Download Solutions to Math 234 Homework: Vector Calculus and Line Integrals (Spring 2008) and more Assignments Advanced Calculus in PDF only on Docsity!
Solutions to Homework from Sections 13.3-13.
Math 234, Spring 2008
Section 13.
F (x, y) = 〈P (x, y), Q(x, y)〉 = 〈1 + 2xy + ln x, x
2 〉. So
∂P
∂y
= 2x =
∂Q
∂x
, which means
F is conservative.
To find the potential function, we use partial integration:
f (x, y) =
P (x, y) dx = x + x
2 y + (x ln x − x) + g(y) = x
2 y + x ln x + g(y)
Differentiating with respect to y, we find:
Q(x, y) = fy (x, y) = x
2
′ (y) =⇒ x
2 = x
2
′ (y) =⇒ g
′ (y) = 0
So g(y) is a constant, which we can choose to be 0. Then f (x, y) = x
2 y + x ln x.
F (x, y, z) = 〈e
y , xe
y , (z + 1)e
z 〉, C is parametrized by ~r(t) = 〈t, t
2 , t
3 〉, 0 ≤ t ≤ 1.
(a) We find the potential function f (x, y, z) such that
F = ∇f by partial integration of the three components of
F :
f (x, y, z) =
e
y dx = xe
y
f (x, y, z) =
xe
y dy = xe
y
(x, z)
f (x, y, z) =
(z + 1)e
z dz = ze
z
(x, y)
Combining these gives f (x, y, z) = xe
y
z .
(b) By the Fundamental Theorem of Line Integrals:
C
F · d~r = f (~r(1)) − f (~r(0)) = f (1, 1 , 1) − f (0, 0 , 0) = (e + e) − (0 + 0) = 2e
- (a) Consider a vector field
F (~r) =
c~r
|~r|
3.^ We first observe that^
F is a conservative vector field. Let f (x, y, z) =
c √
x
2 +y
2 +z
2
. Then:
∇f (x, y, z) = 〈fx, fy , fz 〉 = −c
〈−x, −y, −z〉
(x
2
2
2 )
3 / 2
c〈x, y, z〉
(x
2
2
2 )
3 / 2
c~r
|~r|
3
F (~r)
So we can use the Fundamental Theorem for Line Integrals to find the work done by
F in moving an object from
a point P 1 a distance d 1 from the origin to a point P 2 a distance d 2 from the origin along some path C. The work
is: (^) ∫
C
F · d~r = f (P 2 ) − f (P 1 ) =
−c
d 2
−c
d 1
= c
d 1
d 2
(b) The work done by the gravitational field when the Earth moves from aphelion to perihelion is (converting all
distances to meters, to match the units of the gravitational constant G):
−(mM G)
1. 52 × 10
11
1. 47 × 10
11
= −(5. 97 × 10
24 )(1. 99 × 10
30 )(6. 67 × 10
− 11 )(
− 11 )
= −(79. 24 × 10
32 )(− 0 .0224) ≈ 1. 77 × 10
32 N · m
(c) The work done by the electric field is:
qQ
− 12
− 12
= (8. 985 × 10
9
)(− 1. 6 × 10
− 19
)(1)(− 10
12
) ≈ 1440
Section 13.
- We want to compute
C
xe
− 2 x dx + (x
4
2 y
2 )dy, where C is the boundary of the region between the circles x
2 +y
2 = 1
and x
2
2 = 4. So P (x, y) = xe
− 2 x and Q(x, y) = x
4
2 y
2 .
∂Q
∂x
∂P
∂y
= (4x
3
2 ) − 0 = 4x
3
2
So, applying Green’s Theorem (and converting to polar coordinates):
C
xe
− 2 x dx + (x
4
2 y
2 )dy =
D
(4x
3
2 ) dA =
2 π
0
2
1
(4r
3 sin
3 θ + 4r
3 sin θ cos
2 θ) r dr dθ
2 π
0
2
1
4 r
4 sin θ(sin
2 θ + cos
2 θ) dr dθ =
2 π
0
2
1
4 r
4 sin θ dr dθ
2 π
0
sin θ dθ
2
1
4 r
4 dr
= (cos(2π) − cos(0))
2
1
4 r
4 dr
F (x, y) = 〈x, x
3
2 〉, so P (x, y) = x and Q(x, y) = x
3
2
. Then:
∂Q
∂x
∂P
∂y
= (3x
2
2 ) − 0 = 3(x
2
2 )
So: ∮
C
F · d~r =
D
3(x
2
2
) dA =
π
0
2
0
3 r
2
r dr dθ = π
2
0
3 r
3
dr =
3 π
4
− 0
4
) = 12π
- The coordinates of the centroid are given by:
x =
A
D
x dA and y =
A
D
y dA
Define vector fields
F 1 = 〈P 1 , Q 1 〉 = 〈 0 ,
1
2
x
2 〉 and
F 2 = 〈P 2 , Q 2 〉 = 〈−
1
2
y
2 , 0 〉. Then
∂Q 1
∂x
∂P 1
∂y
= x − 0 = x and
∂Q 2
∂x
∂P 2
∂y
= 0 − (−y) = y. So, by Green’s Theorem:
x =
A
D
x dA =
A
C
F
1
· d~r =
A
C
x
2 dy =
2 A
C
x
2 dy
y =
A
D
y dA =
A
C
F
2
· d~r =
A
C
y
2 dx = −
2 A
C
y
2 dx
Section 13.
F (x, y, z) = 〈 3 z
2 , cos y, 2 xz〉. So the curl of
F is:
∇ ×
F =
i
j
k
∂
∂x
∂
∂y
∂
∂z
3 z
2 cos y 2 xz
∂y
(2xz) −
∂z
(cos y),
∂z
(3z
2 ) −
∂x
(2xz),
∂x
(cos y) −
∂y
(3z
2 )
= 〈 0 − 0 , 6 z − 2 z, 0 − 0 〉 = 〈 0 , 4 z, 0 〉 6 = 〈 0 , 0 , 0 〉
So
F is not conservative.
- Show that curl(f ·
F ) = f curl
F + (∇f ) ×
F. Let
F (x, y, z) = 〈P (x, y, z), Q(x, y, z), R(x, y, z)〉.
curl(f
F ) =
i
j
k
∂
∂x
∂
∂y
∂
∂z
f · P f · Q f · R
∂y
(f · R) −
∂z
(f · Q),
∂z
(f · P ) −
∂x
(f · R),
∂x
(f · Q) −
∂y
(f · P )
= 〈f y
· R + f · R y
− f z
· Q − f · Q z
, f z
· P + f · P z
− f x
· R − f · R x
, f x
· Q + f · Q x
− f y
· P − f · P y
= 〈f · Ry − f · Qz , f · Pz − f · Rx, f · Qx − f · Py 〉 + 〈fy · R − fz · Q, fz · P − fx · R, fx · Q − fy · P 〉
= f · curl
F + (∇f ) ×
F
