2, 0), M(0, 1, 2), and N(1, 0, 2).
(a) Show that K, L, M, and N are coplanar.
(b) Let C be made up of line
segments: C1 from K to L, C2 from L to M, C3 from M to N, and C4
from N to K. Evaluate C y dx + x2 dy + y dz directly.
(c) Evaluate the line integral in part (b) by Stokes’ theorem.
Stokes’ Theorem Supplementary Problems - SOLUTION KEY
z
F · dr where F = x + y + z, y + z, z
1. Calculate the integral
C
C
and C is the intersection of the plane x = y and the cylinder
y2 + z2 = 1
n
(a) directly
x
(b) by Stokes’ theorem.
y
S
Part (a): directly,
The curve of intersection of x = y and y2 + z2 = 1 can be parameterized using t going from 0 to 2π :
x = cos t, y = cos t, z = sin t
(we are taking a standard parameterization around the circle y2 + z2 = 1, then set x = y).
Therefore,
dx
= − sin t dt, dy = − sin t dt, dz = cos
t dt.
F · dr = (x + y + z)dx + (y + z)dy + zdz = (y + z)dx + zdy
C
C
C
(contributions
from xdx, ydy, and zdz on a closed path are all zero).
2π 2π
F · dr = 0 (cos t + sin t)(− sin t) − sin2 t dt = 0 (− sin t cos t − 2 sin2 t)dt
C
=
2π
0
(− sin t cos t − 1 + cos 2t)dt = − 12 sin2 t − t +
1
2
2π
sin 2t 0 = −2π .
Part (b): by Stokes’ theorem.
The interior of the curve C from part (a) can be projected onto the unit disc in the yz - plane.
The original surface S has an equation x = y. A normal vector to the plane x − y = 0 is 1, −1, 0 .
Our unit normal is n = ± √12 1, −1, 0 .
Choose “+”
for consistency with the orientation of C chosen in part (a).
i
j
k ∂/∂y ∂/∂z curl F = ∂/∂x
x+y+z y+z
z ∂
∂
∂
∂
∂
∂
= ∂y z − ∂z (y + z) i − ( ∂x z − ∂z
(x + y + z))j + ( ∂x
(y + z) − ∂y
(x + y + z))k = −1, 1, −1
(curl F · n) dS =
S
1/|n1 |
(−2)
√
R
2
√
2 dydz = −2 R 1dydz = −2(area of R) = −2π
since R is the unit disc in the yz-plane.
z
2. Verify Stokes’ theorem on the triangle with
vertices
(2, 0, 0) , (0, 2, 0), (0, 0, 2) and
F = x2 + y2 , y2 , x2 + z2 .
S
2
n
C3
x
2
2
C1
LHS = C F · dr = C (x2 + y2 )dx + y2 dy + (x2 + z2 )dz = C y2 dx + x2 dz
since contributions from x2 dx, y2 dy, and z2 dz on a closed path are all zero.
C = C1 ∪ C2 ∪ C3
C1 segment from (2, 0, 0) to (0, 2, 0) can be parameterized using t :
x = 2 − 2t; y = 2t; z = 0; dx = −2dt; dy = 2dt; dz = 0; t goes from 0 to 1 :
1
−8
3 1
y2 dx + x2 dz = 0 (2t)2 (−2)dt = −8
3 [t ]0 = 3
C1
C2 segment from (0, 2, 0) to (0, 0, 2) can be parameterizes using t :
y = 2 − 2t; z = 2t; dx = 0; dy = −2dt; dz = 2dt; t goes from 0 to 1:
x = 0;
2
y
dx
+ x2 dz = 0
C2
C3 segment from (0, 0, 2) to (2, 0, 0) can be parameterized using t :
x = 2t; y = 0; z = 2 − 2t; dx = 2dt; dy = 0; dz = −2dt; t goes from 0 to 1 :
1
−8
3 1
y2 dx + x2 dz = 0 (2t)2 (−2)dt = −8
3 [t ]0 = 3
C3
C
F · dr =
RHS =
−8
3
+0−
8
3
= − 16
.
3
(curl F · n) dS
i
j
k
∂/∂z = (0) i − (2x)j + (−2y)k = 0, −2x, −2y .
curl F = ∂/∂x ∂/∂y
x2 + y2
y2
x2 + z 2 The
plane
containing
the
three
points has a normal vector
i j k −2 2 0 = (4) i − (−4)j + (4)k = 4, 4, 4 or 1, 1, 1 .
−2 0 2 The plane equation is(x − 2) + y +
z = 0, i.e. x + y + z = 2.
1
1
1
The unit normal is ± √3 , √3 , √3 .
For consistency with the orientation of C chosen in part (a), we select the “+” sign.
−2
curl F · n = √
(x + y).
3
Projecting onto the xy-plane we obtain
S
1/|n 3 |
√
2 2−x
−2
√
(curl
F
·
n)
dS
=
(x
+
y)
3 dydx = −2 0 0 (x + y)dydx =
0 0
S
3
2
2
2
2 y=2−x
dx
= −2 0 xy + y2
dx = −2 0 x(2 − x) + (2−x)
2
y=0
3
2
2
= 0 x2 − 4 dx = x3 − 4x = 83 − 8 = − 16
3
2 2−x
0
C2
y
z
3. Verify Stokes’ theorem for v = y2 − xyz, 3x + 2x2 , x3 + y
and C is the curve of intersection of the sphere
x2 + y2 + z2 = 25 and the plane z = −4.
x
S
−4
−5
LHS = C v · dr
The sphere intersects the plane along the circle with radius 3 centered at (0, 0, −4).
This curve C can be parameterized using θ going from 0 to 2π :
x
sin θdθ; dy = 3 cos θdθ; dz = 0;
= 3 cos θ; y = 32 sin θ; z = −4; dx = −3
2
v
·
dr
=
(y
−
xyz)dx
+
(3x
+
2x
)dy
+ (x3 + y)dz
C
C
2π = 0 −27 sin3 θ − 108 sin2 θ cos θ + 27 cos2 θ + 54 cos3 θ dθ = 27π
RHS =
(curl v · n) dS
i
j
k
∂/∂y
∂/∂z = (1) i − (3x2 + xy)j + (3 + 4x − 2y + xz)k
curl v = ∂/∂x
y2 − xyz 3x + 2x2 x3 + y = 1, −3x2 − xy, 3 + 4x − 2y + xz
Taking S to be the disc z = −4, x2 + y2 ≤ 9, the unit normal to S is n = k
(orientation chosen to be consistent with the orientation of C)
curl v · k = 3 + 4x − 2y + xz = 3 − 2y since z = −4.
Using cylindrical coordinates,
r=3
2π 3
2π 3r2 2 3
(curl
v
·
n)
dS
=
(3
−
2r
sin
θ)r
dr
dθ
=
−
r
sin
θ
dθ
2
3
0
0
0
S
r=0
27
2π
2π 27
= 0
2 − 18 sin θ dθ = 2 θ + 18 cos θ 0 = 27π
S
y
n
C
z
4. Verify Stokes’ theorem
y, −x, yz over the part
for F =
of the paraboloid z = 2 x2 + y2 for which z ≤ 12 .
n
C
S
x
LHS = C F · dr
Substituting z = 12 into the paraboloid equation yields 12 = 2 x2 + y2 , i.e., x2 + y2 = 14 .
The curve C is a circle that can be parameterized using θ going from 0 to 2π :
x = 12 cos θ; y = 12 sin θ; z = 12 ; dx = −1
sin θ dθ; dy = 12 cos θ dθ; dz = 0
2
2π 2π
F · dr = C ydx − xdy + yzdz = 0 −1
sin2 θ − 14 cos2 θ dθ = −1
dθ = −1
[θ]2π
0 =
0
C
4
4
4
RHS =
y
−π
2
(curl F · n) dS
i
j
k curl F = ∂/∂x ∂/∂y ∂/∂z = (z) i − (0)j + (−2)k = z, 0, −2 .
y
−x
yz The paraboloid can be considered a level surface of g(x, y, z) = 2x2 + 2y2 − z.
√±4x,4y,−1
n = ±∇g
- we must pick the “−” sign for consistency with the orientation of C.
|∇g| =
2
2
S
curl F · n = √
16x +16y +1
1
(−4xz
16x2 +16y 2 +1
− 2)
Project the surface onto the disc D : x2 + y2 ≤
S
(curl F · n) dS =
D
√
1
16x2 +16y 2 +1
1
4
in the xy-plane:
1/|n 3 |
2
2
2
2
− 2) 16x + 16y + 1dx dy
(−4x 2 x + y
Switch to polar coordinates:
r=1/2
2π −8r5
2π 1/2
3
2
(−8r
cos
θ
−
2)r
dr
dθ
=
cos
θ
−
r
dθ
0
0
0
5
r=0
2π 2π
2π
θ
− 14 dθ = −1
[sin θ]0 − 14 [θ]0 = −π
.
= 0 − cos
20
20
2
z
2ydx − zdy + ydz with
5. Calculate the integral
C
C the intersection of the plane 2x + y + z = 0
and the cylinder x2 + y2 = 2y
(a) directly
C
S
x
2x+y+z=0
(b) by Stokes’ Theorem
y
n
2
2
x +y =2y
Part (a): directly
Let us begin by completing the square in the cylinder’s equation:
x2 + y2 = 2y
x + y2 − 2y = 0
x2 + y2 − 2y + 1 = 1
x2 + (y − 1)2 = 1
2
The cylinder’s generating curve is a circle in the xy-plane centered at (0, 1, 0), radius 1.
The intersection of the cylinder and a plane is an ellipse that can be parameterized using
t going from 0 to 2π :
x = cos t
y = 1 + sin t
z = −2 cos t − 1 − sin t (obtained from the plane equation, z = −2x − y)
Therefore, dx = − sin
t dt; dy = cos t dt; dz = (2 sin t − cos t) dt
LHS = C F · dr = 2ydx − zdy + ydz
C
2π
= 0 [2(1 + sin t)(− sin t) − (−2 cos t − 1 − sin t) cos t + (1 + sin t)(2 sin t − cos t)] dt
2π = 0 −2 sin t − 2 sin2 t + 2 cos2 t + cos t + sin t cos t + 2 sin t + 2 sin2 t − cos t − sin t cos t dt
2π
2π
2π
= 0 2 cos2 t dt = 0 (1 + cos 2t) dt = θ + 12 sin 2θ 0 = 2π
Part (b): by
Stokes’ theorem.
RHS = S (curl F · n) dS
i
j
k curl F = ∂/∂x ∂/∂y ∂/∂z = (2) i − (0)j + (−2)k = 2, 0, −2 .
2y
−z
y A normal vector to the plane 2x + y + z = 0 is 2, 1, 1 . A unit normal is n = ± √16 2, 1, 1 .
Choose “+” for consistency with the orientation of C chosen in part (a).
curl F · n = √26
Project S onto the circular disc D : x2 + (y − 1)2 ≤ 1 in the xy-plane:
S
(curl F · n) dS =
√2 dS
S 6
=
1/|n 3 |
√2
D 6
√
6 dx dy = 2 D 1 dx dy = 2(area of D) = 2π
6. Let S be the part of the surface z = y1 with 0 ≤ x ≤ 3
and 1 ≤ y ≤ 2 oriented upwards and let F = z, x2 , x .
Calculate S curl F · n dS
z
C1
(a) directly,
x
(b) by Stokes’ theorem.
3
1
S
n
1
C2
Part (a): directly
i
j
k curl F = ∂/∂x ∂/∂y ∂/∂z = (0) i − (0)j + (2x)k = 0, 0, 2x
z
x2
x S is a level surface of g(x, y, z) = yz (with g = 1); a unit normal vector to S is
C3
±∇g
|∇g|
C4
2
±0,z,y
=√
;
2
2
choose “+” for upward orientation: n = √0,z,y
; curl F · n = √ 2xy
2
2
2
z +y 2
1/|n 3 |
z +y
2
2
z +y
dx dy = 2x dx dy
Project onto the xy-plane so that curl F · n dS = √ 2xy
z 2 +y 2
y
23
2 3 2
S curl F · n dS = 1 0 2x dx dy = x 0 [y]1 = 9
Part (b): by Stokes’theorem
LHS = C F · dr = C zdx + x2 dy + xdz; C = C1 ∪ C2 ∪ C3 ∪ C4
C1 segment from (0, 1, 1) to (3, 1, 1) can be parameterized using x going from 0 to 3 :
y = 1; z = 1; dy = 0; dz = 0;
3
2
3
C1 zdx + x dy + xdz = 0 1 dx = [x]0 = 3
C2 can be parameterized using y going from 1 to 2 :
x = 3; z = 1/y; dx = 0; dz = −1/y2 dy;
2
zdx + x2 dy + xdz = 1 9 − 3y−2 dy = [9y + 3/y]21 =
C2
15
2
C3 segment from (3, 2, 12 ) to (0, 2, 12 ) can be parameterized using x going from 3 to 0 :
y = 2; z = 12 ; dy = 0; dz = 0;
0
zdx + x2 dy + xdz = 3 21 dx = 12 [x]03 = −3
C3
2
C4 can be parameterized using y going from 2 to 1 :
x = 0; z = 1/y; dx = 0; dz = −1/y2 dy;
1
2
C4 zdx + x dy + xdz = 2 0 dy = 0
C
F · dr = 3 +
15
2
−
3
2
+0= 9 .
y
z +y
7. Verify Stokes’ theorem for F = y, −x, xy over the part of
the sphere x2 + y2 + z2 = 1 for which x ≥ 0, z ≥ 0.
z
n
1
S
x
C2
1
C1
LHS =
C
F · dr =
C
ydx − xdy + xydz; C = C1 ∪ C2
C1 can be parameterized using θ
π
x = cos θ, y = sin θ, z = 0, dx = − sin θ dθ, dy = cos θ dθ, dz = 0; θ goes from −π
2 to 2 ;
π/2 π/2
π/2
2
ydx − xdy + xydz = −π/2 − sin θ − cos2 θ dθ = −π/2 −1dθ = [−θ]−π/2 = −π
C1
C2 has x = 0 and dx = 0 therefore
C
C2
ydx − xdy + xydz = 0
F · dr = −π + 0 = −π
RHS =
(curl F · n) dS
i
j
curl F = ∂/∂x ∂/∂y
y
−x
S
= (x) i − (y)j + (−2)k = x, −y, −2
k
∂/∂z
xy
S is a level surface of g(x, y, z) = x2 + y2 + z2 (with g = 1);
ñ2x,2y,2z
a unit normal vector to S is ±∇g
= ± x, y, z ;
|∇g| =
2
2
2
4x +4y +4z
choose “+” for orientation to agree with the curve C: n = x, y, z ; curl F · n = x2 − y2 − 2z
Integrate using spherical coordinates with ρ = 1, dS = 12 sin φ dφ dθ
π/2 π/2 (cos θ sin φ)2 − (sin θ sin φ)2 − 2 cos φ sin φ dφ dθ
(curl F · n) dS = −π/2 0
S
π/2
π/2
π/2
π/2
= −π/2 cos(2θ)dθ
sin3 φdφ − 2 −π/2 dθ 0 cos φ sin φ dφ
0
=
π/2
sin(2θ)
2
−π/2
su bstitu te u=cos φ
π/2
cos3 φ
3
− cos φ
0
π/2
− 2 [θ]−π/2
π/2
sin2 φ
2
0
= −π
1
y
8. Verify Stokes’ theorem for F = 0, 0, x2 − y2 over
the part of the surface x2 + y2 = z2 inside the first octant
with 1 ≤ z ≤ 2 .
z
C3
2
C4
x 2 1
LHS =
C
F · dr =
n
C1
1
C2
S
2
2
x − y2 dz; C = C1 ∪ C2 ∪ C3 ∪ C4
C
C1 can be parameterized using θ going from 0 to π/2
x
θ, y =
=cos
sin θ, z = 1; dx = − sin θ dθ, dy = cos θ dθ, dz = 0
2
2
x
dz = 0
−
y
C1
C2 can be parameterized using y going from 1 to 2
x = 0, z = y; dx = 0, dz = dy
3 2
2
2 2
−y
2
−
y
y
dy
=
= −7
x
dz
=
−
1
C2
3
3
1
C3 can be parameterized using θ going from π/2 to 0
x
θ, y =
=cos
sin θ, z = 2; dx = − sin θ dθ, dy = cos θ dθ, dz = 0
2
2
x
dz = 0
−
y
C3
C4 can be parameterized using x going from 2 to 1
y = 0, z = x; dy = 0, dz = dx
3 1
2
1 2
x
2
x
dz
=
−
y
x
dx
=
= −7
2
C4
3
3
2
C
F · dr = 0 −
RHS =
7
3
+0−
7
3
=
−14
3
S (curl F
· n) dS
i
j
k
∂/∂z = (−2y) i − (2x)j + (0)k = −2y, −2x, 0
curl F = ∂/∂x ∂/∂y
0
0
x2 − y2 S is a level surface of g(x, y, z) = x2 + y2 − z2 (with g = 0);
a unit normal vector to S is ±∇g
= √±2x,2y,−2z
;
|∇g|
2
2
2
4x +4y +4z
choose “+” for orientation to agree with the curve C: n = √2x,2y,−2z
;
2
2
2
curl F · n = √
−8xy
4x2 +4y2 +4z 2
=
−4xy
ρ
4x +4y +4z
Integrate using spherical coordinates with φ = π4 , dS = ρ sin π4 dρ dθ = √ρ2 dρ dθ;
x = ρ sin π4 cos θ = √ρ2 cos θ; y = ρ sin π4 sin θ = √ρ2 sin θ
π/2
π/2 2√2 −4 √ρ2 cos θ √ρ2 sin θ ρ
2√2 2
√
√
√ dρ dθ = −4
√
(curl
F
·
n)
dS
=
cos
θ
sin
θ
dθ
0
S
2
ρ
2 ρ dρ
2
2 2 0
2 π/2 3 2√2
√
√
ρ
−2 sin θ
1
16 2
√
= −2
− 2 3 2 = − 14
=√
2
3 √
3
3
2
2 2
0
2
y
z
9. Verify Stokes’ theorem for F = x − y, y − z, z − x over
the part of the surface x2 + y2 = 9 inside the first octant
with z ≤ 4 and y ≤ x.
C3
4
C2
n
S
C4
x
3
3
y
C1
LHS = C F · dr = C (x − y) dx + (y − z)dy + (z − x) dz = C −ydx − zdy − xdz
since contributions from xdx, ydy, and zdz on a closed path are all zero.
C = C1 ∪ C2 ∪ C3 ∪ C4
C1 can be parameterized using θ going from 0 to π4
x = 3 cos θ, y = 3 sin θ, z = 0; dx = −3 sin θ dθ, dy = 3 cos θ dθ, dz = 0
π/4
π/4
π/4
−ydx − zdy − xdz = 0 9 sin2 θdθ = 92 0 (1 − cos(2θ)) dθ = 9θ
− 94 sin(2θ) 0 =
C1
2
9π
8
−
9
4
C2 can be parameterized using z going from 0 to 4
x = √32 , y = √32 , dx = 0, dy = 0
4
√
−ydx − zdy − xdz = − 0 √32 dz = −6 2
C2
C3 can be parameterized using θ going from π4 to 0
x = 3 cos θ, y = 3 sin θ, z = 4; dx = −3 sin θ dθ, dy = 3 cos θ dθ, dz = 0
0 0
−ydx − zdy − xdz = π/4 9 sin2 θ − 12 cos θ dθ = 9θ
− 94 sin(2θ) − 12 sin θ π/4 = − 9π
+
C3
2
8
9
4
√
+6 2
C4 can be parameterized using z going from 4 to 0
x
= 3, y = 0, dx = 0, dy =
00
−ydx
−
zdy
−
xdz
=
−3dz = 12
4
C4
C
F · dr =
RHS =
9π
8
−
9
4
√
−6 2−
S (curl F
9π
8
+
9
4
√
+ 6 2 − 12 = 12
· n) dS
i
j
k curl F = ∂/∂x ∂/∂y ∂/∂z = (1) i − (−1)j + (1)k = 1, 1, 1
x−y y−z z −x S is a level surface of g(x, y, z) = x2 + y2 (with g = 9);
±2x,2y,0
a unit normal vector to S is ±∇g
=√
= ± x3 , y3 , 0 ;
|∇g|
4x2 +4y 2
choose “+” for orientation to agree with the curve C: n = x3 , y3 , 0 ;
curl F · n = x+y
3
Use cylindrical coordinates with r = 3, dS = 3 dz dθ
√
√
π/4 4 3 cos θ+3 sin θ
(curl F · n) dS = 0
3 dz dθ = 3 [sin θ − cos θ]π/4
[z]40 = 3( 22 − 22 −0+1)(4) = 12
0
0
S
3
z
10. Consider the points K(2, 0, 0), L(0, 2, 0), M(0, 1, 2),
and N(1, 0, 2).
C3
(a) Show that K, L, M, and N are coplanar.
N
(b) Let C be made up of line segments: C1 from K to L,
C2 from L to M, C3 from M to N, and C4 from N to K.
Evaluate C y dx + x2 dy + y dz directly.
(c) Evaluate the line integral in part (b) by Stokes’ theorem.
2
M
C4
n
C2
S
1
1
x 2K
L
2
C1
Part (a): begin by forming an equation of the plane passing through K, L, and M :
i
j k −→ −−→ KL × KM = −2 2 0 = 41i − (−4)j + (2)k = 4, 4, 2 yields a plane equation
−2 1 2 4(x − 2) + 4y + 2z = 0, i.e., 2x + 2y + z = 4.
Since N(1, 0, 2) satisfies this equation, the four points are coplanar.
Part
(b) F
·
dr = C ydx + x2 dy + ydz; C = C1 ∪ C2 ∪ C3 ∪ C4
C
C1 can be parameterized using x going from 2 to 0 : y = 2 − x, z = 0; dy = −dx, dz = 0
0
0
x2
x3
2
2
ydx
+
x
dy
+
ydz
=
(2
−
x
−
x
)dx
=
2x
−
−
= 23
2
C1
2
3
2
C2 can be parameterized using y going from 2 to 1 : x = 0, z = 4 − 2y, dx = 0, dz = −2dy
1
1
ydx + x2 dy + ydz = 2 −2ydy = −y2 2 = 3
C2
C3 can be parameterized using x going from 0 to 1 : y = 1 − x, z = 0; dy = −dx, dz = 0
1
1
x2
x3
2
2
ydx
+
x
dy
+
ydz
=
(1
−
x
−
x
)dx
=
x
−
−
= 16
0
C3
2
3
0
C4 can be parameterized using x going from 1 to 2 : y = 0, z = 4 − 2x, dy = 0, dz = −2dx
ydx + x2 dy + ydz = 0
C4
C
F · dr =
2
3
+3+
1
6
23
6
+0 =
(curl F · n) dS
i
j
k curl F = ∂/∂x ∂/∂y ∂/∂z = (1) i − (0)j + (2x − 1)k = 1, 0, 2x − 1
y
x2
2 2 1 y
From part (a), n = 3 , 3 , 3 (chose the orientation to agree with C); curl F · n =
Projecting onto the xz-plane
Part (c):
S
1/|n 2 |
S
(curl F · n) dS =
2 2−z/2
0
0
2x+1 3
dx dz =
3
2
1
2
x=2−z/2
2 2
dz =
x + x x=0
0
23
6
2+2x−1
3
=
2x+1
3
y