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657
CHAPTER 16
Topics in Vector Calculus
EXERCISE SET 16.
(a) III because the vector field is independent of y and the direction is that of the negative x-axis for negative x, and positive for positive (b) IV, because the y-component is constant, and the x-component varies priodically with x
(a) I, since the vector field is constant (b) II, since the vector field points away from the origin
(a) true (b) true (c) true
(a) false, the lengths are equal to 1 (b) false, the y-component is then zero (c) false, the x-component is then zero
x
y (^) 6.
x
y (^) 7.
x
y
x
y (^) 9.
x
y (^) 10.
x
y
- (a) ∇φ = φxi + φy j = y 1 + x^2 y^2 i^ +^
x 1 + x^2 y^2 j^ =^ F, so^ F^ is conservative for all^ x, y (b) ∇φ = φxi + φy j = 2xi − 6 yj + 8zk = F so F is conservative for all x, y
- (a) ∇φ = φxi + φy j = (6xy − y^3 )i + (4y + 3x^2 − 3 xy^2 )j = F, so F is conservative for all x, y
(b) ∇φ = φxi + φy j + φz k = (sin z + y cos x)i + (sin x + z cos y)j + (x cos z + sin y)k = F, so F is conservative for all x, y
- div F = 2x + y, curl F = zi
- div F = z^3 + 8y^3 x^2 + 10zy, curl F = 5z^2 i + 3xz^2 j + 4xy^4 k
658 Chapter 16
- div F = 0, curl F = (40x^2 z^4 − 12 xy^3 )i + (14y^3 z + 3y^4 )j − (16xz^5 + 21y^2 z^2 )k
- div F = yexy^ + sin y + 2 sin z cos z, curl F = −xexy^ k
- div F =
√^2
x^2 + y^2 + z^2
, curl F = 0
- div F =
x
- xzexyz^ + x x^2 + z^2 , curl F = −xyexyz^ i + z x^2 + z^2 j + yzexyz^ k
- ∇ · (F × G) = ∇ · (−(z + 4y^2 )i + (4xy + 2xz)j + (2xy − x)k) = 4x
- ∇ · (F × G) = ∇ · ((x^2 yz^2 − x^2 y^2 )i − xy^2 z^2 j + xy^2 zk) = −xy^2
- ∇ · (∇ × F) = ∇ · (− sin(x − y)k) = 0
- ∇ · (∇ × F) = ∇ · (−zeyz^ i + xexz^ j + 3ey^ k) = 0
- ∇ × (∇ × F) = ∇ × (xzi − yzj + yk) = (1 + y)i + xj
- ∇ × (∇ × F) = ∇ × ((x + 3y)i − yj − 2 xyk) = − 2 xi + 2yj − 3 k
- Let F = f i + gj + hk ; div( kF) = k ∂f ∂x +^ k
∂g ∂y +^ k
∂h ∂z =^ k^ div^ F
- Let F = f i + gj + hk ; curl (kF) = k
∂h ∂y
∂g ∂z
i + k
∂f ∂z
∂h ∂x
j + k
∂g ∂x
∂f ∂y
k = k curl F
- Let F = f (x, y, z)i + g(x, y, z)j + h(x, y, z)k and G = P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k, then
div( F + G) =
∂f ∂x
+ ∂P
∂x
∂g ∂y
+ ∂Q
∂y
∂h ∂z
+ ∂R
∂z
∂f ∂x
∂P
∂x
+ ∂Q
∂y
+ ∂R
∂z
= div F + div G
- Let F = f (x, y, z)i + g(x, y, z)j + h(x, y, z)k and G = P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k, then
curl (F + G) =
[
∂y (h + R) −
∂z (g + Q)
]
i +
[
∂z (f + P ) −
∂x (h + R)
]
j
[
∂x (g + Q) −
∂y (f + P )
]
k;
expand and rearrange terms to get curl F + curl G.
- Let F = f i + gj + hk ;
div( φF) =
φ ∂f ∂x
∂φ ∂x f
φ ∂g ∂y
∂φ ∂y g
φ ∂h ∂z
∂φ ∂z h
= φ
∂f ∂x
∂g ∂y
∂h ∂z
∂φ ∂x f + ∂φ ∂y g + ∂φ ∂z h
= φ div F + ∇φ · F
660 Chapter 16
- dy/dx = x, y = x^2 /2 + K
x
y
- dy/dx = 1/x, y = ln x + K
x
y
- dy/dx = −y/x, (1/y)dy = (− 1 /x)dx, ln y = − ln x + K 1 , y = eK^1 e−^ ln^ x^ = K/x
x
y
EXERCISE SET 16.
- (a)
0
dy = 1 because s = y is arclength measured from (0, 0)
(b) 0, because sin xy = 0 along C
- (a)
C
ds = length of line segment = 2 (b) 0, because x is constant and dx = 0
- (a) ds =
dx dt
dy dt
dt, so
0
(2t − 3 t^2 )
4 + 36t^2 dt = −
ln(
(b)
0
(2t − 3 t^2 )2 dt = 0 (c)
0
(2t − 3 t^2 )6t dt = −
- (a)
0
t(3t^2 )(6t^3 )^2
1 + 36t^2 + 324t^4 dt =
(b)
0
t(3t^2 )(6t^3 )^2 dt =
(c)
0
t(3t^2 )(6t^3 )^26 t dt =
(d)
0
t(3t^2 )(6t^3 )^218 t^2 dt = 162
- (a) C : x = t, y = t, 0 ≤ t ≤ 1;
0
6 t dt = 3
(b) C : x = t, y = t^2 , 0 ≤ t ≤ 1;
0
(3t + 6t^2 − 2 t^3 )dt = 3
Exercise Set 16.2 661
(c) C : x = t, y = sin(πt/2), 0 ≤ t ≤ 1; ∫ (^1)
0
[3t + 2 sin(πt/2) + πt cos(πt/2) − (π/2) sin(πt/2) cos(πt/2)]dt = 3
(d) C : x = t^3 , y = t, 0 ≤ t ≤ 1;
0
(9t^5 + 8t^3 − t)dt = 3
- (a) C : x = t, y = t, z = t, 0 ≤ t ≤ 1;
0
(t + t − t) dt =
(b) C : x = t, y = t^2 , z = t^3 , 0 ≤ t ≤ 1;
0
(t^2 + t^3 (2t) − t(3t^2 )) dt = −
(c) C : x = cos πt, y = sin πt, z = t, 0 ≤ t ≤ 1;
0
(−π sin^2 πt + πt cos πt − cos πt) dt = − π 2
π
0
1 + t 1 + t dt =
0
(1 + t)−^1 /^2 dt = 2 8.
0
1 + 2t 1 + t^2 dt =
5(π/4 + ln 2)
0
3(t^2 )(t^2 )(2t^3 /3)(1 + 2t^2 ) dt = 2
0
t^7 (1 + 2t^2 ) dt = 13/ 20
∫ (^2) π
0
e−t^ dt =
5(1 − e−^2 π^ )/ 4 11.
∫ (^) π/ 4
0
(8 cos^2 t−16 sin^2 t−20 sin t cos t)dt = 1−π
− 1
t − 2 3 t^5 /^3 + t^2 /^3
dt = 6/ 5
- C : x = (3 − t)^2 /3, y = 3 − t, 0 ≤ t ≤ 3;
0
(3 − t)^2 dt = 3
- C : x = t^2 /^3 , y = t, − 1 ≤ t ≤ 1;
− 1
t^2 /^3 −
t^1 /^3 + t^7 /^3
dt = 4/ 5
- C : x = cos t, y = sin t, 0 ≤ t ≤ π/2;
∫ (^) π/ 2
0
(− sin t − cos^2 t)dt = − 1 − π/ 4
- C : x = 3 − t, y = 4 − 3 t, 0 ≤ t ≤ 1;
0
(−37 + 41t − 9 t^2 )dt = − 39 / 2
0
(−3)e^3 tdt = 1 − e^3
∫ (^) π/ 2
0
(sin^2 t cos t − sin^2 t cos t + t^4 (2t)) dt = π^6 192
- (a)
∫ (^) ln 2
0
e^3 t^ + e−^3 t
e^2 t^ + e−^2 t^ dt
ln(4 +
ln
ln(
ln
(b)
∫ (^) π/ 2
0
[sin t cos t dt − sin^2 t dt] =
π 4
Exercise Set 16.2 663
- Represent the circular arc by x = 3 cos t, y = 3 sin t, 0 ≤ t ≤ π/2. ∫
C
x
yds = 9
∫ (^) π/ 2
0
sin t cos t dt = 6
- δ(x, y) = k
x^2 + y^2 where k is the constant of proportionality, ∫
C
k
x^2 + y^2 ds = k
0
et(
2 et) dt =
2 k
0
e^2 t^ dt = (e^2 − 1)k/
C
kx 1 + y^2 ds = 15k
∫ (^) π/ 2
0
cos t 1 + 9 sin^2 t
dt = 5k tan−^1
- δ(x, y, z) = kz where k is the constant of proportionality, ∫
C
kzds =
1
k(
t)(2 + 1/t) dt = 136k/ 3
- C : x = t^2 , y = t, 0 ≤ t ≤ 1; W =
0
3 t^4 dt = 3/ 5
34. W =
1
(t^2 + 1 − 1 /t^3 + 1/t)dt = 92/9 + ln 3 35. W =
0
(t^3 + 5t^6 )dt = 27/ 28
- C 1 : (0, 0 , 0) to (1, 3 , 1); x = t, y = 3t, z = t, 0 ≤ t ≤ 1 C 2 : (1, 3 , 1) to (2, − 1 , 4); x = 1 + t, y = 3 − 4 t, z = 1 + 3t, 0 ≤ t ≤ 1 W =
0
(4t + 8t^2 )dt +
0
(− 11 − 17 t − 11 t^2 )dt = − 37 / 2
- Since F and r are parallel, F · r = ‖F‖‖r‖, and since F is constant, ∫ F · dr =
C
d(F · r) =
C
d(‖F‖‖r‖) =
− 4
2 dt = 16
C
F · r = 0, since F is perpendicular to the curve
- C : x = 4 cos t, y = 4 sin t, 0 ≤ t ≤ π/ 2 ∫ (^) π/ 2
0
sin t + cos t
dt = 3/ 4
- C 1 : (0, 3) to (6, 3); x = 6t, y = 3, 0 ≤ t ≤ 1 C 2 : (6, 3) to (6, 0); x = 6, y = 3 − 3 t, 0 ≤ t ≤ 1 ∫ (^1)
0
36 t^2 + 9 dt +
0
36 + 9(1 − t)^2 dt =
tan−^1 2 −
tan−^1 (1/2)
- Represent the parabola by x = t, y = t^2 , 0 ≤ t ≤ 2. ∫
C
3 xds =
0
3 t
1 + 4t^2 dt = (
- Represent the semicircle by ∫ x = 2 cos t, y = 2 sin t, 0 ≤ t ≤ π.
C
x^2 yds =
∫ (^) π
0
16 cos^2 t sin t dt = 32/ 3
664 Chapter 16
- (a) 2 πrh = 2π(1)2 = 4π (b) S =
C
z(t) dt
(c) C : x = cos t, y = sin t, 0 ≤ t ≤ 2 π; S =
∫ (^2) π
0
(2 + (1/2) sin 3t) dt = 4π
- C : x = a cos t, y = −a sin t, 0 ≤ t ≤ 2 π, ∫
C
x dy − y dx x^2 + y^2
∫ (^2) π
0
−a^2 cos^2 t − a^2 sin^2 t a^2 dt =
∫ (^2) π
0
dt = 2π
45. W =
C
F · dr =
0
(λt^2 (1 − t), t − λt(1 − t)) · (1, λ − 2 λt) dt = −λ/12, W = 1 when λ = − 12
- The force exerted by the farmer is F =
z
k =
4 π t
k, so
F · dr =
z
dz, and W =
0
z
dz = 10,020. Note that the functions x(z), y(z) are irrelevant.
EXERCISE SET 16.
- ∂x/∂y = 0 = ∂y/∂x, conservative so ∂φ/∂x = x and ∂φ/∂y = y, φ = x^2 /2 + k(y), k′(y) = y, k(y) = y^2 /2 + K, φ = x^2 /2 + y^2 /2 + K
- ∂(3y^2 )/∂y = 6y = ∂(6xy)/∂x, conservative so ∂φ/∂x = 3y^2 and ∂φ/∂y = 6xy, φ = 3xy^2 + k(y), 6xy + k′(y) = 6xy, k′(y) = 0, k(y) = K, φ = 3xy^2 + K
- ∂(x^2 y)/∂y = x^2 and ∂(5xy^2 )/∂x = 5y^2 , not conservative
- ∂(ex^ cos y)/∂y = −ex^ sin y = ∂(−ex^ sin y)/∂x, conservative so ∂φ/∂x = ex^ cos y and ∂φ/∂y = −ex^ sin y, φ = ex^ cos y + k(y), −ex^ sin y + k′(y) = −ex^ sin y, k′(y) = 0, k(y) = K, φ = ex^ cos y + K
- ∂(cos y + y cos x)/∂y = − sin y + cos x = ∂(sin x − x sin y)/∂x, conservative so ∂φ/∂x = cos y + y cos x and ∂φ/∂y = sin x − x sin y, φ = x cos y + y sin x + k(y), −x sin y + sin x + k′(y) = sin x − x sin y, k′(y) = 0, k(y) = K, φ = x cos y + y sin x + K
- ∂(x ln y)/∂y = x/y and ∂(y ln x)/∂x = y/x, not conservative
- (a) ∂(y^2 )/∂y = 2y = ∂(2xy)/∂x, independent of path (b) C : x = −1 + 2t, y = 2 + t, 0 ≤ t ≤ 1;
0
(4 + 14t + 6t^2 )dt = 13 (c) ∂φ/∂x = y^2 and ∂φ/∂y = 2xy, φ = xy^2 + k(y), 2xy + k′(y) = 2xy, k′(y) = 0, k(y) = K, φ = xy^2 + K. Let K = 0 to get φ(1, 3) − φ(− 1 , 2) = 9 − (−4) = 13
- (a) ∂(y sin x)/∂y = sin x = ∂(− cos x)/∂x, independent of path
(b) C 1 : x = πt, y = 1 − 2 t, 0 ≤ t ≤ 1;
0
(π sin πt − 2 πt sin πt + 2 cos πt)dt = 0
(c) ∂φ/∂x = y sin x and ∂φ/∂y = − cos x, φ = −y cos x + k(y), − cos x + k′(y) = − cos x, k′(y) = 0, k(y) = K, φ = −y cos x+K. Let K = 0 to get φ(π, −1)−φ(0, 1) = (−1)−(−1) = 0
666 Chapter 16
- Let f (x, y, z) = yz, g(x, y, z) = xz, h(x, y, z) = yx^2 , then ∂f /∂z = y, ∂h/∂x = 2xy = ∂f /∂z, thus
by Exercise 25, F = f i+gj+hk is not conservative, and by Theorem 16.3.2,
C
yz dx+xz dy+yx^2 dz is not independent of the path.
∂y (h(x)[x sin y + y cos y]) = h(x)[x cos y − y sin y + cos y]
∂ ∂x (h(x)[x cos y − y sin y]) = h(x) cos y + h′(x)[x cos y − y sin y],
equate these two partial derivatives to get (x cos y − y sin y)(h′(x) − h(x)) = 0 which holds for all x and y if h′(x) = h(x), h(x) = Cex^ where C is an arbitrary constant.
- (a)
∂y
cx (x^2 + y^2 )^3 /^2
3 cxy (x^2 + y^2 )−^5 /^2
∂x
cy (x^2 + y^2 )^3 /^2 when (x, y) = (0, 0),
so by Theorem 16.3.3, F is conservative. Set ∂φ/∂x = cx/(x^2 + y^2 )−^3 /^2 , then φ(x, y) = −c(x^2 + y^2 )−^1 /^2 + k(y), ∂φ/∂y = cy/(x^2 + y^2 )−^3 /^2 + k′(y), so k′(y) = 0. Thus φ(x, y) = − c (x^2 + y^2 )^1 /^2 is a potential function.
(b) curl F = 0 is similar to Part (a), so F is conservative. Let
φ(x, y, z) =
cx (x^2 + y^2 + z^2 )^3 /^2 dx = −c(x^2 + y^2 + z^2 )−^1 /^2 + k(y, z). As in Part (a),
∂k/∂y = ∂k/∂z = 0, so φ(x, y, z) = −c/(x^2 + y^2 + z^2 )^1 /^2 is a potential function for F.
- (a) See Exercise 28, c = 1; W =
∫ Q
P
F · dr = φ(3, 2 , 1) − φ(1, 1 , 2) = −
(b) C begins at P (1, 1 , 2) and ends at Q(3, 2 , 1) so the answer is again W = − √^1 14
+ √^1
(c) The circle is not specified, but cannot pass through (0, 0 , 0), so Φ is continuous and differ- entiable on the circle. Start at any point P on the circle and return to P , so the work is Φ(P ) − Φ(P ) = 0. C begins at, say, (3, 0) and ends at the same point so W = 0.
- (a) F · dr =
y dx dt − x dy dt
dt for points on the circle x^2 + y^2 = 1, so
C 1 : x = cos t, y = sin t, 0 ≤ t ≤ π,
C 1
F · dr =
∫ (^) π
0
(− sin^2 t − cos^2 t) dt = −π
C 2 : x = cos t, y = − sin t, 0 ≤ t ≤ π,
C 2
F · dr =
∫ (^) π
0
(sin^2 t + cos^2 t) dt = π
(b) ∂f ∂y = x
(^2) − y 2 (x^2 + y^2 )^2 , ∂g ∂x = − y
(^2) − x 2 (x^2 + y^2 )^2 = ∂f ∂y (c) The circle about the origin of radius 1, which is formed by traversing C 1 and then traversing C 2 in the reverse direction, does not lie in an open simply connected region inside which F is continuous, since F is not defined at the origin, nor can it be defined there in such a way as to make the resulting function continuous there.
Exercise Set 16.4 667
- If C is composed of smooth curves C 1 , C 2 ,... , Cn and curve Ci extends from (xi− 1 , yi− 1 ) to (xi, yi)
then
C
F · dr =
∑^ n
i=
Ci
F · dr =
∑^ n
i=
[φ(xi, yi) − φ(xi− 1 , yi− 1 )] = φ(xn, yn) − φ(x 0 , y 0 )
where (x 0 , y 0 ) and (xn, yn) are the endpoints of C.
C 1
F · dr +
−C 2
F · dr = 0, but
−C 2
F · dr = −
C 2
F · dr so
C 1
F · dr =
C 2
F · dr, thus ∫
C
F · dr is independent of path.
- Let C 1 be an arbitrary piecewise smooth curve from (a, b) to a point (x, y 1 ) in the disk, and C 2 the vertical line segment from (x, y 1 ) to (x, y). Then
φ(x, y) =
C 1
F · dr +
C 2
F · dr =
∫ (^) (x,y 1 )
(a,b)
F · dr +
C 2
F · dr.
The first term does not depend on y; hence ∂φ ∂y
∂y
C 2
F · dr =
∂y
C 2
f (x, y)dx + g(x, y)dy.
However, the line integral with respect to x is zero along C 2 , so ∂φ ∂y
∂y
C 2
g(x, y) dy.
Express C 2 as x = x, y = t where t varies from y 1 to y, then ∂φ ∂y
∂y
∫ (^) y
y 1
g(x, t) dt = g(x, y).
EXERCISE SET 16.
R
(2x − 2 y)dA =
0
0
(2x − 2 y)dy dx = 0; for the line integral, on x = 0, y^2 dx = 0, x^2 dy = 0;
on y = 0, y^2 dx = x^2 dy = 0; on x = 1, y^2 dx + x^2 dy = dy; and on y = 1, y^2 dx + x^2 dy = dx,
hence
C
y^2 dx + x^2 dy =
0
dy +
1
dx = 1 − 1 = 0
R
(1 − 1)dA = 0; for the line integral let x = cos t, y = sin t,
∮
C
y dx + x dy =
∫ (^2) π
0
(− sin^2 t + cos^2 t)dt = 0
− 2
1
(2y − 3 x)dy dx = 0 4.
∫ (^2) π
0
0
(1 + 2r sin θ)r dr dθ = 9π
∫ (^) π/ 2
0
∫ (^) π/ 2
0
(−y cos x + x sin y)dy dx = 0 6.
R
(sec^2 x − tan^2 x)dA =
R
dA = π
R
[1 − (−1)]dA = 2
R
dA = 8π 8.
0
∫ (^) x
x^2
(2x − 2 y)dy dx = 1/ 30
Exercise Set 16.4 669
- C 1 : (0, 0) to (a, 0); x = at, y = 0, 0 ≤ t ≤ 1 C 2 : (a, 0) to (a cos t 0 , b sin t 0 ); x = a cos t, y = b sin t, 0 ≤ t ≤ t 0 C 3 : (a cos t 0 , b sin t 0 ) to (0, 0); x = −a(cos t 0 )t, y = −b(sin t 0 )t, − 1 ≤ t ≤ 0
A =
C
−y dx + x dy =
0
(0) dt +
∫ (^) t 0
0
ab dt +
− 1
(0) dt =
ab t 0
- C 1 : (0, 0) to (a, 0); x = at, y = 0, 0 ≤ t ≤ 1 C 2 : (a, 0) to (a cosh t 0 , b sinh t 0 ); x = a cosh t, y = b sinh t, 0 ≤ t ≤ t 0 C 3 : (a cosh t 0 , b sinh t 0 ) to (0, 0); x = −a(cosh t 0 )t, y = −b(sinh t 0 )t, − 1 ≤ t ≤ 0
A =
C
−y dx + x dy =
0
(0) dt +
∫ (^) t 0
0
ab dt +
− 1
(0) dt =
ab t 0
21. W =
R
y dA =
∫ (^) π
0
0
r^2 sin θ dr dθ = 250/ 3
- We cannot apply Green’s Theorem on the region enclosed by the closed curve C, since F does not have first order partial derivatives at the origin. However, the curve Cx 0 , consisting of y = x^30 / 4 , x 0 ≤ x ≤ 2; x = 2, x^30 / 4 ≤ y ≤ 2; and y = x^3 / 4 , x 0 ≤ x ≤ 2 encloses a region Rx 0 in which Green’s Theorem does hold, and
W =
C
F · dr = lim x 0 → 0 +
Cx 0
F · dr = lim x 0 → 0 +
Rx 0
∇ · F dA
= lim x 0 → 0 +
x 0
∫ (^) x^3 / 4
x^30 / 4
x−^1 /^2 − 1 2 y−^1 /^2
dy dx
= lim x 0 → 0 +
x^30 + x^30 / 2 +^3 14 x^70 / 2 − 3 10 x^50 /^2
C
y dx − x dy =
R
(−2)dA = − 2
∫ (^2) π
0
∫ (^) a(1+cos θ)
0
r dr dθ = − 3 πa^2
- x¯ =
A
R
x dA, but
C
2 x
(^2) dy =
R
x dA from Green’s Theorem so
x¯ =
A
C
x^2 dy =
2 A
C
x^2 dy. Similarly, ¯y = −
2 A
C
y^2 dx.
25. A =
0
∫ (^) x
x^3
dy dx =
; C 1 : x = t, y = t^3 , 0 ≤ t ≤ 1 ,
C 1
x^2 dy =
0
t^2 (3t^2 ) dt =
C 2 : x = t, y = t, 0 ≤ t ≤ 1;
C 2
x^2 dy =
0
t^2 dt =^1 3
C
x^2 dy =
C 1
C 2
=^3
, ¯x = 8 15 ∫
C
y^2 dx =
0
t^6 dt −
0
t^2 dt =
, y¯ =
, centroid
670 Chapter 16
26. A =
a^2 2 ; C 1 : x = t, y = 0, 0 ≤ t ≤ a, C 2 : x = a − t, y = t, 0 ≤ t ≤ a; C 3 : x = 0, y = a − t, 0 ≤ t ≤ a; ∫
C 1
x^2 dy = 0,
C 2
x^2 dy =
∫ (^) a
0
(a − t)^2 dt = a
3 3
C 3
x^2 dy = 0,
C
x^2 dy =
C 1
C 2
C 3
= a
3 3 , x¯ = a 3
C
y^2 dx = 0 −
∫ (^) a
0
t^2 dt + 0 = − a^3 3 , y¯ = a 3 , centroid
( (^) a 3
a 3
- ¯x = 0 from the symmetry of the region,
C 1 : (a, 0) to (−a, 0) along y =
a^2 − x^2 ; x = a cos t, y = a sin t, 0 ≤ t ≤ π C 2 : (−a, 0) to (a, 0); x = t, y = 0, −a ≤ t ≤ a
A = πa^2 / 2 , y¯ = −
2 A
[∫ (^) π
0
−a^3 sin^3 t dt +
∫ (^) a
−a
(0)dt
]
πa^2
4 a^3 3
4 a 3 π ; centroid
4 a 3 π
- A = ab 2 ; C 1 : x = t, y = 0, 0 ≤ t ≤ a, C 2 : x = a, y = t, 0 ≤ t ≤ b; C 3 : x = a − at, y = b − bt, 0 ≤ t ≤ 1; ∫
C 1
x^2 dy = 0,
C 2
x^2 dy =
∫ (^) b
0
a^2 dt = ba^2 ,
C 3
x^2 dy =
0
a^2 (1 − t)^2 (−b) dt = − ba^2 3
C
x^2 dy =
C 1
C 2
C 3
2 ba^2 3 , x¯ = 2 a 3
C
y^2 dx = 0 + 0 −
0
ab^2 (1 − t)^2 dt = − ab
2 3 , y¯ = b 3 , centroid
2 a 3 , b 3
- From Green’s Theorem, the given integral equals
R
(1−x^2 −y^2 )dA where R is the region enclosed
by C. The value of this integral is maximum if the integration extends over the largest region for which the integrand 1 − x^2 − y^2 is nonnegative so we want 1 − x^2 − y^2 ≥ 0, x^2 + y^2 ≤ 1. The largest region is that bounded by the circle x^2 + y^2 = 1 which is the desired curve C.
- (a) C : x = a + (c − a)t, y = b + (d − b)t, 0 ≤ t ≤ 1 ∫
C
−y dx + x dy =
0
(ad − bc)dt = ad − bc
(b) Let C 1 , C 2 , and C 3 be the line segments from (x 1 , y 1 ) to (x 2 , y 2 ), (x 2 , y 2 ) to (x 3 , y 3 ), and (x 3 , y 3 ) to (x 1 , y 1 ), then if C is the entire boundary consisting of C 1 , C 2 , and C 3
A =
C
−y dx + x dy =
∑^3
i=
Ci
−y dx + x dy
[(x 1 y 2 − x 2 y 1 ) + (x 2 y 3 − x 3 y 2 ) + (x 3 y 1 − x 1 y 3 )]
(c) A =
[(x 1 y 2 − x 2 y 1 ) + (x 2 y 3 − x 3 y 2 ) + · · · + (xny 1 − x 1 yn)]
(d) A =
[(0 − 0) + (6 + 8) + (0 + 2) + (0 − 0)] = 8
672 Chapter 16
By symmetry the integrals over σ 1 , σ 2 and σ 3 are equal, as are those over σ 4 , σ 5 and σ 6 , and ∫ ∫
σ 1
(x + y + z)dS =
0
0
(x + y)dx dy = 1;
σ 4
(x + y + z)dS =
0
0
(x + y + 1)dx dy = 2,
thus,
σ
(x + y + z)dS = 3 · 1 + 3 · 2 = 9.
- Let r(φ, θ) = sin φ cos θi + sin φ sin θj + cos φk, 0 ≤ θ ≤ 2 π, 0 ≤ φ ≤ π/2; ‖rφ × rθ ‖ = sin φ, ∫ ∫
σ
(1 + cos φ) dS =
∫ (^2) π
0
∫ (^) π/ 2
0
(1 + cos φ) sin φ dφ dθ
= 2π
∫ (^) π/ 2
0
(1 + cos φ) sin φ dφ = 3π
- R is the circular region enclosed by x^2 + y^2 = 1; ∫ ∫
σ
x^2 + y^2 + z^2 dS =
R
2(x^2 + y^2 )
x^2 x^2 + y^2
2 x^2 + y^2
= lim r 0 → 0 +
R′
x^2 + y^2 dA
where R′^ is the annular region enclosed by x^2 + y^2 = 1 and x^2 + y^2 = r^20 with r 0 slightly larger
than 0 because
x^2 x^2 + y^2 +^
y^2 x^2 + y^2 + 1 is not defined for^ x
(^2) + y (^2) = 0, so
∫ ∫
σ
x^2 + y^2 + z^2 dS = lim r 0 → 0 +
∫ (^2) π
0
r 0
r^2 dr dθ = lim r 0 → 0 +
4 π 3 (1 − r^30 ) =^4 π 3
- Let r(φ, θ) = a sin φ cos θ i + a sin φ sin θ j + a cos φ k,
0 ≤ θ ≤ 2 π, 0 ≤ φ ≤ π; ‖rφ × rθ ‖ = a^2 sin φ, x^2 + y^2 = a^2 sin^2 φ ∫ ∫
σ
f (x, y, z) =
∫ (^2) π
0
∫ (^) π
0
a^4 sin^3 φ dφ dθ =
πa^4
- (a)
0
∫ (^) (12− 2 x)/ 3
0
xy(12 − 2 x − 3 y)dy dx
(b)
0
∫ (^) (12− 4 z)/ 3
0
yz(12 − 3 y − 4 z)dy dz
(c)
0
∫ (^6) − 2 z
0
xz(12 − 2 x − 4 z)dx dz
- (a) a
∫ (^) a
0
∫ √a (^2) −x 2
0
x dy dx (b) a
∫ (^) a
0
∫ √a (^2) −z 2
0
z dy dz
(c) a
∫ (^) a
0
∫ √a (^2) −z 2
0
√^ xz a^2 − x^2 − z^2
dx dz
Exercise Set 16.5 673
- 18
29 / 5 14. a^4 / 3
0
1
y^3 z
4 y^2 + 1 dy dz;
0
1
xz
1 + 4x dx dz
- a
0
∫ (^) a/√ 2
a/√ 5
√^ x^2 y a^2 − y^2
dy dx, a
∫ (^2) a/√ 5
a/√ 2
0
x^2 dx dz 17. 391
- The region R : 3x^2 + 2y^2 = 5 is symmetric in y. The integrand is
x^2 yz dS = x^2 y(5 − 3 x^2 − 2 y^2 )
1 + 36x^2 + 16y^2 dy dx, which is odd in y, hence
σ
x^2 yz dS = 0.
- z =
4 − x^2 , ∂z ∂x = − √ x 4 − x^2
, ∂z ∂y
σ
δ 0 dS = δ 0
R
x^2 4 − x^2
0
0
4 − x^2
dx dy =
πδ 0.
- z =^1 2 (x^2 + y^2 ), R is the circular region enclosed by x^2 + y^2 = 8; ∫ ∫
σ
δ 0 dS = δ 0
R
x^2 + y^2 + 1 dA = δ 0
∫ (^2) π
0
0
r^2 + 1 r dr dθ =
πδ 0.
- z = 4 − y^2 , R is the rectangular region enclosed by x = 0, x = 3, y = 0 and y = 3; ∫ ∫
σ
y dS =
R
y
4 y^2 + 1 dA =
0
0
y
4 y^2 + 1 dy dx =^1 4
- R is the annular region enclosed by x^2 + y^2 = 1 and x^2 + y^2 = 16; ∫ ∫
σ
x^2 z dS =
R
x^2
x^2 + y^2
x^2 x^2 + y^2
y^2 x^2 + y^2
R
x^2
x^2 + y^2 dA =
∫ (^2) π
0
1
r^4 cos^2 θ dr dθ =
π.
23. M =
σ
δ(x, y, z)dS =
σ
δ 0 dS = δ 0
σ
dS = δ 0 S
- δ(x, y, z) = |z|; use z =
a^2 −x^2 −y^2 , let R be the circular region enclosed by x^2 +y^2 = a^2 , and σ the hemisphere above R. By the symmetry of both the surface and the density function with respect to the xy-plane we have
M = 2
σ
z dS = 2
R
a^2 − x^2 − y^2
x^2 a^2 − x^2 − y^2
2 a^2 − x^2 − y^2
Rr 0
dA
where Rr 0 is the circular region with radius r 0 that is slightly less than a. But
Rr 0
dA is simply
the area of the circle with radius r 0 so M = lim r 0 →a− 2 a(πr^20 ) = 2πa^3.
Exercise Set 16.6 675
EXERCISE SET 16.
- (a) zero (b) zero (c) positive (d) negative (e) zero (f ) zero
- (a) positive (b) zero (c) zero (d) zero (e) negative (f ) zero
- (a) positive (b) zero (c) positive (d) zero (e) positive (f ) zero
- 0; the flux is zero on the faces y = 0, 1 and z = 0, 1; it is 1 on x = 1 and −1 on x = 0
- (a) n = − cos vi − sin vj (b) inward, by inspection
- (a) −r cos θi − r sin θj + rk (b) inward, by inspection
- n = −zxi − zy j + k,
R
F · n dS =
R
(2x^2 + 2y^2 + 2(1 − x^2 − y^2 )) dS =
∫ (^2) π
0
0
2 r dr dθ = 2π
- With z = 1 − x − y, R is the triangular region enclosed by x + y = 1, x = 0 and y = 0; use upward normals to get ∫ ∫
σ
F · n dS = 2
R
(x + y + z)dA = 2
R
dA = (2)(area of R) = 1.
- R is the annular region enclosed by x^2 + y^2 = 1 and x^2 + y^2 = 4; ∫ ∫
σ
F · n dS =
R
x^2 √ x^2 + y^2
y^2 √ x^2 + y^2
dA
R
x^2 + y^2 dA =
∫ (^2) π
0
1
r^2 dr dθ = 14 π 3
- R is the circular region enclosed by x^2 + y^2 = 4; ∫ ∫
σ
F · n dS =
R
(2y^2 − 1)dA =
∫ (^2) π
0
0
(2r^2 sin^2 θ − 1)r dr dθ = 4π.
- R is the circular region enclosed by x^2 + y^2 − y = 0;
σ
F · n dS =
R
(−x)dA = 0 since the
region R is symmetric across the y-axis.
- With z =^1 2 (6 − 6 x − 3 y), R is the triangular region enclosed by 2x + y = 2, x = 0, and y = 0; ∫ ∫
σ
F · n dS =
R
3 x^2 +
yx + zx
dA = 3
R
x dA = 3
0
∫ (^2) − 2 x
0
x dy dx = 1.
- ∂r/∂u = cos vi + sin vj − 2 uk, ∂r/∂v = −u sin vi + u cos vj, ∂r/∂u × ∂r/∂v = 2u^2 cos vi + 2u^2 sin vj + uk; ∫ ∫
R
(2u^3 + u) dA =
∫ (^2) π
0
1
(2u^3 + u)du dv = 18π
676 Chapter 16
- ∂r/∂u = k, ∂r/∂v = −2 sin vi + cos vj, ∂r/∂u × ∂r/∂v = − cos vi − 2 sin vj; ∫ ∫
R
(2 sin^2 v − e−^ sin^ v^ cos v) dA =
∫ (^2) π
0
0
(2 sin^2 v − e−^ sin^ v^ cos v)du dv = 10π
- ∂r/∂u = cos vi + sin vj + 2k, ∂r/∂v = −u sin vi + u cos vj, ∂r/∂u × ∂r/∂v = − 2 u cos vi − 2 u sin vj + uk; ∫ ∫
R
u^2 dA =
∫ (^) π
0
∫ (^) sin v
0
u^2 du dv = 4/ 9
- ∂r/∂u = 2 cos u cos vi + 2 cos u sin vj − 2 sin uk, ∂r/∂v = −2 sin u sin vi + 2 sin u cos vj; ∂r/∂u × ∂r/∂v = 4 sin^2 u cos vi + 4 sin^2 u sin vj + 4 sin u cos uk; ∫ ∫
R
8 sin u dA = 8
∫ (^2) π
0
∫ (^) π/ 3
0
sin u du dv = 8π
- In each part, divide σ into the six surfaces σ 1 : x = −1 with |y| ≤ 1, |z| ≤ 1, and n = −i, σ 2 : x = 1 with |y| ≤ 1, |z| ≤ 1, and n = i, σ 3 : y = −1 with |x| ≤ 1, |z| ≤ 1, and n = −j, σ 4 : y = 1 with |x| ≤ 1, |z| ≤ 1, and n = j, σ 5 : z = −1 with |x| ≤ 1, |y| ≤ 1, and n = −k, σ 6 : z = 1 with |x| ≤ 1, |y| ≤ 1, and n = k,
(a)
σ 1
F · n dS =
σ 1
dS = 4,
σ 2
F · n dS =
σ 2
dS = 4, and
σi
F · n dS = 0 for
i = 3, 4 , 5 , 6 so
σ
F · n dS = 4 + 4 + 0 + 0 + 0 + 0 = 8.
(b)
σ 1
F · n dS =
σ 1
dS = 4, similarly
σi
F · n dS = 4 for i = 2, 3 , 4 , 5 , 6 so
∫ ∫
σ
F · n dS = 4 + 4 + 4 + 4 + 4 + 4 = 24.
(c)
σ 1
F · n dS = −
σ 1
dS = −4,
σ 2
F · n dS = 4, similarly
σi
F · n dS = −4 for i = 3, 5
and
σi
F · n dS = 4 for i = 4, 6 so
σ
F · n dS = −4 + 4 − 4 + 4 − 4 + 4 = 0.
- Decompose σ into a top σ 1 (the disk) and a bottom σ 2 (the portion of the paraboloid). Then
n 1 = k,
σ 1
F · n 1 dS = −
σ 1
y dS = −
∫ (^2) π
0
0
r^2 sin θ dr dθ = 0,
n 2 = (2xi + 2yj − k)/
1 + 4x^2 + 4y^2 ,
σ 2
F · n 2 dS =
σ 2
y √(2x^2 + 2y^2 + 1) 1 + 4x^2 + 4y^2
dS = 0,
because the surface σ 2 is symmetric with respect to the xy-plane and the integrand is an odd function of y. Thus the flux is 0.
