Solutions to Math 227 Final Exam, April, 2005
1. Let z = f (x, y) be a C 3 function on an open set D ⊂ IR2 , and assume that all partial derivatives of
∂3 z
orders one, two and three equal zero at a point (a, b) ∈ D except for ∂x∂y
2 (a, b) = 6. Determine whether
(a, b) is a local maximum, a local minimum or a saddle point, and give reasons for your answer. Remark:
Your reasons do not need to be completely rigorous but they should be at least heuristically convincing.
Solution. Near (a, b),
1
1!2! 6(x
f (x, y) ≈ f (a, b) +
− a)(y − b)2
As (x − a)(y − b)2 takes both signs arbitrarily close to (a, b), f (x, y) has a saddle point there.
2. Use the method of Lagrange multipliers (no credit will be given for any other method) to find the points
on the ellipse x2 + 4y 2 = 4 which are closest to the point (1, 0). Hint: Minimize the square of the
distance from a point on the ellipse to (1, 0), and be careful to find all four solutions to the equations
specified by the method of Lagrange multipliers.
Solution. We are to minimize (x − 1)2 + y 2 subject to the constraint that x2 + 4y 2 − 4 = 0. So define
L(x, y, λ) = (x − 1)2 + y 2 + λ(x2 + 4y 2 − 4). Then
0 = Lx = 2(x − 1) + 2λx
=⇒ (1 + λ)x = 1
=⇒ y = 0 or λ = − 41
0 = Ly = 2y + 8λy
0 = Lλ = x2 + 4y 2 − 4
If y = 0 then the last equation gives x = ±2.
2
If λ =√− 41 , then the first equation gives x = 43 and the last equation gives y 2√= 1 − 41 43 = 59 or
y = ± 35 . From the table below, we see that there are two points, namely 43 , ± 35 , closest to (1, 0).
(x − 1)2 + y 2
(x, y)
(2, 0)
(−2, 0)
√ 4
5
3 , 3√
4
5
3, − 3
1
9
2
3
2
3
3. Use the change of variables x = u/v, y = v to evaluate the double integral
the region specified by the inequalities y ≤ 3, y ≥ x and 1 ≤ xy ≤ 4.
RR
D
y dxdy where D ⊂ IR2 is
Solution. In terms of u and v, the conditions y ≤ 3, y ≥ x and 1 ≤ xy ≤ 4 become
v≤3
v≥
u
v
1≤u≤4
When v > 0 these conditions are equivalent to
0<v≤3
u ≤ v2
1≤u≤4
and when v < 0, they are equivalent to
v<0
u ≥ v2
1≤u≤4
The figure on the left below shows the domain D in the original (x, y) coordinates and the figure on the
right shows D in the new (u, v) coordinates
1
y
y=3
v=3
v
u=1
D
D
x
y = 1/x
y = 4/x
u=4
y = 4/x
y = 1/x
u
D
D
u = v2
y=x
The inverse change of coordinates is u = xy, v = y and
1
− vu2
∂(x,y)
v
=
det
=
∂(u,v)
0
1
1
|v|
so that the integral over the upper part of D is
Z 4
Z 4 Z 3
Z 4 Z 3
4
√
1
du [3 − u] = 3u − 23 u3/2 1 = 12 −
du √ dv |v|
v=
du √ dv =
u
1
u
1
1
16
3
−3+
2
3
=
13
3
and the integral over the lower part of D (which should be negative since y is negative there) is
Z 4 Z 0
Z 4
Z 4 Z 0
4
√
2
14
1
v=−
du √ dv = −
du u = − 23 u3/2 1 = − 16
du √ dv |v|
3 − 3 =− 3
1
1
− u
− u
1
1
All together, the integral over all of D is − 13
.
4. Let F = ex cos(πy 2 )+ ay − 1 , byex sin(πy 2 )+ x+ y be a vector field in IR2 , where a and b are constants.
(a) Find the values of a and b such that F is conservative.
R
(b) With the values of a and b determined in part (a), evaluate the vector line integral F · dr taken
over any smooth curve starting at (0, 0) and ending at (1, 1).
Solution. (a) Since
∂
dx
byex sin(πy 2 ) + x + y =
∂
dy
x
e cos(πy 2 ) + ay − 1 for all x, y
⇐⇒ byex sin(πy 2 ) + 1 = −2πyex sin(πy 2 ) + a
for all x, y
⇐⇒ a = 1 and b = −2π
the only allowed values are a = 1, b = −2π.
(b) When a = 1 and b = −2π, F = ∇ϕ with ϕ = ex cos(πy 2 ) + xy − x + 21 y 2 . So
Z
(1,1)
F · dr = ex cos(πy 2 ) + xy − x + 12 y 2 (0,0) = (−e + 1 − 1 + 21 ) − (1) = −e −
1
2
5. Find the surface area of the part of the paraboloid z = 2 − x2 − y 2 lying above the xy–plane.
Solution. Write the equation of the paraboloid in the form F (x, y, z) = z + x2 + y 2 − 2 = 0 or in the
form z = f (x, y) = 2 − x2 − y 2 . Then
p
n̂dS = ± 2x, 2y, 1 dxdy =⇒ dS = 1 + 4x2 + 4y 2 dxdy
The part of the parbola lying above the xy–plane has x2 + y 2 ≤ 2. So
ZZ
Z 2π Z
p
dθ
surface area =
1 + 4x2 + 4y 2 dxdy =
x2 +y 2 ≤2
= 2π
1
12 (1
+ 4r2 )3/2
√2
0
0
=
2
π
6 [27
− 1] =
13
3 π
0
√
2
dr r
p
1 + 4r2
3
3
r
6. Let F = krk
is an open set containing the origin
3 be a vector field in IR , where r = (x, y, z). If D ⊂ IR
with a smooth boundary ∂D, then it was shown in class that the Divergence or Gauss’s Theorem does
RRR
RR
not apply directly for F, D and ∂D. That is ∂D F · dS 6= D div(F) dV , where the triple integral must
be interpreted as an improper integral since F is not defined at the origin.
r
(a) Let G = krk
2 with D as above. Does the Divergence Theorem apply directly for G, D and ∂D? More
RRR
RR
precisely, is ∂D G · dS = D div(G) dV ? Here, as above, the triple integral must be interpreted
as an improper integral since G is not defined at the origin. Give precise reasons for your answer.
(b) Verify your answer to part (a) is correct in the special case where D = (x, y, z) x2 +y 2 +z 2 < a2 .
Solution. As a preliminary calculation, we find
∇·G =∇·
h
(x, y, z) i
2(x2 + y 2 + z 2 )
1
3
−
= 2
=
2
2
2
2
2
2
2
2
2
2
x +y +z
x +y +z
[x + y + z ]
x + y2 + z 2
(a) Denote by Bε = (x, y, z) ∈ IR3 x2 + y 2 + z 2 < ε2 the (solid) ball of radius ε centred on the
origin and by Sε = (x, y, z) ∈ IR3 x2 + y 2 + z 2 = ε2 the sphere of radius ε centred on the origin,
with outward pointing normal. Note that
ZZ
ZZ
ZZ
1
r
r
lim
G · dS = lim
dS
=
lim
·
dS = lim 1ε 4πε2 = 0
2
krk
krk
ε
ε→0
ε→0
Sε
ε→0
Sε
ε→0
Sε
If ε is small enough that Bε ⊂ D, then G is a smooth vector field on
D ε = D \ Bε =
(x, y, z) ∈ D
(x, y, z) ∈
/ Bε
and the divergence theorem is directly applicable to G, Dε and ∂Dε = ∂D − Sε . So, using the definition
RRR
of D div(G) dV as an improper integral,
ZZZ
div(G) dV = lim
ε→0
D
ZZZ
div(G) dV = lim
ε→0
Dε
ZZ
G · dS −
∂D
ZZ
Sε
G · dS =
ZZ
∂D
G · dS
so the divergence theorem does apply directly to G, D and ∂D.
(b) Using the notation of part (a),
ZZ
∂D
G · dS =
ZZ
Sa
r
krk2
·
r
krk
dS =
1
a
ZZ
Sa
dS = a1 4πa2 = 4πa
and
ZZZ
D
div(G) dV = lim
ε→0
ZZZ
div(G) dV = lim
ε→0
D\Bε
ZZZ
Ba \Bε
1
krk2
dV = lim
ε→0
Z
a
ε
2
1
R2 4πR
dR
= lim 4π[a − ε] = 4πa
ε→0
7. Verify that Stokes’ Theorem is true in the special case of the vector field F = y, xz, x2 over the triangle
with vertices at (1, 0, 0), (0, 1, 0) and (0, 0, 1).
Solution. Denote by L1 the line segment from (1, 0, 0) to (0, 1, 0), by L2 the line segment from (0, 1, 0)
to (0, 0, 1), by L3 the line segment from (0, 0, 1) to (1, 0, 0), and by C the triangular curve L1 + L2 + L3 .
Denote by S the part of the plane x + y + z = 1 bounded by the triangle C, with upward pointing unit
normal vector n̂ = √13 (ı̂ı + ̂ + k̂).
3
R
Computation of
so that
L1
F · dr: We parametrize L1 by r(t) = (1 − t, t, 0), 0 ≤ t ≤ 1. Then r′ (t) = (−1, 1, 0)
Z
F · dr =
L1
R
Computation of
so that
L2
L1
R
L3
L1
R
C
t, 0, (1 − t)
0
· (−1, 1, 0)dt = −
1
Z
t dt = − 21
0
F · dr =
Z
1
1 − t, 0, 0 · (0, −1, 1)dt = −
0
1
Z
0 dt = 0
0
F · dr: We parametrize L3 by r(t) = (t, 0, 1 − t), 0 ≤ t ≤ 1. Then r′ (t) = (1, 0, −1)
Z
Computation of
2
F · dr: We parametrize L2 by r(t) = (0, 1 − t, t), 0 ≤ t ≤ 1. Then r′ (t) = (0, −1, 1)
Z
Computation of
so that
1
Z
Z
F · dr =
1
2
0, t − t , t
0
2
· (1, 0, −1)dt = −
Z
1
0
t2 dt = − 31
F · dr:
Z
C
F · dr =
Z
L1
F · dr +
Z
L2
F · dr +
Z
L3
F · dr = − 21 + 0 −
1
3
= − 56
RR
Computation of S ∇ × F · n̂ dS (method 1): We parametrize S by r(s, t) = (1, 0, 0) + s(−1, 1, 0) +
t(−1, 0, 1) = (1 − s − t, s, t), s ≥ 0, t ≥ 0, s + t ≤ 1. To see that this works, observe that
◦ r(0, 0) = (1, 0, 0), r(1,0) = (0,1,0), r(0, 1) = (0, 0, 1).
◦ As s runs from 0 to 1, r(s, 0) = (1 − s, s, 0) runs from (1, 0, 0) to (0, 1, 0) along L1 . (Replace t by s
in the parametrization of L1 above.)
◦ As t runs from 0 to 1, r(0, t) = (1 − t, 0, t) runs from (1, 0, 0) to (0, 0, 1) along L3 . (Replace t by
1 − t in the parametrization of L3 above).
◦ If we set s = 1 − t and left t run from 0 to 1, then r(1 − t, t) = (0, 1 − t, t) runs from (0, 1, 0) to
(0, 0, 1) along L2 . (See the parametrization of L2 above.)
◦ r(s, t) lies on the plane x + y + z = 1 for all s and t.
Since
∇ × F = det
ı̂ı
̂
k̂
∂
dx
∂
dy
y
xz
∂
dz
2
= −xı̂ı − 2x̂ + (z − 1)k̂ = (−1 + s + t , −2 + 2s + 2t , t − 1)
x
and
∂r
∂s
= (−1, 1, 0)
∂r
∂t
= (−1, 0, 1)
we have
ZZ
S
Z
1
∇ × F · n̂ dS =
Z
1
=
Z
1
Z
1
ds
0
ds
0
Z
1−s
×
∂r
∂t
= (1, 1, 1)
=⇒
n̂dS = (1, 1, 1) ds dt
dt (−1 + s + t , −2 + 2s + 2t , t − 1) · (1, 1, 1)
dt (−4 + 3s + 4t)
0
0
=
1−s
0
0
=
Z
∂r
∂s
ds [(−4 + 3s)(1 − s) + 2(1 − s)2 )]
ds [−2 + 3s − s2 ] = −2 +
3
2
−
1
3
= − 65
as desired.
RR
Computation of S ∇ × F · n̂ dS (method 2): We parametrize S by using x and y as parameters.
That is, r(x, y) = (x, y, 1 − x − y). As (x, y, z) runs over the triangle with vertices (1, 0, 0), (0, 1, 0),
4
(0, 0, 1), the parameters (x, y) runs over the triangle with vertices (1, 0), (0, 1), (0, 0). So the domain is
(x, y) x ≥ 0, y ≥ 0, x + y ≤ 1 .
Since
∇ × F = det
ı̂ı
̂
k̂
∂
dx
∂
dy
y
xz
∂
dz
2
x
= −xı̂ı − 2x̂ + (z − 1)k̂
we have
ZZ
S
∇ × F · n̂ dS =
Z
1
dx
0
Z
1
=−
Z
1
=−
Z
1
=−
0
n̂dS = (1, 1, 1) dx dy
1−x
dy (−x , −2x , −x − y) · (1, 1, 1)
0
dx
0
0
Z
and
Z
1−x
dy (4x + y)
0
dx [4x(1 − x) + 21 (1 − x)2 ]
dx [ 21 + 3x − 27 x2 ] = −
1
2
+
3
2
− 76 ] = − 65
as desired.
8. Let ω = ω(x,y) = xy dx be a 1–form on IR2 . According to the discussion in your textbook, the exterior
derivative or differential of ω is a 2–form on IR2 deffined by the formula dω = dω(x,y) = (ydx+xdy)∧dx =
x dy ∧ dx = −x dx ∧ dy). At a point (x, y), dω operates on pairs of vectors in IR2 . For example, at the
point (2, 1),
dω(2,1)
2
0
2
0
2 0
,
= −2 dx ∧ dy
,
= −2 det
= −2(2) = −4
3
1
3
1
3 1
The purpose of this problem is to ask you to give a more geometric definition of the exterior derivative
motivated by the generalizedStokes’s
In order to keep everything very concrete, give a
Theorem.
2
0
geometric definition of dω(2,1)
,
, involving a limit h → 0 and an integral of ω over a certain
3
1
2
0
2
closed curve Ch in IR . Then verify by use of your definition that dω(2,1)
,
= −4. Your
3
1
discussion should be accompanied by a picture showing all key aspects of your solution.
Solution. The point here is that, for any constant 2–form a dx∧dy and any two vectors [x1 , y1 ]t , [x2 , y2 ]t
Ω
x1
y1
x2
x1
x2
x
,
= a dx ∧ dy
,
= a det 1
y2
y1
y2
y1
x2
y2
is a times the area of the parallelogram with sides [x1 , y1 ]t and [x2 , y2 ]t .
For each h > 0, let
◦ Ph be the parallelogram in IR2 that has one vertex at (2, 1) and has sides h[2, 3]t and h[0, 1]t . (We
could also choose to have Ph centred on (2, 1).)
L2h
◦ L1h be the line segment from (2, 1) to (2, 1) + h(2, 3),
◦ L2h be the line segment from (2, 1)+h(2, 3) to (2, 1)+h(2, 3)+h(0, 1),
L3h
◦ L3h be the line segment from (2, 1) + h(2, 3) + h(0, 1) to (2, 1) + h(0, 1),
Ph
◦ L4h be the line segement from (2, 1) + h(0, 1) to (2, 1), and
L1h
◦ Ch be the curve, L1h + L2h + L3h + L4h , bounding Ph .
L4h
5
(2, 1)
By the generalized Stokes’ theorem
Z
ω=
ZZ
dω
Ph
Ch
When h is very small, dω is approximately the constant 2–form −2dx ∧ dy on Ph . This constant 2–form
gives
ZZ
2 0 2
(−2dx ∧ dy) = −2Area(Ph ) = −2 det
h
3 1
Ph
So I choose to define
Z
2
0
,
= lim h12
ω
3
1
h→0
Ch
R
Computation of L1h ω: We parametrize L1 by x(t), y(t) = (2, 1) + th(2, 3), 0 ≤ t ≤ 1. Then ω =
(2 + 2th)(1 + 3th)(2h dt) so that
dω(2,1)
Z
ω=
Z
1
0
L1h
2h 2 + 8th + 6t2 h2 dt = 2h[2 + 4h + 2h2 ]
R
Computation of L2h ω: We parametrize L2 by x(t), y(t) = (2, 1) + h(2, 3) + th(0, 1), 0 ≤ t ≤ 1. Then
ω = (2 + 2h)(1 + 3h + th)(0 dt) = 0 so that
Z
ω=0
L2h
R
Computation of L3h ω: We parametrize L3 by x(t), y(t) = (2, 1) + (1 − t)h(2, 3) + h(0, 1), 0 ≤ t ≤ 1.
Then ω = (2 + 2h − 2th)(1 + 4h − 3th)(−2h dt) so that
Z
L3h
Z
1
(−2h) (2 + 2h)(1 + 4h) − 2th(1 + 4h) − 3th(2 + 2h) + 6t2 h2 dt
0
= (−2h) (2 + 2h)(1 + 4h) − h(1 + 4h) − 3h(1 + h) + 3h2
ω=
R
Computation of L4h ω: We parametrize L4 by x(t), y(t) = (2, 1) + (1 − t)h(0, 1), 0 ≤ t ≤ 1. Then
ω = (2)(1 + h − th)(0 dt) = 0 so that
Z
ω=0
L4h
Computation of
1
2
h→0 h
lim
Z
Ch
limh→0 h12
1
2
h→0 h
ω = lim
R
Ch
Z
L
ω:
ω+
Z
L
ω+
Z
ω+
Z
L
L
ω
4h
3h
2h
h 1h
i
2
=
2h[2 + 4h + 2h ] + (−2h) (2 + 2h)(1 + 4h) − h(1 + 4h) − 3h(1 + h) + 3h2
h
i
= lim h2 [2 + 4h + 2h2 ] − (2 + 2h)(1 + 4h) − h(1 + 4h) − 3h(1 + h) + 3h2
h→0
h
i
= lim h2 [2 + 4h] − (2 + 10h) − h − 3h
h→0
h
i
= lim h2 4h − 6h
lim 12
h→0 h
h→0
= −4
as desired.
6