Advanced Calculus Final Exam June 27, 2011
There are 7 questions with total 124 points in this exam.
1. Let A ⊂ Rn.
(a) (6 points) Define what it means to say that A has (n-dimensional) content c(A) zero.
Solution: A set Z ⊂ Rn has n−content zero if ∀² > 0, ∃ a finite set C = {Kj}mj=1 of n−cells such that
(a) Z ⊂ ∪mj=1Kj, (b)
Xm j=1
c(Kj) < ².
(b) (6 points) Define what it means to say that A has (n-dimensional) content.
Solution: A set A ⊂ Rn is said to have content (or it is said to be (Jordan) measurable) if it is bounded and its boundary ∂A has content zero.
2. (a) (6 points) Let A = (0, 1) × (0, 1). Show that A has (2-dimensional) content and find its content.
Solution: (1) For each (x, y) ∈ A, since k(x, y)k ≤√
2, A is bounded.
(2) For any 4 > ² > 0, let K1 = [0, ²
4] × [0, 1], K2 = [1 − ²
4, 1] × [0, 1], K3 = [0, 1] × [0,² 4], and K4 = [0, 1] × [1 − ²
4, 1].
Since ∂A =¡
{0, 1} × [0, 1]¢
∪¡
[0, 1] × {0, 1}¢
⊂ ∪4i=1Ki, and X4
i=1
c(Ki) = ², ∂A has content zero.
By (1) and (2), we conclude that A has content and c(A) = c( ¯A) = 1.
(b) (6 points) Let Z = {(x, y) | |x| + |y| = 1}. Show that Z has (2-dimensional) content zero.
Solution: For each ² > 0, choose n ∈ N such that 1 n < ²
4. For each i = 1, . . . , n, let Ki = [i − 1
n , i
n] × [1 − i
n, 1 − i − 1
n ]. Then Z ∩ {(x, y) | x ≥ 0, y ≥ 0} ⊂ ∪ni=1Ki and Xn
i=1
c(Ki) = 1 n < ²
4. Similarly, we can cover the remaining part of Z by using cells of the same size. Hence, Z has content zero.
3. Let K be a closed cell in Rn, f be a bounded function defined on K to R, and P = {Kj}mj=1 be a partition of K.
(a) (6 points) Define the upper sum UP(f, K), the lower sum LP(f, K), and a Riemann sum SP(f, K) of f corresponding to the partition P over K.
Solution: Since f is bounded on K, supKjf and infKjf exist for each j = 1, . . . , m.
The upper sum of f corresponding to the partition P over K is defined by UP(f, K) = Xm
j=1
¡sup
Kj
f¢ c(Kj);
the lower sum of f corresponding to the partition P over K is defined by LP(f, K) = Xm
j=1
¡inf
Kj
f¢
c(Kj); and
Advanced Calculus Final Exam (Continued) June 27, 2011
SP(f, K) = Xm
j=1
f (xj)c(Kj) for any xj ∈ Kj is called a Riemann sum of f corresponding to the partition P over K.
(b) (4 points) Let Q = {Ii}li=1 be another partition of K. Define what it means to say that Q is a refinement of P.
Solution: We say that Q is a refinement of P if for each Ii ∈ Q, there exists a Kj ∈ P such that Ii ⊆ Kj.
(c) (8 points) Let P, Q be partitions of K such that P ⊂ Q i.e. Q is finer than P. Show that LP(f, K) ≤ LQ(f, K) ≤ SQ(f, K) ≤ UQ(f, K) ≤ UP(f, K).
Solution: For each Ii ∈ Q, since P ⊂ Q, there exists a Kj ∈ P such that Ii ⊆ Kj. Since inf
Kj
f ≤ inf
Ii
f ≤ f (xi) ≤ sup
Ii
f ≤ sup
Kj
f,
and c(Kj) = c¡ Kj∩
[l i=1
Ii¢
= X
Ii∈Q and Ii⊆Kj
c(Ii),
we have LP(f, K) ≤ LQ(f, K) ≤ SQ(f, K) ≤ UQ(f, K) ≤ UP(f, K).
4. (a) (6 points) Let f (x) = (
1 if x ∈ [−1, 0), 2 if x ∈ [0, 1].
Show that f is integrable on [−1, 1] and find R1
−1f (x)dx.
Solution: For each ² > 0, let P = {−1 = x0 < x1 < · · · < xn = 1} be a partition of [−1, 1] such that kP k = max
1≤i≤n|xi− xi−1| < ². Since |UP(f, [−1, 1]) − LP(f, [−1, 1])| < 2², f is integrable on [−1, 1] by Riemann Criterion.
Z 1
−1
f (x)dx = lim
kP k→0UP(f, [−1, 1]) = 1 + 2 = 3.
(b) (6 points) For a < b, let f (x) =
(a if x ∈ Q ∩ [0, 1],
b if x ∈ [0, 1] \ Q.
Show that f is not integrable on [0, 1].
Solution: Let ²0 = b − a
2 > 0 and P be any partition of [0, 1], since |UP(f, [0, 1]) − LP(f, [0, 1])| = b − a > ²0, f is not integrable on [0, 1].
(c) (6 points) Let A be a bounded subset of Rn, and f be a bounded function defined on A to R. Define what it means to say that f is integrable on A.
Solution: Let K be a closed cell in Rn containing A and fK be an extension of f on K defined by fK(x) =
(
f (x) if x ∈ A, 0 if x ∈ K \ A.
We say that f is integrable on A if fK is integrable on K, and define R
Af =R
KfK.
5. (a) (6 points) Let ψ : R2 → R2 be defined by (x, y) = ψ(r, θ) = (r cos θ, r sin θ). Find dψ(r, θ) and det dψ(r, θ).
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Advanced Calculus Final Exam (Continued) June 27, 2011
Solution: det dψ =
¯¯
¯¯cos θ −r sin θ sin θ r cos θ
¯¯
¯¯ = r.
(b) (6 points) Let ψ : R3 → R3 be defined by (x, y, z) = ψ(ρ, φ, θ)
= (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ). Find dψ(ρ, φ, θ) and det dψ(ρ, φ, θ).
Solution: det dψ =
¯¯
¯¯
¯¯
sin φ cos θ ρ cos φ cos θ −ρ sin φ sin θ sin φ sin θ ρ cos φ sin θ ρ sin φ cos θ
cos φ −ρ sin φ 0
¯¯
¯¯
¯¯
= cos φ
¯¯
¯¯ρ cos φ cos θ −ρ sin φ sin θ ρ cos φ sin θ ρ sin φ cos θ
¯¯
¯¯ + ρ sin φ
¯¯
¯¯sin φ cos θ −ρ sin φ sin θ sin φ sin θ ρ sin φ cos θ
¯¯
¯¯
= ρ2sin φ cos2φ(cos2θ + sin2θ) + ρ2sin3φ(cos2θ + sin2θ) = ρ2sin φ.
(c) (6 points) Show that R∞
0 e−x2dx =
√π
2 . Solution: ³R∞
0 e−x2dx
´2
=R∞
0
R∞
0 e−x2−y2dxdy =Rπ/2
0
R∞
0 re−r2drdθ = π 4.
(d) (6 points) Let B = {(x, y, z) ∈ R3 | x2+ y2+ z2 = ρ2} be the ball of radius ρ with the center at origin.
Show that the volume of B is Z Z Z
B
dxdydz = 4πρ3 3 . Solution:
Z Z Z
B
dxdydz = Z 2π
0
Z π
0
Z ρ
0
r2sin φ drdφdθ
=¡ θ|2π0 ¢ ¡
− cos φ|π0¢ ¡r3 3|ρ0¢
= 4πρ3 3 .
6. (a) (6 points) State the Bounded Convergence Theorem with which one can conclude that
n→∞lim Z
K
fn = Z
K
n→∞lim fn.
Solution: Let {fn} be a sequence of integrable functions on a closed cell K ⊂ Rp. Suppose that there exists B > 0 such that kfn(x)k ≤ B for each n ∈ N and for all x ∈ K. If the function f (x) = lim fn(x), x ∈ K, exists and is integrable, thenR
Kf = limR
Kfn.
(b) (6 points) For 0 < a < 2, show that fn(x) = e−nx2 converges uniformly on [a, 2], and show that lim
n→∞
Z 2
a
fn(x) dx = 0.
Solution: For each x ∈ [a, 2], since lim
n→∞fn(x) = 0, and lim
n→∞ sup
x∈[a,2]
|fn(x)−0| ≤ lim
n→∞e−na2 = 0, fn converges uniformly on [a, 2]. Since each fn is continuous on [a, 2], fn is integrable there. Thus, {fn} is a sequence of integrable function that converges uniformly on [a, 2]
and we have lim
n→∞
Z 2
a
fn(x) dx = Z 2
a
n→∞lim fn(x) dx = 0.
(c) (6 points) Show that fn(x) = e−nx2 does not converge uniformly on [0, 2], and use a suitable convergence theorem to show that lim
n→∞
Z 2
0
fn(x) dx = 0.
Solution: The limiting function f is given by f (x) = lim
n→∞fn(x) =
(1 if x = 0 0 if x ∈ (0, 2].
Since each fn is continuous on [0, 2] while f is not continuous at x = 0, the convergence of
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Advanced Calculus Final Exam (Continued) June 27, 2011 fn to f is not uniform on [0, 2].
Note that |fn(x)| = e−nx2 ≤ 1 for each n ∈ N and for all x ∈ [0, 2], and that f is integrable on [0, 2], we can use the Bounded Convergence Theorem to conclude that
n→∞lim Z 2
a
fn(x) dx = Z 2
a
n→∞lim fn(x) dx = Z 2
0
f (x) dx = 0.
7. (a) (8 points) Use the Dominated Convergence Theorem to show that the integral F (t, u) =Z ∞
0
e−txsin uxdx converges uniformly for all t ≥ γ > 0 and for all u ∈ R to u u2+ t2. Solution: Since |e−txsin ux| ≤ e−tx for all x ≥ 0, t ≥ γ > 0, u ∈ R and
Z ∞
0
e−txdx exists for all t ≥ γ > 0, we can use the Dominated Convergence Theorem to conclude that the integral F (t, u) =
Z ∞
0
e−txsin uxdx converges uniformly for all t ≥ γ > 0 and for all u ∈ R.
By using the integration by parts, we have F (t, u) = Z ∞
0
e−txsin uxdx = e−tx
−t sin ux|∞0 − u
−t Z ∞
0
e−txcos uxdx = ue−tx
−t2 cos ux|∞0 −−u2
−t2 Z ∞
0
e−txsin uxdx = u t2−u2
t2F (t, u) = u u2+ t2. (b) (6 points) Use the Dirichlet’s test to show that the integral G(t, u) =
Z ∞
0
e−txsin ux x dx converges uniformly for all t ≥ γ > 0 and for all u ∈ R.
Solution: Since | Z c
0
sin ux dx| ≤ 2 for all c ≥ 0, u ∈ R, and the function e−tx
x is monotone decreasing for x > 0, and lim
x→∞
e−tx
x = 0, we can use the Dirichlet’s test to conclude that G(t, u) =
Z ∞
0
e−txsin ux
x dx converges uniformly for all t ≥ γ > 0 and for all u ∈ R.
(c) (8 points) Let G(t) = R∞
0 e−x2−t2/x2dx for t > 0. Show that G0(t) = −2G(t) and G(t) =
√π
2 e−2t.
Solution: G0(t) = Z ∞
0
−2t
x2 e−x2−t2/x2dx By setting z = t
x, we have dz = − t
x2dx and G0(t) = 2
Z 0
∞
e−t2/z2−z2dz = −2G(t)
By separating the variables in the differential equation and integrating from 0 to t, we get log G(t)
G(0) = −2t ⇒ G(t) = G(0) e−2t=¡Z ∞
0
e−x2dx¢
e−2t=
√π 2 e−2t.
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