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6. Vector Integral Calculus in Space
6A. Vector Fields in Space
6A-1 a) the vectors are all unit vectors, pointing radially outward.
b) the vector at P has its head on the y-axis, and is perpendicular to it
1
6A-2 2(−xi −yj −z k )
6A-3 ω(−z j + yk )
6A-4 A vector field F = M i + N j + P k is parallel to the plane 3x − 4y + z = 2 if it is
perpendicular to the normal vector to the plane, 3i − 4j + k : the condition on M;N;P
therefore is 3M − 4N + P = 0, or P = 4N − 3M.
The most general such field is therefore F = M i +N j +(4N −3M)k , where M and N
are functions of x;y;z.
6B. Surface Integrals and Flux
6B-1 Wehave n = xi + yj + z k ; therefore F· n = a.
a
� � 3
Flux through S = S F · n dS = a(area of S) = 4π a :
6B-2 Since k is parallel to the surface, the field is everywhere tangent to the cylinder,
hence the flux is 0. 1
i + j + k 1 n
6B-3 √ is a normal vector to the plane, so F · n = √ . 2
3 3 S
√ √ √ 1
1 1 3
area of region 2(base)(height) 2( 2)( 2 2) 1 1
Therefore, flux = √ = √ = √ = 2 .
3 3 3
xi + yj + z k y2
6B-4 n = a ; F· n = a : Calculating in spherical coordinates,
flux = � � y2 dS = 1 � π � π a 4 sin3 φ sin2 θ dφdθ = a 3 � π � π sin3 φsin2 θ dφdθ.
S a a 0 0 0 0
Inner integral: sin2 θ(−cosφ+ 1 cos3 φ)�π = 4 sin2 θ;
3 0 3
4 3 1 1 �π 2 3
Outer integral: 3a (2θ − 4 sin2θ) 0 = 3πa .
1
2 E. 18.02 EXERCISES
i + j + k z
6B-5 n = √3 ; F · n = √3 .
� � z dxdy 1 � � dxdy � 1 � 1−y
flux = √ |n· k | = √ (1 −x −y) √ = (1 −x −y)dxdy.
S 3 3 S 1= 3 0 0
1 2 �1−y 1 2 1
Inner integral: = x − 2x −xy 0 = 2(1 −y) : n
� 1 1 1 1 �1 1
Outer integral: = (1 −y)2dy = · − · (1 − y)3 = : S
0 2 2 3 0 6
1
1 x+y=1
6B-6 z = f(x;y) = x2 + y2 (a paraboloid). By (13) in Notes V9,
dS = (−2xi −2yj + k )dxdy.
(This points generally “up”, since the k component is positive.) Since F = xi +yj +z k ,
� � S F · dS = � � R (−2x 2 − 2y 2 + z)dxdy ,
where R is the interior of the unit circle in the xy-plane, i.e., the projection of S onto the
2 2
xy-plane). Since z = x + y , the above integral
= − � � (x 2 + y 2)dxdy = − � 2π � 1 r 2 · r dr dθ = −2π · 1 = − π :
R 0 0 4 2
The answer is negative since the positive direction for flux is that of n, which here points
into the inside of the paraboloidal cup, whereas the flow xi + y j + z k is generally from
the inside toward the outside of the cup, i.e., in the opposite direction.
xi + yj y2
6B-8 On the cylindrical surface, n = a ; F· n = a :
In cylindrical coordinates, since y = asinθ, this gives us F · dS = F· n dS = a2 sin2 θ dz dθ.
Flux = � π=2 � k a 2 sin2 θ dz dθ = a 2h � π=2 sin2 θ dθ = a 2h� θ − sin2θ�π=2 = π a2h :
−π=2 0 −π=2 2 4 −π=2 2
6B-12 Since the distance from a point (x;y;0) up to the hemispherical surface is z,
�� S z dS
average distance = �� S dS :
In spherical coordinates, � � z dS = � 2π � π=2 acosφ · a 2 sinφdφdθ.
S 0 0
3 � π=2 3 sin2 φ �π=2 a3 a3 � 2π 3
Inner: = a sinφcosφdφ = a ( 2 = 2 : Outer: = 2 dθ = πa :
0 0 0
Finally, � � dS = area of hemisphere = 2πa2, so average distance = πa3 = a :
2πa2 2
S
6. VECTOR INTEGRAL CALCULUS IN SPACE 3
6C. Divergence Theorem
6C-1a div F = Mx + Ny + Pz = 2xy + x + x = 2x(y + 1):
6C-2 Using the product and chain rules for the first, symmetry for the others,
n n−1 x n n n−1 y n n n−1 z n
(ρ x)x = nρ ρ x + ρ ; (ρ y)y = nρ ρ y + ρ ; (ρ z)z = nρ ρ z + ρ ;
x2 + y2 + z2
n−1 n n
adding these three, we get div F = nρ ρ + 3ρ = ρ (n +3):
Therefore, div F = 0 ⇔ n = −3.
6C-3 Evaluating the triple integral first, we have div F = 3, therefore
� � � 2 3 3
div F dV = 3(vol.of D) = 3 3πa = 2πa :
D
To evaluate the double integral over the closed surface S1 + S2, the normal vectors are:
n = xi + yj + z k (hemisphere S ); n = −k (disc S );
1 a 1 2 2
using these, the surface integral for the flux through S is
� � F · dS = � � x2 + y2 + z2 dS + � � −zdS = � � adS;
S S1 a S2 S1
2 2 2 2 2
since x +y +z = ρ = a on S1, and z = 0 on S2. So the value of the surface integral is
2 3
a(area of S1)= a(2πa ) = 2πa ,
which agrees with the triple integral above. 1
6C-5 The divergence theorem says � � S F · dS = � � � D div F dV: 1
Here div F = 1, so that the right-hand integral is just the volume of the
tetrahedron, which is 1(base)(height)= 1(1)(1) = 1. 1
3 3 2 6
6C-6 The divergence theorem says � � S F · dS = � � � D div F dV:
Here div F = 1, so the right-hand integral is the volume of the solid cone, which has
height 1 and base radius 1; its volume is 1(base)(height)= π=3.
3
6C-7a Evaluating the triple integral first, over the cylindrical solid D, we have
div F = 2x + x = 3x; � � � D 3xdV = 0,
since the solid is symmetric with respect to the yz-plane. (Physically, assuming the density
is 1, the integral has the value x¯(mass of D), where x ¯ is the x-coordinate of the center of
mass; this must be in the yz plane since the solid is symmetric with respect to this plane.)
To evaluate the double integral, note that F has no k -component, so there is no flux
across the two disc-like ends of the solid. To find the flux across the cylindrical side,
n = xi + yj ; F· n = x3 + xy2 = x3 + x(1 − x2) = x;
4 E. 18.02 EXERCISES
since the cylinder has radius 1 and equation x2 + y2 = 1. Thus
� � xdS = � 2π � 1 cosθ dz dθ = � 2π cosθ dθ = 0:
S 0 0 0
6C-8 a) Reorient the lower hemisphere S2 by reversing its normal vector; call the reori-
ented surface S′ . Then S = S1 + S′ is a closed surface, with the normal vector pointing
2 2
outward everywhere, so by the divergence theorem,
� � � � � � � � � S
F · dS = F · dS + F · dS = div F dV = 0; 1
S S1 S′ D
2
since by hypothesis div F = 0. The above shows
� � � � � � S
2
S F ·dS = − S′ F · dS = S F · dS;
1 2 2
since reversing the orientation of a surface changes the sign of the flux through it.
b) The same statement holds if S1 and S2 are two oriented surfaces having the same
boundary curve, but not intersecting anywhere else, and oriented so that S1 and S′ (i.e., S2
2
with its orientation reversed) together make up a closed surface S with outward-pointing
normal.
6C-10 If div F = 0, then for any closed surface S, we have by the divergence theorem
� � S F · dS = � � � D div F dV = 0.
Conversely: � � S F · dS = 0 for every closed surface S ⇒ div F = 0.
For suppose there were a point P0 at which (div F)0 6= 0 — say (div F)0 > 0. Then
by continuity, div F > 0 in a very small spherical ball D surrounding P0, so that by the
divergence theorem (S is the surface of the ball D),
� � S F · dS = � � � D div F dV > 0.
But this contradicts our hypothesis that � � S F · dS = 0 for every closed surface S.
6C-11 flux of F = � � S F · dn = � � � D div F dV = � � � D 3dV = 3(vol. of D):
6D. Line Integrals in Space
6D-1 a) C : x = t; dx = dt; y = t2; dy = 2tdt; z = t3; dz = 3t2 dt;
� � 1 2 3 2
C y dx + z dy − xdz = 0 (t )dt + t (2tdt) − t(3t dt)
� 1 �1
2 4 3 t3 2t5 3t4 1 2 3 1
= 0 (t +2t −3t )dt = 3 + 5 − 4 0 = 3 + 5 − 4 = −60 :
b) C : x = t; y = t; z = t; � ydx+ zdy −xdz = � 1 tdt = 1 :
C 0 2
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