Practice Quiz 2

Daniel Seleznev

March 2014

1. Given a transformation Φ

(uv

), the integral over the original space (now

in parameter space) is ∫R2

f ◦ Φ|det[DΦ]||dudv|

which in this case will equal to∫R2

f

(uv

u− uv

)det[DΦ]||dudv|

Here, [DΦ] =

[v u

1− v −u

], and thus |det[DΦ]| = u Hence the answer is

E

2. Let A be the region enclosed by the graph of r2 = sin θ The area of thisregion is then ∫∫

A

dxdy =

∫∫A

rdrdθ

, after making the transformation to polar coordinates. Then,∫∫rdrdθ =

∫ π

0

∫ √sin θ

−√sin θ

rdrdθ =

∫ π

0

sin θdθ = 2

The answer is E

3. If T = [ ~v1| ~v2| ~v3] then vol3P ( ~v1, ~v2, ~v3) =√

det[TTT ].

TTT =

1 1 1 11 −1 −1 11 1 0 0

1 1 11 −1 11 −1 01 1 0

=

4 0 20 4 02 0 2

√

det[TTT ] =√

16 = 4

1

4. Because no part of the ball is priveleged, the density function µ(~x) is con-stant. The integral of the density function over the northern hemispheremust equal one, so µ(~x) = 3

2π Therefore,

E(z) =

∫∫∫zµ

xyz

dxdydz =3

2π

∫∫∫zdxdydz

Converting to spherical coordinates, this integral becomes

3

2π

∫∫∫r3 sinφ cosφdrdθdφ =

3

2π

∫ π/2

0

∫ 2π

0

∫ 1

0

r3 sinφ cosφdrdθdφ =

3

2π

∫ π/2

0

sinφ cosφdφ

∫ 2π

0

dθ

∫ 1

0

r3dr =3

2π· 1

2· 2π · 1

4=

3

8

In cylindrical coordinates, θ ∈ [0, 2π), and r ∈ (0, 1). The equation of aunit sphere is x2 + y2 + z2 = 1. Since we are dealing with a unit half ball,z2 < 1− x2 − y2 ⇒ z <

√1− r2 but is greater than 0. Therefore,

3

2π

∫∫∫zdxdydz =

3

2π

∫∫∫zr dzdrdθ =

3

2π

∫ 2π

0

∫ 1

0

∫ √1−r2

0

zrdzdrdθ =

=3

8

5. Let the parametrization be the mapping γ :

(rθ

)→

r cos θr sin θ

2r

where

[Dγ] =

cos θ −r sin θsin θ r cos θ

2 0

. Then, [Dγ]T [Dγ] =

[5 00 r2

]and the determi-

nant of that matrix is 5r2. At z = 1, x2 + y2 = 14 . Hence, we integrate∫∫

C

r√

5 drdθ =

∫ 2π

0

∫ 12

0

r√

5 drdθ = 2π√

5

∫ 12

0

rdr =

√5

4π

A simple check by geometry shows that this is true: the radius of the baseof the cone, as determined earlier, is 1

2 , while the height is z = 1. The

slant height l of the cone is equal to√

( 12 )2 + 1 =

√52 . The area of the

curved surface of the cone is then πrl =√54 π.

6. (a) ∫ ∞0

e−ax√xdx =

2√a

∫ ∞0

e−u2

du

2

. To make this statement, we used the transformation u =√ax ⇒

du =√a

2√xdx⇒ 2√

adu = dx√

x. Therefore,

2√a

∫ ∞0

e−u2

du =2√a

√π

2=

√π

a

(b) Let Ik be the signed area of the function in (π(k − 1), πk). Wesee on the graph of the function, that ∀k, |Ik| > |Ik+1| and thatlimk→∞ |Ik| = 0, as the non signed areas of the regions decrease onthe graph. Therefore the sequence is convergent by the alternatingseries test.Let fk(x) = sin x√

xwhen x ∈ (π(k − 1), πk) and 0 otherwise. Then

Ik =

∫ πk

π(k−1)| sinx√

x|dx >

∫ πk

π(k−1)| sinxx|dx > 1

πk

∫ πk

π(k−1)| sinx|dx =

2

πk∑∞k=1

2πk is divergent, so therefore, by the comparison test,

∞∑k=1

Ik >

∞∑k=1

2

πk

Since the infinite series of Ik is divergent, our function is not L-integrable.

7. Knowing that the fact that Q∩ [0, 1] is a countably infinite set, enumeratethe rationals in that set (e.g. x1 = 1, x2 = 1

2 , x3 = 13 ...) . Now consider

an infinite sequence {fk(x1)} ∈ [−R,R]. This is a compact set, so bythe Bolzano-Weirtstrass theorem, there exists a convergent subsequence{f1,k(x1)} whose limit, defined to be f(x1), lies in [−R,R].Now consider {f1,k(x2)} ∈ [−R,R] which has a subsequence, {f2,k(x2)},by the Bolzano-Weirstrass theorem, that converges to a specified f(x2).Repeat this procedure for all the rationals in [0, 1]. Then use a ”diagonaltrick” by considering the sequence {fn,n(x)}. For any rational xm, allthe members of this subsequence with m > n converge at xm, because allthese member belong to a subsequence {fn,i(x)} which converge to f(x),by construction. Thus, we have made a sequence that converges to f(x),∀x ∈ Q ∩ [0, 1].To prove that this sequence converges at irrational numbers as well, wewill prove that for irrational x, {fn,n(x)} is a Cauchy sequence. ∀ε > 0,we may find a rational y that is close to an irrational x (|y − x| < δ) ∀n:|fn,n(x) − fn,n(y)| < ε

3 . This is due to the fact that it is given that thissequence is equicontinuous. By the triangle inequality

|fn,n(x)−fm,m(x)| < |fn,n(x)−fn,n(y)|+|fn,n(y)−fm,m(y)|+|fm,m(y)−fm,m(x)|

For the middle term of this expression choose an N large enough that thedifference is less than ε

3 for chosen rational y. Then, choose δ that works

3

∀m,n to make the first and last terms be less than ε3 Then the inequality

is less than ε3 + ε

3 + ε3 = ε, and thus {fn,n} is Cauchy and convergent to

f(x) (as all Cauchy sequences of real numbers converge).Now we prove the continuity of f(x). Because fn,n(x) is equicontinuous,∀ε > 0, |y − x| < δ ⇒ |fn,n(x) − fn,n(x)| < ε

3 , ∀n. Now, by the triangleinequality,

|f(y)− f(x)| < |f(y)− fn,n(y)|+ |fn,n(y)− fn,n(x)|+ |fn,n(x)− f(x)|

The first term in this expression can be made less than ε3 by choosing

n > N1, as it is a convergent sequence. Similarly, the last expression can bemade less than ε

3 by choosing n > N2, as it is a convergent sequence as well.To ensure both of these expressions are less than ε

3 , let n = max{N1, N2}.Then the inequality is less than ε

3 + ε3 + ε

3 = ε. Therefore, ∀ε > 0,|y − x| < δ ⇒ |f(y)− f(x)| < ε.

4