Math 227 Problem Set IX Solutions
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1. Evaluate, both by direct integration and by Stokes’ Theorem, C (z dx + x dy + y dz) where C is the circle
x + y + z = 0, x2 + y 2 + z 2 = 1. Orient C so that its projection on the xy–plane is counterclockwise.
Solution. Direct integration: The plane x + y + z = 0 passes through the centre, (0, 0, 0), of the sphere
x2 +y 2 +z 2 = 1, so the circle C is centred on (0, 0, 0) and has radius 1. It lies in a plane with unit (upward
pointing) normal vector k̂′ = √13 (1, 1, 1). The point √12 (1, −1, 0) is on C, so the vector ı̂ı′ = √12 (1, −1, 0)
lies in the plane of C. Another vector which lies in the plane of C, but which is perpendciular to ı̂ı′ is
̂′ = k̂′ × ı̂ı′ = √16 (1, 1, −2). Hence we may parametrize the curve using
r(θ) = cos θ ı̂ı′ + sin θ ̂′
which gives

√1
√1

 x(θ) = 2 cos θ + 6 sin θ
y(θ) = − √12 cos θ + √16 sin θ

 z(θ) = − √2 sin θ
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and
 ′
√1
√1

 x (θ) = − 2 sin θ + 6 cos θ
y ′ (θ) = √12 sin θ + √16 cos θ

 z ′ (θ) = − √2 cos θ
6
The check that motion, projected on the xy–plane, is counterclockwise, observe that at θ = 0, (x, y) =
√1 (1, −1) (in the fourth quadrant) and dx , dy = √1 (1, 1), which is up and to the right. The integral
dθ dθ
2
6
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is of the form C F · dr where F = zı̂ı + x̂ + y k̂ and C is curve parametrized above. Hence
I
C
2π
F · dr =
Z
2π
=
Z
Z
2π
0
0
=
0
=
F r(θ) · r′ (θ) dθ
h
−
h
√3
12
sin θ + √16 cos θ
+ √12 cos θ + √16 sin θ √12 sin θ + √16 cos θ
i
+ − √12 cos θ + √16 sin θ − √26 cos θ dθ
√3 2π
12
=
√2
6
sin θ −
cos2 θ +
√
√1
2
√3
12
sin2 θ + −
2
6
+
1
6
+
1
2
−
2
6
i
sin θ cos θ dθ
3π
Stokes’ Theorem: Choose as S the portion of the plane x + y + z = 0 interior to the sphere. Then
n̂ = √13 (ı̂ı + ̂ + k̂) and ∇ × F = ı̂ı + ̂ + k̂ so, by Stokes’ Theorem,
I
F · dr =
C
ZZ
S
∇ × F · n̂ dS =
ZZ
S
(ı̂ı + ̂ + k̂) ·
√1 (ı̂ı + ̂
+
3
k̂) dS =
√ ZZ
√
3
dS = 3π
S
since S is a circle of radius 1.
2. Let C be the intersection of x+2y−z = 7 and x2 −2x+4y 2 = 15. The curve C is oriented counterclockwise
when viewed from high on the z–axis. Let
2
F = ex + yz ı̂ı + cos(y 2 ) − x2 ̂ + sin(z 2 ) + xy k̂
Evaluate
H
C
F · dr.
1
Solution. We apply Stokes’ Theorem with S being the part of the plane x + 2y − z = 7 that is inside
the ellipse x2 − 2x + 4y 2 = 15 (which can also be written (x − 1)2 + 4y 2 = 16.) We can parametrize S
by z = −7 + x + 2y, with (x − 1)2 + 4y 2 ≤ 16, so that n̂dS = (−1, −2, 1) dxdy (upward normal). So, by
Stokes’ Theorem and the observation that ∇ × F = xı̂ı − (z + 2x)k̂, and using R to denote the interior
of the ellipse (x − 1)2 + 4y 2 = 16,
I
C
F · dr =
=
=
ZZ
Z ZS
Z ZR
R
∇ × F · n̂ dS
[xı̂ı − (z + 2x)k̂]z=−7+x+2y · (−1, −2, 1) dx dy
[7 − 4x − 2y] dx dy
= (area of ellipse with semi–axes a = 4, b = 2)[7 − 4x̄ − 2ȳ]
= π × 4 × 2[7 − 4 × 1 − 2 × 0] = 24π
Here we have used that, for example, the average value of x over R is x̄ =
RR
R
x dx dy/
RR
R
dx dy.
RR
∇ × F) · n̂ dS where S is the portion of the sphere x2 + y 2 + z 2 = 1 that obeys x + y + z ≥ 1,
3. Consider S (∇
n̂ is the upward pointing normal to the sphere and F = (y − z)ı̂ı + (z − x)̂ + (x − y)k̂. Find another
RR
RR
RR
∇ × F) · n̂ dS.
∇ × F) · n̂ dS and evaluate S ′ (∇
∇ × F) · n̂ dS = S ′ (∇
surface S ′ with the property that S (∇
Solution. Let S ′ be the portion of x + y + z = 1 that is inside the sphere x2 + y 2 + z 2 = 1. Then
∂S = ∂S ′ , so, by Stokes’ Theorem, (with n̂ always the upward pointing normal)
ZZ
S′
(∇
∇ × F) · n̂ dS =
I
∂S ′
As ∇ × F = −2(ı̂ı + ̂ + k̂) and, on S ′ , n̂ =
ZZ
S′
(∇
∇ × F) · n̂ dS =
F · dr =
√1 (ı̂ı + ̂
+
3
I
∂S
F · dr =
ZZ
S
(∇
∇ × F) · n̂ dS
k̂)
√
√ − 2 3 dS = −2 3 × Area(S ′ )
ZZ
S′
S ′ is a circular disk. It’s center (xc , yc , zc ) has to obey xc + yc + zc = 1. By symmetry, xc = yc = zc , so
xc = yc = zc = 13 . Any point, like (0, 0, 1), which satisfies both x + y + z = 1 and x2 + y 2 + z 2 = 1 is on
q
the boundary of S ′ . So the radius of S ′ is 1 , 1 , 1 − (0, 0, 1) = 1 , 1 , − 2 = 2 . So the area of S ′
3 3 3
is 32 π and
ZZ
S′
3 3
3
3
√
(∇
∇ × F) · n̂ dS = −2 3 × Area(S ′ ) = − √43 π
H
∇φ · dr for any continuously differentiable scalar fields φ and
∇ψ · dr = − C ψ∇
4. Verify the identity C φ∇
ψ and curve C that is the boundary of a piecewise smooth surface.
H
Solution. Suppose that C = ∂S. Then, by Stokes’ Theorem
I
ZZ
[φ∇
∇ψ + ψ∇
∇φ] · dr =
∇ × [φ∇
∇ψ + ψ∇
∇φ] · n̂ dS
C
But
S
∇ × [φ∇
∇ψ + ψ∇
∇φ] = ∇φ × ∇ψ + φ∇
∇ × (∇
∇ψ) + ∇ ψ × ∇φ + ψ∇
∇ × (∇
∇φ)
= ∇φ × ∇ψ + ∇ψ × ∇φ
=0
2
since φ∇
∇ × (∇
∇ψ) = ψ∇
∇ × (∇
∇φ = 0
so
H
C
[φ∇
∇ψ + ψ∇
∇φ] · dr = 0.
5. Use Green’s Theorem to show that
ZZ
R
∇ · F dA =
I
∂R
F · n̂ ds
where ∂R is the boundary of the plane domain R, with the usual orientation, F = F1 ı̂ı + F2 ̂, n̂ is the
outward normal to ∂R and s is the arclength along ∂R.
Solution. If dr = dxı̂ı + dy ̂ is an infinitesmal piece of ∂R, then n̂ ds = dr × k̂ = dy ı̂ı − dx ̂. (Note that
dr × k̂ has the same length as dr, lies in the xy–plane and is perpendicular to dr. Use the right hand
rule to check that dr × k̂ is n̂ ds rather than −n̂ ds.) So, by Green’s Theorem
n̂ ds
ZZ
I
I
ZZ
dr
∂
∂
[−F2 dx + F1 dy] =
F · n̂ ds =
∇ · F dA
∂x F1 − ∂y (−F2 ) dA =
R
∂R
∂R
R
R
dr
n̂ ds
∂R
x dy−y dx
1
counterclockwise around
6. Integrate 2π
C x2 +y 2
(a) the circle x2 + y 2 = a2
(b) the boundary of the square with vertices (−1, −1), (−1, 1), (1, 1) and (1, −1)
(c) the boundary of the region 1 ≤ x2 + y 2 ≤ 2, y ≥ 0
H
Solution. (a) Parametrize the circle by x = a cos θ, y = a sin θ, 0 ≤ θ ≤ 2π. Then dx = −a sin θ dθ and
dy = a cos θ dθ so
1
2π
I
C
x dy−y dx
x2 +y 2
=
1
2π
Z
2π
a2 cos2 θ dθ+a2 sin2 θ dθ
a2 cos2 θ+a2 sin2 θ
0
1
2π
=
Z
2π
dθ = 1
0
(b) From the figure on the right,
1
2π
I
C
x dy−y dx
x2 +y 2
=
1
2π
Z
1
dx
x2 +1
1
2π
+
−1
1
= 4 2π
1
tan−1 x =
−1
Z
1
dy
1+y 2
+
1
2π
−1
2
π
h
π
4
+
π
4
i
Z
−1
1
−dx
x2 +1
+
1
2π
Z
1
−1
(−1, 1)
−dy
1+y 2
= 1
dy=0
y=1
dx=0
x=−1
(−1, −1)
(1, 1)
dx=0
x=1
dy=0
y=−1
(1, −1)
(c) The outer semicircle gives 12 , as in part (a). The inner semicircle gives − 21 ,
as in part (a). The two flat pieces each give zero, since on them y = 0 and
H x dy−y dx
1
=0 .
dy = 0. So 2π
C x2 +y 2
7. Show that
∂
∂x
x
x2 +y 2
=
∂
∂y
−y
x2 +y 2
for all (x, y) 6= (0, 0). Discuss the connection between this result and the results of the last question.
Solution.
∂
∂x
∂
∂y
x
x2 +y 2
−y
x2 +y 2
=
(x2 +y 2 )−x(2x)
(x2 +y 2 )2
=
y 2 −x2
(x2 +y 2 )2
=
−(x2 +y 2 )−(−y)(2y)
(x2 +y 2 )2
=
y 2 −x2
(x2 +y 2 )2
The two are well–defined and equal everywhere except at the origin (0, 0). Were it not for the singularity
R
at (0, 0), the vector field of the last problem would be conservative and the integral F~ · d~r around any
3
closed curve would be zero. But as we saw in parts (a) and (b) of the last problem, this is not the case.
On the other hand, by Green’s Theorem (or Stokes’ Theorem), the integral around the boundary of any
region that does not contain (0, 0) is zero, as happened in part (c).
8. Find a continuously differentiable simple, closed, counterclockwise oriented curve, C, in the xy–plane
H
for the which the value of the line integral C (y 3 − y) dx − 2x3 dy is a maximum among all C 1 simple,
closed, counterclockwise oriented curves. “Simple” means that the curve does not intersect itself.
Solution. By Green’s Theorem
ZZ h
I
3
∂
(y 3 − y) dx − 2x3 dy =
∂x (−2x ) −
C
R
3
∂
∂y (y
ZZ h
i
i
1 − 6x2 − 3y 2 dx dy
− y) dx dy =
R
where R is the region in the xy–plane whose boundary is C. Observe that the integrand 1 − 6x2 − 3y 2
is positive in the elliptical region 6x2 + 3y 2 ≤ 1 and negative outside of it. To maximize the integral
RR 1−6x2 −3y 2 dx dy we should choose R to contain all points (x, y) with the integrand 1−6x2 −3y 2 ≥ 0
R
and to exclude all points (x, y) with the integrand 1 − 6x2 − 3y 2 < 0. So we choose
R=
The corresponding C is 6x2 + 3y 2 = 1 .
(x, y) 6x2 + 3y 2 ≤ 1
4