Utrecht University
Mathematics
Infinitesimaalrekening C (WISB234)
Block 3, 2012
Test, May 30, 2012
JUSTIFY YOUR ANSWERS
Allowed: material handed out in class and handwritten notes (your handwriting )
Problem 1. (10 pts.) A wire has the form of the parabola y = x2 , with −1 ≤ x ≤ 1. The density at
each point (x, y) of the wire is equal to |x|. Calculate the total mass of the wire (remember that the mass
is the path integral of the density).
√
→
→0
Answer: The most natural parametrization is c (x) = (x, x2 ), x ∈ R. Hence k c (x)k = 1 + 4x2 and
Z 1 p
Z 1
p
3/2 1
1
1 √
x 1 + 4x2 dx =
1 + 4x2
5 5−1 .
Mass =
|x| 1 + 4x2 dx = 2
=
6
6
0
0
−1
→
Problem 2. (10 pts.) Prove that the field F = (−z sin x cos y , −z cos x sin y , cos x cosy) is conservative
→
→
and find f such that F =∇f .
→
→ →
→
→
Answer: It is easy to verify that ∇ × F = 0 . Integrating or guessing one obtains F =∇ f with f =
z cos x sin y.
Problem 3. Consider the ellipse
x2 y 2
+ 2 = 1.
a2
b
(a) (5 pts.) Find a parametrization of the curve obtained when traversing the ellipse once in a counterclockwise direction, starting from the point (a, 0).
√
√
(b) (5 pts.)
Show
that
the
line
tangent
to
the
ellipse
at
(a/
2,
−b/
2) intersects the x-axis at x =
√
2a/ 2.
→
(c) (10 pts.) Find the line integral of F (x, y) = (−y, x) along the curve.
(d) (10 pts.) Use Green’s theorem to compute the area inside the ellipse.
Answers:
(a) ~c(θ) = a cos θ b sin θ , θ ∈ [0, 2π].
√
√ →
→0
(b) First, observe that the condition a/ 2, −b/ 2 = c (θ) implies θ = 2π − (π/4). Second, c (θ) =
√
√ →0
−a sin θ , b cos θ , hence the line is in the direction of c (7π/4) = a/ 2 , b/ 2 . The parametric
equation of the tangent is, therefore,
→ `(t)
=
a
a(1 + t) b(t − 1) a
b b √ , −√ + t √ , √
√
=
, √
.
2
2
2
2
2
2
√
The y-coordinate is zero if and only if t = 1, in which case the x-coordinate takes the value 2a/ 2.
(c)
Z
2π
−b sin θ , a cos θ · −a sin θ , b cos θ dθ = ab
0
Z
2π cos2 θ + sin2 θ dθ = 2πab .
0
(d)
1
Area =
2
Z
1
xdy − ydx =
2
~c
Z
2π
−b sin θ , a cos θ · −a sin θ , b cos θ dθ = πab .
0
1
Problem 4. Consider the upper cone
x2 + y 2 − z 2 = 0
z≥0,
the field
→
F (x, y, z) = (−x, −y, 2z)
and the sphere
√
x2 + y 2 + z 2 = 2 2 z .
(a) (5 pts.) Draw the sphere and the cone on the same set of coordinate axis.
(b) (5 pts.) Determine a parametrization of the cone.
(c) (5 pts.) For which point does not the cone have a tangent plane? Justify your assertion.
→
(d) (10 pts.) Determine the flux of the field F towards the outside of the cone, for 0 ≤ z ≤ 1.
→
(e) (10 pts.) Determine the outward flux of F out of the closed surface formed by the portion of the
cone inside the sphere capped by the portion of the sphere inside the cone.
Answers:
→
(a) Φ(ρ, θ) = ρ cos θ , ρ sin θ , ρ , θ ∈ [0, 2π] and ρ ≥ 0.
(b) The outward normal is
→
→
∂θ
∂θ
×
=
∂θ
∂ρ
ρ cos θ , ρ sin θ , −ρ .
The only point without tangent plane is the only point where the normal becomes the zero vector,
namely ρ = 0 or (x, y, z) = (0, 0, 0).
(c)
Z
→
→
Fd S
2π
Z
=
S
Z
dθ
0
1
dρ −ρ cos θ , −ρ sin θ , 2ρ · ρ cos θ , ρ sin θ , −ρ
0
Z
1
= 2π
dρ (−3ρ2 )
0
= −2π .
→ →
(d) ∇ · F = 0, hence by Gauss the outward flux is zero.
→
→
Problem 5. Let v = (v1 , v2 , v3 ) be a constant vector and denote r = (x, y, z).
→
→
→
(a) (5 pts.) Compute ∇ ×( v × r ).
(b) (10 pts.) Show that for every simple surface S,
Z
Z Z
→
1
→
→
→
→
v ·d S .
(v × r ) · d s =
2 ∂S
S
Answers:
→
→
→
→
(a) ∇ ×( v × r ) = 2 v .
(b) Use Stokes.
2