Boğaziçi University
Department of Mathematics
Math 102 Calculus II
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Math 102 Number :
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Summer 2008 – Final Examination
August 2, 2008
09:10-11:30
IMPORTANT
1. Write your name, surname on top of each page. 2. The exam consists of 7 questions some of which have more than
one part. 3. Read the questions carefully and write your answers neatly under the corresponding questions. 4. Show all
your work. Correct answers without sufficient explanation might not get full credit. 5. Calculators are not allowed.
1. Suppose the vectors u, v and w satisfy u + v + w = 0. Show that u × v = v × w = w × u.
Solution: We have
u × v = u × (−u − w) = −u × u − u × w = −u × w = w × u .
The identity u × v = v × w is proved similarly.
2. Determine whether the limit
lim(x,y)→(0,0)
2x2 y
x4 + y 2
exists. If so, find its value.
Solution: Considering, for instance, the curves
C1 : x = t, y = 0
and
we have
lim
C1
(x,y)−→(0,0)
C2 : x = t, y = t2
2x2 y
0
= lim 4 = 0
4
2
t→0 t
x +y
and, on the other hand,
lim
C2
(x,y)−→(0,0)
2t4
2x2 y
=
lim
= 1.
x4 + y 2 t→0 2t4
Therefore the given limit does not exist.
F (x, y) = x2 − xy + y 4.
3. Locate and classify the extrema for the function
Solution: First
∇F (x, y) = 2x − y, −x + 4y 3 = 0
if and only if
(x, y) ∈
√ !
2
2
,
,−
(0, 0) , −
8
4
(
√
√
√ !)
2 2
.
,
8 4
Now,
so that
F (x, y) Fxy (x, y)
D(x, y) = xx
Fyx (x, y) Fyy (x, y)
D(0, 0) = −1 < 0
⇒
2
−1
=
−1 12y 2
= 24y 2 − 1
saddle point at (0,0) .
On the other hand,
√ !
2
2
=D
,−
D −
8
4
√
√
√ !
2 2
= 24
,
8 4
√ !2
2
−1=2>0
4
and
Fxx ≡ 0
implies that F has relative minimum at the points
4. Use Lagrange multipliers to investigate the function
curve
1
1
1
+ 2 = 2
2
x
y
a
√
√ !
2
2
−
and
,−
8
4
f (x, y) =
1 1
+
x y
√
√ !
2 2
.
,
8 4
for extrema over the
where a is a positive constant.
Solution: For g(x, y) =
1
1
1
+ 2 − 2 , ∇f (x, y) = λ∇g(x, y) holds if and only if
2
x
y
a
1 1
1 1
−
,
= −2λ
,
x2 y 2
x3 y 3
yielding
x = y = 2λ .
√
The constraint equation g(x, y) = 0, in turn, gives λ = ∓a/ 2. Finally, the computation
√
√
√ √
√
√
2
2
f ( 2a, 2a) =
and
f (− 2a, − 2a) = −
a
a
√ √
2a, 2a),
shows that, subject to
the
constraint
g(x,
y)
=
0,
f
attains
its
maximum
at
(
√
√
and minimum at (− 2a, − 2a).
5. Find the volume V of the solid bounded by the plane z = 0, and the surfaces x2 + y 2 = a2 and
az = a2 − x2 .
Solution:
a2 −x2
a
√
a2 −x2
a2 −x2
a
√
Z a
3/2
4
a2 − x2
dx
a2 − x2 dz dx =
dy dz dx = 4
V =4
a 0
0
0 0
0 0
2
Z π/2
Z π/2
4 Z π/2
cos 2θ + 1
(∗) 4a
4
3
3
=
cos θ dθ = 4a
dθ = a
cos2 2θ + 2 cos 2θ + 1 dθ
a 0
2
0
0
θ=π/2
Z π/2
3π 3
sin 4θ
3
cos 4θ + 1
3
3
=
+ 2 cos 2θ + 1 dθ = a
+ sin 2θ + θ
a
=a
2
8
2 θ=0
4
0
Z aZ
Z
Z aZ
where in (∗) we used the substitution x = a sin θ.
x y z
x2 y 2 z 2
6. The plane + + = 1 (a, b, c are positive constants) divides the solid ellipsoid 2 + 2 + 2 ≤ 1
a b c
a
b
c
into two unequal pieces. In this problem, you will find the volume V of the smaller piece G as
follows.
(a) (4 pts.) Let G1 be the image of the solid G under the change of variables u = x/a, v = y/b,
and w = z/c; and let V1 be its volume. Find a relation between V and V1 .
Solution: The Jacobian of the change of variables u = x/a, v = y/b, and w = z/c is
given by
∂x ∂x ∂x ∂u ∂v ∂w a 0 0 ∂(x, y, x) ∂y ∂y ∂y = 0 b 0 = abc
=
∂(u, v, w) ∂u ∂v ∂w ∂z ∂z ∂z 0 0 c ∂u ∂v ∂w
so that
ZZZ
ZZZ
ZZZ ∂(x, y, x) dVu,v,w = abc V1 .
V =
dVx,y,z =
dVu,v,w = abc
G1
G
G1 ∂(u, v, w)
(b) (4 pts.) Let G
√2 be the the spherical cap bounded above by the unit sphere, and below by the
plane z = 1/ 3; and let V2 be its volume. Find a relation between V1 and V2 .
Solution: Since n = ha, b, ci := h1, 1, 1i is a normal to the plane M : u + v + w = 1,
the distance of M to the origin (u0 , v0 , w0 ) := (0, 0, 0) is
D=
1
|au0 + bv0 + cw0 − 1|
√
=√ .
a2 + b2 + c2
3
Accordingly, the solid G1 = {(u, v, w) ∈ R3 : u2 + v 2 + w 2 ≤ 1 & u + v + w ≥ 1} is
2
2
2
the (smaller
√ ) spherical cap obtained by cutting the unit sphere u + v + w = 1 via
a plane 1/ 3 units away from the origin. Therefore V1 = V2 .
(c) (4 pts.) Find the volume V2 of the spherical cap G2 .
Solution:
V2 =
Z 2πZ √2/3Z
0
0
√
1−r 2
dw r dr dθ =
√
1/ 3
Z 2πZ √2/3 √
0
r=√2/3
Z 2π 2
1
r
=
− (1 − r 2 )3/2 − √
3
2 3 r=0
0
√
2π
= √ 3 3−4
9 3
1
r dr dθ
1 − r2 − √
3
0
1
1
1
dθ = 2π − √ − √ +
9 3 3 3 3
7. (a) (6 pts.) Using Green’s theorem, find a simple closed curve C with counterclockwise orientation
that maximizes the value of
Z
x3
y3
dy
dx + x −
I=
3
C 3
and explain your reasoning.
Solution: Using Green’s theorem, we have
3 ZZ ZZ
∂
x3
∂
y
I=
x−
−
dA =
1 − x2 − y 2 dA
3
∂y 3
R ∂x
R
where R is the region enclosed by the simple closed curve C. The region that maximizes this double integral is therefore
R = (x, y) ∈ R2 : 1 − x2 − y 2 ≥ 1 = the closed unit disk .
Thus the curve C that maximizes I is the unit circle oriented in the counterclockwise
direction, that is
C : x = cos t, y = sin t, 0 ≤ t ≤ 2π .
(b) (4 pts.) Find the maximum value of I by evaluating the double integral you found in part (a).
Solution: Passing to polar coordinates we get
Imax =
ZZ
2
x2 +y 2 ≤1
1−x −y
2
dA =
Z 2πZ
0
0
1
1−r
2
r2 r4
−
r dr dθ = 2π
2
4
r=1
r=0
=
π
.
2
(c) (2 pts.) The double integral you evaluated in part (b) corresponds to the volume of a solid
above the xy-plane. Describe this solid.
Solution: This is the solid bounded above by the elliptic paraboloid z = 1 − x2 − y 2
and below by the plane z = 0.