MATH 227 MIDTERM 1
February 8, 2006
1. Prove that if f is a C 1 function and F is a C 1 vector field on R3 , then
∇ · (f F) = f ∇ · F + (∇f ) · F.
If F = (F1 , F2 , F3 ), then
∇ · (f F) =
∂
∂
∂
(f F1 ) +
(f F2 ) + (f F3 )
∂x
∂y
∂z
∂f
∂F1 ∂f
∂F2 ∂f
∂F3
F1 + f
+
F2 + f
+
F3 + f
∂x
∂x
∂y
∂y
∂z
∂z
∂f
∂f
∂f
∂F1 ∂F2 ∂F3 F1 +
F2 +
F3 + f
+
+
=
∂x
∂y
∂z
∂x
∂y
∂z
=
= (∇f ) · F + f ∇ · F.
( x − 1)dA, if D is the region in the first quadrant bounded by the hyperbolas
2. Evaluate the integral
D y
x2 − y 2 = 1, x2 − y 2 = 4, and the lines x = 2y, x = 3y.
RR
Let u = x2 − y 2 , v = x/y, then the region of integration is given by 1 ≤ u ≤ 4, 2 ≤ v ≤ 3. To find the
Jacobian, we first compute
x2
∂(u, v) 2x
−2y =
−2
+ 2 = 2 − 2v 2 ,
=
1/y −x/y 2 ∂(x, y)
y2
so that
∂(x, y)
1
.
=
∂(u, v)
2 − 2v 2
Hence
Z 4Z 3
1 x
( − 1)dA =
(v − 1)
dvdu
2
2 − 2v
D y
1
2
Z 4Z 3
Z 4Z 3
v−1
1
=
dvdu =
dvdu
2
1
2 2(v − 1)
1
2 2(v + 1)
Z 4
Z 4
ln(v + 1) 3
ln 4 − ln 3
=
du
du =
2
2
2
1
1
3
= (ln 4 − ln 3).
2
2
2
We used that |2 − 2v | = 2v − 2 for v ∈ [2, 3].
RRR
3. Write out the integral
xy dV as an iterated integral in cylindrical coordinates, if W is the threeW
p
dimensional solid inside the cylinder x2 + y 2 = 1, bounded from below by the cone z = x2 + y 2 and from
above by the sphere x2 + y 2 + z 2 = 4. (Do not try to evaluate the integral.)
Z Z
In cylindrical coordinates, the cylinder, cone and sphere have equations
r = 1, z = r, r2 + z 2 = 1. The
√
inequalities describing the solid are 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π, r ≤ z ≤ 4 − r2 . The Jacobian for cylindrical
coordinates is r. Also, xy = r2 cos θ sin θ. Hence the integral is
Z
2π
1
Z
0
√
Z
0
4−r2
r3 cos θ sin θ dz dr dθ.
r
4. Find the mass of the solid bounded from below by the cone φ = π/3 and from above by the sphere ρ = 2,
if the density at a point with spherical coordinates (ρ, θ, φ) is ρ. (Evaluate the integral.)
The solid (shaped like an ice-cream cone) is described in spherical coordinates by the inequalities 0 ≤ ρ ≤ 2,
0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π/3. The Jacobian is ρ2 sin φ. Hence the mass is
Z
2
π/3
Z
Z
2π
Z
2
2
Z
ρ · ρ sin φ dθ dφ dρ =
0
0
0
2
Z
=
0
π/3
2πρ3 (− cos φ ) dρ =
Z
0
0
Z
=
2
πρ3 dρ =
0
0
2
π/3
2πρ3 sin φ dφ dρ
0
1
2πρ3 (− + 1) dρ
2
4 2
ρ 16
= 4.
=
4 0
4
√
5. Let x(t) = ( 2t3/2 , (t + 2)3/2 , (t − 2)3/2 ) for t ∈ (2, ∞).
(a) Find the unit tangent and principal normal vectors at t = 2.
(b) Find the curvature κ(t) (for all t > 2), and prove that it has limit 0 as t → ∞.
3√2
3
3
t1/2 , (t + 2)1/2 , (t − 2)1/2 ,
2
2
2
1/2
√
3
3√
kx0 (t)k =
2t + (t + 2) + (t − 2)
=
4t = 3 t,
2
2
r
1 √ r
√
2
2
x0 (t)
3 √ √
√
T(t) = 0
=
2t, t + 2, t − 2 =
2, 1 + , 1 −
,
kx (t)k
2
t
t
2·3 t
√ √
1 √ √ 2 2 T(2) =
2, 2, 0 =
,
,0 .
2
2 2
Strictly speaking, this is a “one-sided” unit tangent because the path is only defined for t ∈ (2, ∞). Next,
r
r
1
−2
2
1
1
1 t
t 0
T (t) =
0, q
· 2 , q
· 2 = 2 0,
,
.
2
t
t
2t
t
+
2
t
−
2
2
2
2 1+ t
2 1− t
0
x (t) =
0
T (2)
The last component goes to infinity as t → 2, so N(2) cannot be computed as kT
0 (2)k . This is quite enough
for the purpose
of
the
midterm.
But
actually
there
is
a
way
to
circumvent
the
problem,
by taking the limit
T0 (t)
of N(t) = kT0 (t)k as t → 2:
kT0 (t)k =
1 t
t 1/2
1 t2 − 2t + t2 + 2t 1/2
+
=
2t2 t + 2 t − 2
2t2
t2 − 4
1 2t2 1/2
1
p
,
=
2t2 t2 − 4
t 2(t2 − 4)
p
r
r
t 2(t2 − 4) T0 (t)
t
t N(t) =
=
·
0,
,
kT0 (t)k
2t2
t+2
t−2
r
r
r
(t + 2)(t − 2) t
t =
·
0,
,
2t2
t+2
t−2
r
r
t−2
t + 2
,
,
= 0,
2t
2t
lim+ N(t) = (0, 0, 1).
=
t→2
With the work done so far, the easiest way to compute the curvature is
κ(t) =
As t → ∞, t3/2 and
1
kT0 (t)k
1
1
p
p
√
·
.
=
=
kx0 (t)k
t 2(t2 − 4) 3 t
3t3/2 2(t2 − 4)
p
2(t2 − 4) both go to infinity, hence κ(t) → 0.