math 2020 b solution

8/19/2019 MATH 2020 B Solution

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MATH2020B Advanced Calculus II

Tutorial Solution

by Chan Chi Ming

This note contains all the solutions to the tutorial questions given by an excellent tutor,John, LAM Yi-Chun, in his MATH2020B class in 2013-2014. When I was a tutor in theMath Clinic (http://mathclinic.math.cuhk.edu.hk/), the questions are found useful forthe students to familiar with the definitions and theorems. Therefore, I decide to write allthe solutions for the coming MATH2020 students. Since definitions and theorems are notprovided in this note, students are advised to use this note, as a list of worked examples,after learning them. Although this note covers most of the topic in the syllabus, it shouldNOT be used to substitute any lectures and tutorials.

The solutions are modified from the material given by John after discussing with mystudents, with emphasis placed on the use of definition and theorems but not tediouscomputation. Although many solutions are short, the students are highly advised towork out ALL the steps (smartly). Unlike the notation in the questions, a, b is usedto denote the standard inner product of the two vectors a and b in Rn instead of a · b.Sometimes a parametrization of a curve is written in γ while a parametrization of a surfaceis written in Φ instead of the common notation r to avoid confusion.

Finally, I would like to thank John, LAM Yi-Chun for providing the source so that thisnote can be completed. Also, I am grateful to Kelvin, CHAN Cheuk-Kit for verifyingevery step in the solution and many helpful discussions. If there are any mistakes, pleasenotify me by email: . Any help would be appreciated.

Change Log:Version: Date: Comment:

1 2013-09-03 First release.

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MATH2020B Tutorial 1

By John, Lam Yi Chun

(1) Review:

(a) Give a parametrization for the following surfaces:

Sphere, cylinder, cone, ellipsoid and paraboloid.

(b) Parametrize the surfaces given by the equations

(i) x2

a2 + y

2

b2 − z

2

c2 = 1 (Hyperboloid of one sheet)

(ii) x2

a2 − y

2

b2 − z

2

c2 = 1 (Hyperboloid of two sheets)

(c) Determine the surface given by the parametrization

P (u, v) = (u + v, u − v,uv), 0 < u, v


3/45


Tutorial 1 Solution

by Chan Chi Ming

September 3, 2013

*Please email to me if there are mistakes.

1. Google yourself.

2. The region ∆ can be parameterized to be0 ≤ y ≤ x, 0 ≤ x ≤ 1

0 ≤ y ≤ 2 − x, 1 ≤ x ≤ 2.

Therefore,

∆

xeydA = 1

0 x

0

xeydydx+ 2

1 2−x

0

xeydydx = 1

0

x(ex−1)dx+ 2

1

x(e2−x−1)dx = 2e−4.

3. 10

10

xmax{x, y}dydx =

10

x0

xmax{x, y}dy +

1x

xmax{x, y}dy

dx

=

10

x0

x2dy +

1x

xydy

dx

=

10

x3dx + 1

2

10

x(1 − x2)dx

= 3

8.

4. (a) Interchanging the order of integration,

60

2x/3

ey2

dydx =

20

3y0

ey2

dxdy = 3

20

yey2

dy = 3

2e4 −

3

2.

(b) Interchanging the order of integration,

40

√ yy/2

eyxdxdy = −

40

y/2√ y

eyxdydx = −

20

2xx2

eyxdydx = −

20

(xe2−xex)dx = 1−e2.

5. (a) By a change of variable u = x + y, 10

10

x − y

(x + y)3dydx =

10

1+xx

2x − u

u3 dudx =

10

1+xx

2x

u3 −

1

u2

dudx =

1

2.

(b) By a change of variable v = x + y,

10

10

x − y

(x + y)3dxdy =

10

1+yy

v − 2y

v3 dvdy =

10

1+yy

1

v2 −

2y

v3

dvdy = −

1

2.

1


4/45

6. In polar coordinates, the region can be parameterized to be

0 ≤ r ≤ sec(θ), 0 ≤ θ ≤ π

4

0 ≤ r ≤ secπ

2 − θ

,

π

4 ≤ θ ≤

π

2.

Therefore,

(x2 + y2)dA =

π4

0

sec(θ)0

r2 · rdrdθ +

π2

π2

secπ2−θ

0r2 · rdrdθ

= 1

4

π4

0sec4(θ)dθ +

1

4

π2

π2

sec4π

2 − θ

dθ

= 1

2

π4

0sec4(θ)dθ

= 2

3

.

2


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Review on Section 13.1 - 13.5

(1) Evaluate

R

f , where f (x, y) = xy and R is the first-quadrant circle bounded by

x2 + y2 = 1 and the coordinate axes.

(2) Evaluate

1

0

1

y

1

1 + x4dxdy.

(3) Using double integration, verify the formula for the volume of a sphere of radius r.

(4) Find the volume of the first octant part of the solid bounded by the cylindersx2 + y2 = 1 and y2 + z 2 = 1.

(5) Evaluate

2

1

√ 2x−x2

0

dydx x2 + y2

.

(6) Find the volume of the solid bounded by x2 + y2 + z 2 = a2 and z =

x2 + y2.

(7) Find the volume and surface area of the solid torus obtained by revolving the disk

r ≤ a around the line x = b > a.

(8) Find the mass and centroid of the following region: The region inside r = 2sin θ

and outside r = 1 with density ρ(x, y) = y.

(9) A uniform rectangular plate with base length a, height b, and mass m is centered

at the origin. Show that is polar moment of inertia is I 0 = 1

12m(a2 + b2).

(10) (Extra) Using polar coordinates, evaluate

f dA, where f (x, y) = x2 + y2 and

= {(x, y) ∈ R2 : 0 ≤ x, y ≤ 1}.

(11) (Extra) Let α > 0 and 0 < < 12

. Evaluate S

gdA, where

g(x, y) = (x2 + y2)−α

log

1 x2 + y2

2

and

S =

(x, y) ∈ R2 : ≤

x2 + y2 ≤

1

2

.

Determine for what values of α does lim→0 S

gdA exists.

1


6/45


Tutorial 2 Solution

by Chan Chi Ming

September 3, 2013


1. In polar coordinates, the region R can be parameterized to be

0 ≤ r ≤ 1, 0 ≤ θ ≤ π2

.

Therefore, R

xydA =

π2

0

10

r cos(θ)r sin(θ) · rdrdθ = π

2

0cos(θ)sin(θ)

10

r3drdθ = 1

8.

2. Interchanging the order of integration,

10

y1

dxdy

1 + x4 = 10

x0

dydx

1 + x4 = 10

x

1 + x4 dx = 1

2 10

d(x2)

1 + (x2)2 = π

8 .

3. Consider the sphere given by {(x,y ,z) ∈ R3 : x2 + y2 + z2 = ρ2}, where ρ > 0. LetD ≡ {(x, y) ∈ R2 : x2 + y2 = ρ2}. Then the volume of the sphere is given by, with the useof polar coordinate,

V =

D

ρ2 − x2 − y2dA =

2π0

r0

ρ2 − r2 · rdrdθ = 2π · 2

3ρ3 =

4

3πρ3.

4. Choose the base to be R ≡ {(x, y) ∈ R2 : x2 + y2 = 1, 0 ≤ x, y ≤ 1} together with theheight function (z =)f (x, y) = 1 −

y2 on R. The volume is then given by

V =

D

f (x, y)dA =

10

√ 1−y20

1 − y2dxdy =

10

(1 − y2)dy = 23

.

5. In polar coordinates, the region can be parameterized to be

sec θ ≤ r ≤ 2cos(θ), 0 ≤ θ ≤ π4

.

Therefore, 21

√ 2x−x20

dydx

x2 + y2=

π4

0

2cos(θ)sec(θ)

rdrdθ

r =

π4

0 2cos(θ)−sec(θ)

dθ =

√ 2−log(

√ 2+1).

6. For the choice of the base, we solve the equationsx2 + y2 + z2 = a2

z =

x2 + y2.

Therefore, we choose the base to be R ≡ {(x, y) ∈ R2 : x2 + y2 = a2/2} together withthe height function (z =)f (x, y) =

a2 − x2 − y2 −

x2 + y2 on R. With the help of the

polar coordinate, the volume is then given by

V = D f (x, y)dA = 2π

0 a√ 2

0( a

2 − r2 − r) · rdrdθ = (2 −√

2)

3 πa3.

1


7/45

7. Observe that the differential volume is given by dV = 2π(b − x)dA. With the help of thepolar coordinate, the volume is then given by

V =

2π0

a0

2π

b − r cos(θ) · rdrdθ = 2π0

πa2b − 2

3πr3 cos(θ)

· rdrdθ = 2π2a2b.By Pappus’s second (centroid) theorem, the surface area is given by

A = 2πa · 2πb = 4πab.

8. For the range of the region, we solve the equationsr = 2sin(θ)

r = 1.

Therefore, the region is R ≡

r cos(θ), r sin(θ)

: 1 ≤ r ≤ 2sin(θ), π/6 ≤ θ ≤ 5π/6

. The

mass is given by

M =

RδdA =

5π6π6

2 sin(θ)1

r sin(θ) · rdrdθ = 8π + 3

√ 3

12 .

Also,

M x =

R

y · δdA = 5π

6

π6

2sin(θ)1

r3 sin2(θ)drdθ = 12π + 11

√ 3

16 ,

M y =

R

x · δdA = 5π

6

π6

2 sin(θ)1

r3 sin2(θ) cos(θ)drdθ = 0.

The centroid is then given by 0, 36π + 33

√ 3

32π + 12

√ 3.

9. Since the rectangular plate is uniform, the density is uniformly given by δ = m/(ab).Hence,

I y =

a2

−a2

b2

− b2x2 · δdydx = δ

a2

−a2bx2dx =

1

12δa3b =

1

12ma2.

Similarly, I x = 1

12mb2.

10. See Tutorial 1, Q.6.

11. In polar coordinates, the region S ε can be parameterized to be

ε ≤ r ≤ 12

, 0 ≤ θ ≤ 2π.

Therefore, using the polar coordinates,

S ε

gdA =

2π0

12

ε

r−2α

log

1

r

2· rdrdθ

= 2π

12

ε

r1−2α log2(r)dr

= −π

4(α − 1)3 r2−2α 2(α −

1) log(r) + 12 + 112

ε

.

2


8/45

Therefore, S ε

gdA = π

4(α − 1)3

ε2−2α

2(α−1) log(ε)+12+1−22α−21−2(α−1) log(2)2+1.For the existence of the limit lim

ε→

0 S ε

gdA, it suffices to consider the limit

limε→0

ε2−2α

2(α − 1) log(ε) + 12 + 1.The limit exists if 0 < α


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By John, Lam Yi Chun

(1) (Continue)Double Integrals

(a) Using polar coordinates, evaluate

f dA, where f (x, y) = x2 + y2 and =

{(x, y) ∈ R2 : 0 ≤ x, y ≤ 1}.

(b) Evaluate

D

dxdy

(x2 + y2)2, where D = {(x, y) ∈ R2 : |x| + |y| ≥ 1}.

(c) Show that ∞

0

e−x2

dx =

√ π

2 .

(d) Evaluate

∞

0

tan−1(πx) − tan−1 xx

dx.

(Remark: This is an example of Frullani integral.)

(2) Triple Integrals and Volumes

(a) Find the volume of the solid bounded by the plane z = y +4 and the paraboloid

z = x2 + y2 + y.

(b) Find the volume of the solid bounded by the spheres x2

+ y2

+ z 2

= 4 and(x − 1)2 + y2 + z 2 = 4.

(c) Let a > 0. Find the volume of solid bounded by the sphere x2 + y2 + z 2 = a2

and excluding the solid bounded by the surface x2 + y2 = a|x|.(d) (Extra Question):

Let V n(R) be the volume of the n dimensional ball of radius R (i.e. the set

{(x1,...,xn) ∈ Rn : x21 + ... + x2n ≤ R2}). Show that V 2n(R) = πn

n! R2n and

V 2n−1(R) = 4nπn−1n!

(2n)! R2n−1.

(3) Change of Variable Formula

(a) Evaluate B

e−ρ3

dV , where B is the ball centered at the origin with radius R

and ρ =

x2 + y2 + z 2.

(b) Evaluate R

x2 + y2 + z 2dV , where R is the region bounded by the plane

z = 3 and the cone z =

x2 + y2.

(c) Evaluate R

(√

x +√

y)1

2 dA, where R is the region in the first quadrant which

is bounded by the coordinate axes and the curve √ x + √ y = 1.

1


10/45


Tutorial 3 Solution

by Chan Chi Ming

September 3, 2013


1. (a) See Tutorial 1, Q.6.

(b) By the symmetry of the region and the integrant, it suffices to consider the regionD ≡ D ∩ {(x, y) ∈ R2, x , y ≥ 0}. In polar coordinates, this new region D can beparameterized to be

1

cos(θ) + sin(θ) ≤ r < ∞, 0 ≤ θ ≤ π

2.

Therefore, using the polar coordinates,

D

dxdy

(x2 + y2)2 = 4

D

dxdy

(x2 + y2)2 = 4 π20

∞1

cos(θ)+sin(θ)

rdrdθ

r4 = π + 2.

(c) Let I ≡ ∞0

e−x2

dx > 0. Then by the polar coordinates,

I 2 =

∞0

∞0

e−x2−y2dxdy =

π2

0

∞0

e−r2 · rdrdθ = π

4.

So I =√

π/2.

(d) Instead of evaluating the integral, we show the following result:

∞0

f (bx) − f (ax)x

dx =

limx→∞ f (x) − limx→0 f (x)

log

ba

,

where 0 < a ≤ b and f : (0, ∞) → R is a continuously differentiable function.

To see this, ∞0

f (bx) − f (ax)x

dx =

∞0

ba

f (yx)dydx

=

ba

∞0

f (yx)dxdy

= b

a


1y

dy

=


log

b

a

.

In particular,

∞0

arctan(πx) − arctan(x)x

dx = 1

2π log(π).

2. (a) For the choice of the base, we solve the equations

z = x2 + y2 + y

z = y + 4.

1


11/45

Therefore, we choose the base to be R ≡ {(x, y) ∈ R2 : x2 + y2 = 4}. the volume isthen given by

V =

R

y+4x2+y2+y

dzdxdy =

2π0

20

(4 − r2)rdrdθ = 8π.

(b) For the choice of the base, we solve the equations x2 + y2 + z2 = 4

(x − 1)2 + y2 + z2 = 4.

Therefore, we choose the base to be R ≡ {(y, z) ∈ R3 : y2 + z2 = 15/4}. The volumeis then given by

V =

R

√ 4−y2−z21−

√ 4−y2−z2

dxdydz =

2π0

√ 152

0(2

4 − r2 − 1)rdrdθ = 274

π.

(c) For the choice of the base, note that the base is always the projection of the solidonto some planes, we solve the equations (here we choose the plane z = 0)x2 + y2 = a2

x2 + y2 = a|x|.Therefore, we choose the base to be

R ≡


: a| cos(θ)| ≤ r ≤ a, 0 ≤ θ ≤ 2π

.

By the symmetry of the region and the integrant, the volume is then given by

V =

R

√ a2−r2−√ a2−r2

rdzdrdθ = 4 π

2

0

a

a| cos(θ)|2r

a2 − r2drdθ = 83

a3 π

2

0sin3(θ)dθ = 16

9 a3.

(d) Observe that the formula for V j(R) is given by

V j(R) =

R−R

√ R2−x21

−√

R2−x21

· · · √ R2−x2

1−...−x2

n−1

−√

R2−x21−...−x2

n−1

dxn . . . dx2dx1,

then the result follows from an induction on j ∈ N.3. (a) Using spherical coordinates, that is,

x = ρ cos(θ)sin(φ), y = ρ sin(θ) sin(φ), z = ρ cos(φ),

where ρ > 0, 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ π, we have

∂ (x,y ,z)

∂ (ρ,θ,φ) =

cos(θ) sin(φ) − sin(θ)sin(φ) cos(θ)cos(φ)sin(θ)sin(φ) cos(θ)cos(φ) sin(θ) cos(φ)

cos(φ) 0 sin(φ)

= ρ2 sin(φ).Therefore,

B e−ρ3dV =

π

0 2π

0 R

0e−ρ

3 · ρ2 sin(φ)dρdθdφ = 43

π(1 − e−R3).

2


12/45

(b) Using cylindrical coordinates, that is,

x = r cos(θ), y = r sin(θ), z = z,

where r > 0 and 0 ≤ θ ≤ 2π, we have

∂ (x,y ,z)∂ (r,θ,z)

=cos(θ)

−r sin(θ) 0

sin(θ) r cos(θ) 00 0 1

= r.The region R can be parameterized to be

0 ≤ r ≤ z, 0 ≤ z ≤ 3, 0 ≤ θ ≤ 2π.

Therefore, R

x2 + y2 + z2dV =

30

2π0

z0

r2 + z2 · rdrdθdz = 27

3 (2

√ 2 − 1)π.

(c) From the change of variableu =

√ x, v =

√ y,

we have∂ (x, y)

∂ (u, v) =

2u 00 2v = 4uv

and the region R can be parameterized to be

0 ≤ u ≤ 1 − v, 0 ≤ v ≤ 1.

Therefore, R

(√

x +√

y)12 dA =

1

0

1−v

0(u + v)

12 · 4uvdudv

= 4

10

1−v0

v(u + v)

32 − v2(u + v)12 dudv

= 4

10

2

5v(1 − v 52 ) − 2

3v2(1 − v 32 )

dv

= 4

27.

3


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Change of Variable Formula

(1) Evaluate B e−ρ

3

dV , where B is the ball centered at the origin with radius R and

ρ =

x2 + y2 + z 2.

(2) Evaluate R

x2 + y2 + z 2dV , where R is the region bounded by the plane z = 3

and the cone z =

x2 + y2.

(3) Evaluate R

(√ x +

√ y)

1

2dA, where R is the region in the first quadrant which is

bounded by the coordinate axes and the curve √ x +√ y = 1.(4) Evaluate

R

(x2 +y2)dA, where R is the region in the first quadrant that is boundedby the hyperbolas xy = 1, xy = 3, x2 − y2 = 1, and x2 − y2 = 4.

(5) Evaluate

A

ex−y

x+y dxdy, where A is the first quadrant region bounded by the

coordinate axes and the line x + y = 1.

(6) Using the transformation u = 2xx2+y2

, v = 2yx2+y2

, evaluate

A

dxdy

(x2 + y2)2, where A

is the region in the first quadrant of the xy-plane which is bounded by the circlesx2 + y2 = 6x, x2 + y2 = 4x, x2 + y2 = 8y, x2 + y2 = 2y.

(7) Using the transformation

x = r

t cos θ, y =

r

t sin θ, z = r2,

find the volume of the region bounded by the paraboloids z = x2+y2, z = 4(x2+y2)

and also between the planes z = 1 and z = 4.

(8) (Extra Question) Let A be an n×n symmetric positive definite matrix, and q (x) =xt

Ax. Show that Rn

e−q(x)dV = π

n

2

(detA)1

2

.

1


14/45


Tutorial 4 Solution

by Chan Chi Ming

September 3, 2013


1. See Tutorial 3, Q.3(a).

2. See Tutorial 3, Q.3(b).

3. See Tutorial 3, Q.3(c).

4. From the change of variableu = xy, v = x2 − y2,

we have

∂ (x, y)

∂ (u, v) =

∂ (u, v)

∂ (x, y)

−1

= y x2x −2y

−1

=

1

2(x2 + y2) =

1

2√ v2 + 4u2 ,and the region R can be parameterized to be

1 ≤ u ≤ 3, 1 ≤ v ≤ 4.

Therefore,

R

(x2 + y2)dA =

4

1

3

1

v2 + 4u2 · 1

2√

v2 + 4u2dudv =

1

2

4

1

3

1

dudv = 3.

5. From the change of variable

u = x − y, v = x + y,we have

∂ (x, y)

∂ (u, v) =

∂ (u, v)

∂ (x, y)

−1

=

1 −11 1−1

= 1

2,

and the region A can be parameterized to be

−v ≤ u ≤ v, 0 ≤ v ≤ 1.

Therefore,

A

ex−yx+y dxdy =

1

0 v

−v

euv

· 1

2dudv =

1

2 1

0

v(e−

e−1)dudv = 1

4(e−

e−1).


u = 2x

x2 + y2, v =

2y

x2 + y2,

we have

∂ (x, y)

∂ (u, v) =

∂ (u, v)

∂ (x, y)

−1

=

−2(x2 − y2)

(x2 + y2)2 − 4xy

(x2 + y2)2

−

4xy

(x2

+ y2

)2

2(x2 − y2)

(x2

+ y2

)2

−1

= 1

4(x2 + y2)2,

1


15/45

and the region A can be parameterized to be

1

3 ≤ u ≤ 1

2,

1

4 ≤ v ≤ 1.

Therefore,

A

dxdy

(x2 + y2)2 =

1

4 1

14

1

2

13

dudv = 1

32.


x = r cos(θ)

t , y =

r sin(θ)

t , z = r2,

where r > 0, t > 0 and 0 ≤ θ ≤ 2π, we have

∂ (x, y , z)

∂ (r,θ,t) =

cos(θ)

t −r cos(θ)

t2 −r sin(θ)

tsin(θ)

t −r sin(θ)

t2r cos(θ)

t2r 0 0

= 2r3

t3

and the region is transformed to be

1 ≤ r ≤ 2, 0 ≤ θ ≤ 2π, 1 ≤ t ≤ 2.Therefore, the volume is given by

V =

2

1

2π

0

2

1

2r3

t3 drdθdt =

45

8 π.

8. Since A is a symmetric positive definite matrix, there is some orthogonal matrix Q suchthat QT AQ = D, where

D =

λ1 0 00 . . . 00 0 λn

is a diagonal matrix whose diagonal entries are the eigenvalues λ1, . . . , λn(> 0) of A. In

particular, det(A) =n

j=1

λ j . Form the change of variable

x = Qu,

we have∂ (x1, . . . , xn)

∂ (u1, . . . , un)

= Q = 1and the transformed region is still Rn. Therefore, Rn

e−xT Axdx =

Rn

e−uT Du·1du =

Rn

e−n

j=1 λju2

j du =

∞

−∞

e−λ1u2

1 · · · ∞

−∞

e−λnu2ndun . . . du1.

By Tutorial 3, Q.1(c),

∞

0

e−t2

dt =√

π. So,

Rn

e−xT Axdx =

n j=1

∞

−∞

e−λju2

j du j =n

j=1

π

λ j =

πn2

det(A).

2


16/45


Surface Area

(1) Find the surface areas of a torus and a ball of radius R in R3.

(2) Find the surface area of an Enneper’s surface.

(3) Let S be the graph of the differentiable function f : D → R, where D ⊂ R2 is adomain. Show that

Area(S ) =

D

1 +

∂f

∂x

2+

∂f

∂y

2dxdy.

(4) Let f : [a, b] → R+ be differentiable. Let S be the surface obtained by revolvingthe graph of y = f (x) about the x-axis. Show that

Area(S ) = 2π

ba

f (x)

1 + (f (x))2dx.

(5) Let r(u, v) be a regular surface defined on a domain D ⊂ R2. Show that its surfacearea is given by

D√

EG − F 2dudv,

where E F

F G

=

∂ r∂u · ∂ r∂u ∂ r∂u · ∂ r∂v

∂ r∂u · ∂ r

∂v∂ r∂v · ∂ r

∂v

.

(6) Find the surface area of the part of the plane x + y + z = 1 that lies in the first

octant.

(7) Find the surface area of the part of z = xy that lies in the cylinder given by

x2 + y2 = 1.

(8) Find the area of the surface obtained by revolving the graph of y = cosh(x) about

the x-axis for 0 ≤ x ≤ 1.

(9) Find the surface area of the region common to the intersecting cylinders x2+y2 = a2

and x2 + z 2 = a2.

(10) Find the surface area of the ellipsoid x2

a2 + y

2

a2 + z

2

b2 = 1, where a > b > 0.

1


17/45


Tutorial 5 Solution

by Chan Chi Ming

September 3, 2013


1. (a) Consider the surface of a ball given by S ≡ {(x,y,z) ∈ R3 : x2 + y2 + z2 = R2} andparameterize it by the function Φ given by

Φ(θ, φ) ≡ R cos(θ)sin(φ), R sin(θ) sin(φ), R cos(φ)for any (θ, φ) ∈ D ≡ [0, 2π] × [0, π]. Then

Φθ × Φφ =

e1 e2 e3−R sin(θ)sin(φ) R cos(θ) sin(φ) 0R cos(θ)cos(φ) R sin(θ)cos(φ) −R sin(φ)

= −R2 sin(φ) · cos(θ) sin(φ), sin(θ)sin(φ), cos(φ)

and so Φθ × Φφ = R2 sin(φ) in D . The surface area of S is given by

A =

S

dS =

D

Φθ × Φφdθdφ = π0

2π0

R2 sin(φ)dθdφ = 4πR2.

(b) Consider the surface of a torus, with major radius R and minor radius r respectively,

given by S ≡ {(x,y,z) ∈ R3 : (R− x2 + y2)2 + z2 = r2} and parameterize it by the

function Φ given by

Φ(θ, φ) ≡R + r cos(φ)

cos(θ),

R + r cos(φ)

sin(θ), r sin(φ)

for any (θ, φ) ∈ D ≡ [0, 2π] × [0, 2π]. Then

Φθ × Φφ =

e1 e2 e3−R + r cos(φ) sin(θ) R + r cos(φ) cos(θ) 0

−r sin(φ)cos(θ) −r sin(φ)sin(θ) −r cos(φ)

= −rR + r cos(φ) · cos(φ)cos(θ), cos(φ) sin(θ), sin(φ)

and so Φθ × Φφ = rR + r cos(φ)

in D. The surface area of S is given by

A =

S

dS =

D

Φθ × Φφdθdφ = 2π0

2π0

rR + r cos(φ)

dθdφ = 4π2rR.

(This verifies the Pappus’s second (centroid) theorem.)

2. Note that an Enneper’s surface S can be parameterized by the function Φ given by

Φ(u, v) ≡u

3

1 − u

2

3 + v2

,−v

3

1 − v

2

3 + u2

, 1

3(u2 − v2)

for any (u, v) ∈ D ≡ [−1, 1] × [−1, 1]. Then

Φu × Φv =

e1 e2 e3

1

3(1 − u2 + v2) −2

3uv

2

3u

2

3vu −1

3(1 − v2 + u2) −2

3v

=

2

9

u(u2 + v2 + 1), 2

9

v(u2 + v2 + 1), 1

9(u2 + v2)2 −

11


18/45

and so Φu × Φv = (1 + u2 + v2)2/9 in D . The surface area of S is given by

A =

S

dS =

D

Φu × Φvdudv = 1−1

1−1

1

9(1 + u2 + v2)2dudv =

532

405.

3. Note that S = {(x,y,f (x, y) ∈ R3 : (x, y) ∈ D)} and it can be parameterized by thefunction Φ given by

Φ(u, v) ≡ u,v,f (u, v)for any (u, v) ∈ D. Then

Φu × Φv =e1 e2 e31 0 f u0 1 f v

= (−f u,−f v, 1)and so Φu × Φv =

1 + f 2u + f

2v in D. The surface area of S is given by

A = S

dS = D

Φu ×

Φvdudv =

D 1 + f 2u + f 2v dudv.

4. Note that S can be parameterized by the function Φ given by

Φ(u, v) ≡ u, f (u)cos(v), f (u) sin(v)for any (u, v) ∈ D ≡ [a, b] × [0, 2π]. Then

Φu × Φv =

e1 e2 e31 f cos(v) f sin(v)0 −f sin(v) f cos(v)

=ff ,−f cos(v),−f sin(v)

and so Φu × Φv = |f | 1 + (f )2 in D . The surface area of S is given byA =

S

dS =

D

Φu × Φvdudv = 2π ba

|f (u)|

1 +f (u)

2du.

5. For every (u, v) ∈ D, denote by θ the angle between ru and rv. Note that

ru×rv2 = ru2rv2 sin2(θ) = ru2rv2−ru2rv2 cos2(θ) = ru2rv2−ru, rv2 .

in D . The surface area is given by

r(D)

dS = D

ru ×

rvdudv =

D EG − F 2dudv.

6. (Method I):By some simple coordinate geometry, the surface area can be found to be

A = 1

2 ·

√ 2 ·

√ 2 · sin

π3

=

√ 3

2 .

(Method II):The surface S ≡ {(x,y,z) ∈ R3 : x + y + z = 1} can be parameterized by the function Φgiven by

Φ(u, v) ≡

(u,v, 1−u

−v)

2


19/45

for any (u, v) ∈ D ≡ {(x, y) ∈ R2 : x + y ≤ 1,x,y ≥ 0}. Then

Φu × Φv =e1 e2 e31 0 −10 1 −1

= (1, 1, 1)and so

Φu

×Φv

=

√ 3 in D . The surface area of S is given by

A =

S

dS =

D

Φu × Φvdudv =√

3

D

dudv =√ 3

2 .

(Method III):Let f (x, y) ≡ 1 − x− y in D ≡ {(x, y) ∈ R2 : x + y ≤ 1,x,y ≥ 0}. By Q.3 the surface areais given by

A =

D

√ 3dxdy =

√ 3

2 .

7. Let f (x, y) ≡ xy in D ≡ {(x, y) ∈ R2 : x2 + y2 ≤ 1}. By Q.3 the surface area is given by

A =

D

1 + y2 + x2dxdy =

2π0

10

1 + r2rdrdθ =

2(2√

2−

1)

3 π.

8. Let f (x) ≡ cosh(x) in [0, 1]. By Q.4 the surface area is given by

A = 2π

10

cosh(x)

1 + sinh2(x)dx = 2π

10

cosh2(x)dx = π

10

1+cosh(2x)

dx =

1 +

sinh(2)

2

π.

9. Denote the surface by S . By the symmetry of the surface, it suffices to consider the

surface S ≡ S ∩ {(x,y,z) ∈ R3 : x, y, z ≥ 0}. Let f (x, y) ≡ a2 − x2 in D ≡ {(x, y) ∈

R2 : x2 + y2 ≤ 1,x,y ≥ 0}. By Q.3 the surface area is given by

A = 16

D

1 + x

2

a2 − x2dxdy = 16a π

2

0

a0

r a2 − r2 cos2(θ)drdθ = 16a

2.

10. Denote the surface by S and parameterize it by the function Φ given by

Φ(θ, φ) ≡ a cos(θ)sin(φ), a sin(θ)sin(φ), b cos(φ)for any (θ, φ) ∈ D ≡ [0, 2π] × [0, π]. Then

Φθ × Φφ =

e1 e2 e3−a sin(θ)sin(φ) a cos(θ)sin(φ) 0a cos(θ)cos(φ) a sin(θ)cos(φ) −b sin(φ)

= −ab cos(θ)sin2(φ), ab sin(θ)sin2(φ), a2 sin(φ)cos(φ)and so Φθ × Φφ = a sin(φ)

a2 cos2(φ) + b2 sin2(φ) in D . The surface area of S is given

by

A =

S

dS =

π0

2π0

a| sin(φ)| a2 cos2(φ) + b2 sin2(φ)dθdφ.

By a change of variable t = a2/b2 − 1cos(φ) and symmetry, we have

A = 4πab2√ a2 − b2

a2b2

−1

0

1 + t2dt = 2π

a +

ab2√ a2 − b2 log

a +

√ a2 − b2b

.

3


20/45


Line Integrals

(1) Let U ⊂ R3 be an open set. Let f, g : U → R be smooth functions, F, G : U → R3

be smooth vector fields. Show the following formulae:

(a) ∇(fg) = f ∇g + g∇f

(b) ∇ · (f F) = ∇f · F + f ∇ · F

(c) ∇ · (F × G) = ∇× F · G − F · ∇ × G

(d) ∇ ×∇f = 0

(e) ∇ · ∇ × F = 0

(2) Suppose U ⊂ R3 is a convex open set and F : U → R3 is a smooth vector field.

Show that if ∇ × F = 0, then F = ∇u for some smooth function u.

(3) Evaluate

C

xy2ds, where C is the right half of the circle x2 + y2 = 16.

(4) Evaluate

C

4x3ds, where C is the line segment from (−2,−1) to (1, 2).

(5) Evaluate C

4xds, where C is the curve consisting of two parts:

C 1 : y = −1 from (−2,−1) to (0,−1), and

C 2 : y = x3 − 1 from (0,−1) to (1, 0).

(6) Evaluate

C

xyzds, where C is the helix given by r(t) = (cos(t), sin(t), t), where

t ∈ [0, 1].

(7) Evaluate

∆

(2x− y + 4)dx + (5y + 3x− 6)dy, where ∆ is a triangle in the xy-plane

with vertices at (0, 0), (3, 0) and (3, 2) with counterclockwise orientation.

(8) Evaluate 1

2

C

xdy − ydx, where C is the ellipse x2

a2 + y

2

b2 = 1 oriented counterclock-

wisely.

(9) Show that

C

∇f · dr = f (r(b)) − f (r(a)), where C is a smooth curve given by

r(t), a ≤ t ≤ b and f is a smooth scalar function on C .

1


21/45


Tutorial 6 Solution

by Chan Chi Ming

September 3, 2013


1. Google yourself.

2. Let (x0, y0, z0) ∈ U and write F = (F 1, F 2, F 3). Define a function u : U → R by

u(x,y,z) ≡ xx0

F 1(t, y0, z0)dt +

yy0

F 2(x0, t ,z0)dt +

zz0

F 3(x0, y0, t)dt

for any (x,y,z) ∈ U . Since ∇ × F = 0, u is well defined. One can check that F = ∇u inU .

3. Parameterize C by the function γ given by

γ (t) ≡ 4cos(t), 4sin(t)for any t ∈ [−π/2, π/2]. Then γ (t) = 4 and so

C

xy2ds =

π2

−π2

4cos(t)

4sin(t)

2 · 4dt = 5123 .


γ (t) ≡ (−2 + 3t,−1 + 3t)

for any t ∈ [0, 1]. Then γ (t) = 3√ 2 and so C

4x3ds =

1

0

4(−2 + 3t)3 · 3√

2dt = −15√

2.

5. Parameterize C 1 by the function γ 1 given by

γ 1(t) ≡ (t,−1)

for any t ∈ [−2, 0]. Then γ 1(t) = 1.Parameterize C 2 by the function γ 2 given by

γ 2(t) ≡ (t, t3 − 1)

for any t ∈ [0, 1]. Then γ 2(t) =

1 + 9t4. So

C

4xds =

C 1

+

C 2

4xds =

0

−2

4(t)·1dt+ 1

0

4(t)·

1 + 9t4dt = −8+√

10+1

3 log(3+

√ 10).

6. Since r(t) =√

2, we have

C

xyzds =

1

0 cos(t)

sin(t)

(t) ·

√ 2ds =

√ 2

8 sin(2) − 2 cos(2)

.

1


22/45

7. Denote the straight from (0, 0) to (3, 0) by C 1, the straight from (3, 0) to (3, 2) by C 2 andthe straight from (3, 2) to (0, 0) by C 3. So ∆ = C 1 ∪ C 2 ∪ C 3.Parameterize C 1 by the function γ 1 given by

γ 1(t) ≡ (t, 0)

for any t ∈ [0, 3].Parameterize C 2 by the function γ 2 given by

γ 2(t) ≡ (3, t)

for any t ∈ [0, 2].Parameterize C 3 by the function γ 3 given by

γ 3(t) ≡ (3 − 3t, 2 − 2t)

for any t ∈ [0, 1]. Therefore,

∆

(2x− y + 4)dx + (5y + 3x− 6)dy

=

C 1

+

C 2

+

C 3

(2x− y + 4)dx + (5y + 3x− 6)dy

=

3

0

(2t− 0 + 4) · 1 + (0 + 3t− 6) · 0dt +

2

0

(6 − t + 4) · 0 + (5t + 9 − 6) · 1dt

+

1

0

2(3 − 3t) − (2 − 2t) + 4 · (−3) + 5(2 − 2t) + 3(3 − 3t) − 6 · (−2)dt

= 12.


γ ≡ a cos(t), b sin(t)

for any t ∈ [0, 2π]. Then1

2

C

xdy − ydx = 12

C

a cos(t) · b cos(t) − b sin(t) · − a sin(t)dt = πab.

9. For every t ∈ [a, b], write r(t) = x(t), y(t), z(t). Then by chain rule,

C ∇

f, dr

= b

a∂f ∂x ·

dx

dt +

∂f

∂y · dy

dt +

∂f

∂z · dz

dt dt =

b

a

d

dtf r(t)dt = f r(b)−f r(a).

2


23/45


Line Integral for Vector Fields and Conservative Vector Fields

(1) Let F : R3 → R3 be a smooth vector field. Suppose C r is a circle of radius r

centered at the origin with unit normal vector n. Show that

(∇ × F) (0, 0, 0) · n = limr→0

1

πr2

C r

F · dr.

(2) Evaluate

(2x − y + 4)dx + (5y + 3x − 6)dy around a triangle in the xy -plane with

vertices at (0,0), (3,0) and (3,2) transversed in a counterclockwise direction.

(3) Let F = (3x − 2y)i + (y + 2z ) j − x2k. Evaluate

C

F · dr from (0, 0, 0) to (1, 1, 1),

where C is the curve x = t, y = t, z = t3.

(4) Let F : R2 \ {0} → R2 be defined by

F(x, y) =

−y

x2 + y2,

x

x2 + y2

.

(a) Let C be a parametric curve be defined by r : [0, 2π] → R2

, where

r(t) = (cos nt, sin nt) and n ∈ Z.

Find

C

F · dr.

(b) Let C be the ellipse defined by r : [0, 2π] → R2, where

r(t) = (cos t, 2sin t).

Find C

F · dr.

(c) Does there exist a smooth function f : R2 \ {0} → R such that F = ∇f ?

(5) Let F be a continuously differentiable vector field on a domain D in R2 or R3. Show

that the following statements are equivalent:

(a)

C

F · dr is independent of path.

(b) F is conservative.

(c) C

F · dr = 0 for all closed curves C contained in D.

1


24/45


Tutorial 7 Solution

by Chan Chi Ming

September 3, 2013


1. Parameterize C r by the function r given by

r(t) ≡

r cos(t), r sin(t), 0

for any t ∈ [0, 2π]. Then for every r > 0,

1

πr2

C r

F, dr = 1

πr

2π

0

−F 1


sin(t)+F 2


cos(t)

dt.

Fix t ∈ (0, 2π). By the mean value theorem, for every r > 0 there are some 0 < δ 1,r, δ 2,r < r

such thatF 1


= F 1(0, 0, 0) +

∇F 1

δ 1,r cos(t), δ 1,r sin(t), 0

,

cos(t), sin(t), 0

· r,

F 2


= F 2(0, 0, 0) +

∇F 2


,

cos(t), sin(t), 0

· r.

Therefore, for every r > 0,

1

πr2

C r

F, dr

= − 1

πr

2π

0

F 1(0, 0, 0) sin(t) +

∂ F 1∂x


cos(t)sin(t) · r

+ ∂F 1

∂y δ 1,r cos(t), δ 1,r sin(t), 0 sin2(t) · r dt+

1

πr

2π

0

F 2(0, 0, 0) cos(t) +

∂ F 2∂x


cos2(t) · r

+ ∂F 2

∂y


sin(t)cos(t) · r

dt

= −1

π

2π

0

∂F 1∂x


cos(t) sin(t) +

∂ F 1∂y


sin2(t)

dt

+ 1

π

2π

0

∂F 2∂x


cos2(t) +

∂ F 2∂y


sin(t) cos(t)

dt.

1


25/45

Since F is smooth, we have

limr→0

1

πr2

C r

F, dr

= limr→0

−

1

π

2π

0

∂F 1∂x


cos(t)sin(t) +

∂ F 1∂y


sin2(t)

dt

+ 1π 2π0

∂F 2∂x

δ 1,r cos(t), δ 1,r sin(t), 0 cos2(t) + ∂ F 2∂y

δ 1,r cos(t), δ 1,r sin(t), 0 sin(t) cos(t) dt= −

1

π

2π

0

∂F 1∂x

(0, 0, 0) cos(t) sin(t) + ∂ F 1

∂y (0, 0, 0)sin2(t)

dt

+ 1

π

2π

0

∂F 2∂x

(0, 0, 0)cos2(t) + ∂ F 2

∂y (0, 0, 0) sin(t)cos(t)

dt

= ∂F 2

∂x (0, 0, 0) −

∂ F 1∂y

(0, 0, 0)

= (∇ × F)(0, 0, 0), n .


3. Parameterize C by thew function r given by

r(t) ≡ (t,t,t3)

for any t ∈ [0, 1]. Then C

F, dr =

1

0

(3t − 2t) · 1 + (t + 2t3) · 1 + (−t2) · 3t2

dt =

9

10.

4. (a) By direction computation, C

F, dr =

2π

0

− sin(nt)

cos2(nt) + sin2(nt) ·

− n sin(nt)

+ cos(nt)

cos2(nt) + sin2(nt) · n cos(nt)

dt = 2nπ.

(b) By direction computation, C

F, dr =

2π

0

−2sin(t)

cos2(t) + 4 sin2(t) ·

− sin(t)

+ cos(t)

cos2(t) + 4 sin2(t) · 2cos(t)

dt = 2π.

(c) There is no such function f . Suppose on the contrary there were such a functionf : R2 \ {0} → R. Then let C be the curve defined in (a), we have

C

F, dr = C

∇f, dr = f

r(2π)

− f

r(0)

= 0,

which is impossible.

5. Suppose D ⊂ R2.(a) ⇒ (b)

Let (x0, y0) ∈ D. Define a function f : D → R by

f (x, y) ≡

C

F, dr

2


26/45

for any (x, y) ∈ D, where C is a path joining (x, y) and (x0, y0). f is well-defined becausethe integral is independent of path. Let (x, y) ∈ D, h > 0 (be small enough) so that(x + h, y) ∈ D and C be the straight line joining (x, y) and (x + h, y). Then

f (x + h, y) − f (x, y)

h =

1

h

C

F, dr = 1

h

h0

F 1(x + h, y)dt

By the fundamental theorem of Calculus,

∂f

∂x(x, y) = lim

h→0

f (x + h, y) − f (x, y)

h = lim

h→0

1

h

h0

F 1(x + h, y)dt = F 1(x, y).

Similarly, (∂f/∂y)(x, y) = F 2(x, y) and so F = ∇f .

(b) ⇒ (c)

Since there is a function f : D → R such that F = ∇f , suppose C is parameterized byr : [a, b] → D, then by the fundamental theorem of Calculus (see See Tutorial 6, Q.9.),

C

F, dr = C

∇f, dr = f r(b) − f r(a) = 0.

(c) ⇒ (a)

Let (x, y), (x0, y0) ∈ D and C , C be paths in D joining (x, y) and (x0, y0). Then C − C

forms a closed curve and so C

F, dr −

C

F, dr =

C −C

F, dr = 0.

Hence

C F, dr =

C F, dr and the integral is independent of path.

3


27/45


Review on Sections 13.6 − 14.2

(1) Find the volume of the solid bounded above by the sphere ρ = a and below by

φ = π3

.

(2) Find the moment of inertia around the z −axis of a solid sphere of radius a centered

at the origin with constant density δ .

(3) Find the mass of the solid ellipsoid x2

a2 + y

2

b2 + z

2

c2 ≤ 1, where its density at each point

(x,y,z ) is given by δ (x,y,z ) = 1 − x2

a2 − y

2

b2 − z

2

c2.

(4) Find the surface area of the sphere x2 + y2 + z 2 = 4 that lies between the planes

x + y + z = 1 and x + y + z = −1.

(5) Evaluate

1

0

1

0

1

1 − x2y2dxdy using the substitution x = sinu

cosv, y = sinv

cosu. Use this

result to evaluate

1

0

1

0

dxdy

1 − xy.

(6) Find the average distance D = 1

s

C

D(x, y)ds from the origin to the spiral C

parametrized by x = e−t

cos t, y = e−t

sin t from t = 0 to infinity, where s is thelength of C and D(x, y) is the distance from the origin to the point (x, y) on the

spiral C .


F(x, y) =

−y

x2 + y2,

x

x2 + y2

.

(a) Does there exist a smooth function f : R2 \ {0} → R such that F = ∇f ?

(b) Does there exist a smooth scalar function f defined on the upper-half plane

such that F = ∇f on the upper-half plane?

(c) Does there exist a smooth scalar function f defined on the right-half plane such

that F = ∇f on the right-half plane?

(d) Show that

C a,b

F · dr = 2π, where C a,b is the ellipse defined by x(t) =

a cos t, y(t) = b sin t, 0 ≤ t ≤ 2π.

1


28/45


Tutorial 8 Solution

by Chan Chi Ming

September 3, 2013


1. In spherical coordinates, the region is parameterized to be(ρ,θ,φ) : 0 ≤ ρ ≤ a, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π

3

.

Then the volume is given by

V =

π3

0

2π

0

a0

ρ2 sin(φ)dρdθdφ = 1

3πa3.

2. In spherical coordinates, the region is parameterized to be

R ≡ {(ρ,θ,φ) : 0 ≤ ρ ≤ a, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π} .The mass is given by M = 4/3 · πa3δ . Then the moment of inertia around the z-axis isgiven by

I z =

R

(x2+y2)·δdxdydz = δ π0

2π

0

a0

ρ2 sin2(φ)·ρ2 sin(φ)dρdθdφ = 25

4

3πa3δ

a2 =

2

5ma2.


x = aρ cos(θ)sin(φ) y = bρ sin(θ)sin(φ), z = cρ cos(φ),

we have

∂ (x,y ,z)

∂ (ρ,θ,φ) =a cos(θ) sin(φ) −aρ sin(θ) sin(φ) aρ cos(θ)cos(φ)b sin(θ)sin(φ) bρ cos(θ) sin(φ) bρ sin(θ)cos(φ)

c cos(φ) 0 −cρ sin(φ)

= abcρ2 sin(φ),and the region can be parameterized to be

0 ≤ ρ ≤ 1, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π.Therefore, the mass is given by

M =

π0

2π

0

1

0

δ ·abcρ2 sin(φ)dρdθdφ = π0

2π

0

1

0

(1−ρ2)·abcρ2 sin(φ)dρdθdφ = 815

πabc.

4. Through a rotation Q (which is an orthogonal matrix and |Q| = 1), the two planesare transformed to be z = 1/

√ 3 and z = −1/

√ 3 respectively while the sphere remains

unchanged. The region is then parameterized to be(ρ,θ,φ) : 0 ≤ ρ ≤ 2, 0 ≤ θ ≤ 2π, − arccos

1

2√

3

≤ φ ≤ arccos

1

2√

3

.

By Tutorial 5, Q.1(a), the surface area is given by

A =

− arccos 12√ 3

− arccos

1

2√ 3

2π

0

(2)2 sin(φ)dθdφ = 8√

3π.

1


29/45


x = sin(u)

cos(v), y =

sin(v)

cos(u),

we have

∂ (x, y)

∂ (u, v) =

cos(u)cos(v)

sin(u) sec(v)tan(v)

sin(v) sec(u)tan(u) cos(v)

cos(u)

= 1 − tan2(u)tan2(v)

and the region [0, 1] × [0, 1] can be parameterized to be

0 ≤ u ≤ π2 − v, 0 ≤ v ≤ π

2.

Therefore,

10 10

1

1 − x2y2dxdy = π2

0

π2−v

0

1

1 − tan2(u)tan2(v) · 1 − tan2(u)tan2(v)dudv = π2

8 .

Let I ≡ 1

0

1

0

1

1 − xy dxdy. Then by changing variables, 1

0

1

0

1

1 − xy − 1

1 + xy

dxdy =

1

0

1

0

2xy

1 − x2y2dxdy = 1

2

1

0

1

0

1

1 − x2y2 d(x2)d(y2) =

I

2.

Also,

1

0 1

0

1

1 + xy

dxdy = 1

0 1

0

1 − xy1 − x

2

y2

dxdy = 1

0 1

0

1

1 − x2

y2

dxdy

− 1

0 1

0

xy

1 − x2

y2

dxdy = π2

8 −I

4

.

Therefore,

I = I

2 −

π2

8 − I

4

,

and so I = π2/6.Actually, due to the uniform convergence of the power series, we have

π2

6 =

1

0

1

0

1

1 − xy dxdy = 1

0

1

0

∞n=0

xnyndxdy =∞n=0

1

0

1

0

xnyndxdy =∞n=1

1

n2.

The result also follows from the fact that ζ (2) = π2/6.

6. Since

s =

∞0

x2t + y

2t dt =

∞0

e−2t

cos(t) + sin(t)

2+ e−2t

cos(t) − sin(t)2dt = ∞

0

√ 2e−tdt =

√ 2,

D = 1

s

C

D(x, y)ds =

∞0

e−t cos(t)

2+

e−t sin(t)2·√ 2e−tdt = 1√

2

∞0

√ 2e−2tdt =

1

2.

7. (a) See Tutorial 7, Q.4(c).

2


30/45

(b) Observe the function f : {(x, y) ∈R2 : y > 0} →R given by

f (x, y) ≡ − arctan

x

y

is well defined and has the property that F = ∇f in {(x, y) ∈R2 : y > 0} (as well as

{(x, y) ∈R2

: y 0. The function f :

{(x, y) ∈ R2 : x


31/45


(A) Path Independence

(1) Let F be a smooth vector field on a domain D in R2 or R3. Show that the

following statements are equivalent:

(a)

C

F · dr is independent of path.

(b) F is conservative.

(c)

C

F · dr = 0 for all closed curves C contained in D.


F(x, y) =

−y

x2 + y2,

x

x2 + y2

.

(a) Show that F is conservative on the right-half plane and upper-half plane.

(b) Show that

C a,b

F · dr = 2π, where C a,b is the ellipse defined by

x(t) = a cos t, y(t) = b sin t, 0 ≤ t ≤ 2π.

(B) Potential Functions and Conservative Vector Fields

(1) Find a potential function for the following vector fields:

(a) F : R2 → R2 defined by F(x, y) = (y cos x + y2, sin x + 2xy − 2y).

(b) G : R3 → R3 defined by G(x,y,z ) = 2xy3z 4i + 3x2y2z 4 j + 4x2y3z 3k.

(c) H : R3 → R3 defined by H(x,y,z ) = (2x cos y−2z 3)i+(3+2yez−x2 sin y) j+

(y2ez − 6xz 2)k.

(2) Evaluate C (2x log y − yz )dx +

x2y − xz

dy − xydz , where C is a curve from

(1, 2, 1) to (2, 1, 1) in the first octant.

(3) Evaluate

C

2xdx + 2ydy + 2zdz

x2 + y2 + z 2 , where C is a curve from (−2, −2, −2) to

(2, 2, 2) not passing through the origin.

1


32/45


Tutorial 9 Solution

by Chan Chi Ming

September 3, 2013




3. Recall that a necessary condition for a vector field F to have a potential function f is∇ × F = (∇ × ∇f ) = 0. So, one should always check whether ∇ × F = 0 before “finding”a potential function.

(a) Suppose f : R2 → R is a function such that F = ∇f . Then we have

f x = y cos(x) + y2, f y = sin(x) + 2xy − 2y.

Therefore, f (x, y) = −y sin(x) + xy2 + g(y) and so

− sin(x) + 2xy + g(y) = f y = sin(x) + 2xy − 2y.

Hence, g(y) = −y2 + C and f (x, y) = −y sin(x) + xy2 − y2 + C in R2.

(b) Suppose f : R3 → R is a function such that G = ∇f . Then we have

f x = 2xy3z4, f y = 3x

2y2z4, f z = 4x2y3z3.

Therefore, f (x,y,z) = x2y3z4 + g1(y, z) and so

3x2y2z4 + (g1)y(y, z) = f y = 3x2y2z4.

Hence, g1(y, z) = g2(z) + C and so

4x2y3z3 + g2(z) = f z = 4x2y3z3.

Thus, g2(z) ≡ C and f (x , y , z) = x2y3z4 + C in R3.

(c) Suppose f : R3 → R is a function such that H = ∇f . Then we have

f x = 2x cos(y) − 2z3, f y = 3 + 2ye

z − x2 sin(y), f z = y2ez − 6xz2.

Therefore, f (x,y,z) = x2

cos(y) − 2xz3

+ g1(y, z) and so

−x2 sin(y) + (g1)y(y, z) = f y = 3 + 2yez − x2 sin(y).

Hence, g1(y, z) = 3y + y2ez + g2(z) and so

−6xz2 + y2ez + g 2(z) = f z = y2ez − 6xz2.

Thus, g2(z) ≡ C and f (x , y , z) = x2 cos(y) − 2xz3 + 3y + y2ez + C in R3.

1


33/45

4. Let F : R3 \ {(0, 0, 0)} → R3 be a vector field given by

F(x,y,z) ≡

2x log(y) − yz,

x2

y − xz, −xy

for any (x , y , z) ∈ R3 \ {(0, 0, 0)}. Then ∇ × F = 0 in R3 \ {(0, 0, 0)}. Suppose f :R3

\ {(0, 0, 0)} →R

is a function such that F = ∇f . Then we have

f x = 2x log(y) − yz, f y = x2

y − xz, f z = −xy.

Therefore, f (x,y,z) = x2 log(y) − xyz + g1(y, z) and so

x2

y − xz + (g1)y(y, z) = f y =

x2

y − xz.

Hence, g1(y, z) = g2(z) and so

−xy + g2

(z) = f z

= −xy.

Thus, g2(z) ≡ C and f (x,y,z) = x2 log(y) − xyz + C in R3 \ {(0, 0, 0)}. Therefore,

C

2x log(y)−yz

dx+

x2

y − xz

dy−xydz =

22 log(1)−2·1·1

−

12 log(2)−1·2·1

= − log(2).

5. It is easy to see that a potential function f of the vector field

F =

2x

x2 + y2 + z2,

2y

x2 + y2 + z2,

2z

x2 + y2 + z2

is given by f (x,y,z) = log(x2

+ y2

+ z2

) in R3

\ {(0, 0, 0)} Therefore, C

2xdx + 2ydy + 2zdz

x2 + y2 + z2

C

∇f, dr = log(12) − log(12) = 0.

2


34/45


Green’s Theorem

(1) Let D ⊂ R2 be the region bounded by a piecewise smooth simple closed curve C withpositive orientation. Show that Area(D) =

1

2

C

xdy − ydx = C

xdy = − C

ydx.

(2) Evaluate the following integrals:

(a)

C

x2ydx−y2xdy, counterclockwise around the region bounded by y = √ a2 − x2

and y = 0.

(b) C

ex sin ydx + ex cos ydy, where C is any smooth simple closed curve with

positive orientation.

(c)

C

−ydx + xdyx2 + y2

, where C is any smooth simple closed curve not passing through

the origin with positive orientation.

(3) Let f : D → R be a C 2 function, where D ⊂ R2 is open. Show that C

∂f

∂xdy −

∂f

∂ydx = 0 for all simple closed curves C in D if and only if f is harmonic in D.

(4) Let D ⊂ R2 be the region bounded by a smooth simple closed curve C . Define∂f

∂n = ∇f · N to be the normal derivative of f , where N is the unit outer normal

vector to ∂D. Show that

(a)

D

(f ∆g + ∇f · ∇g)dxdy = ∂D

f ∂g

∂nds.

(b)

D

(f ∆g − g∆f )dxdy = ∂D

f

∂g

∂n − g ∂f

∂n

ds.

(5) Let f : B(0, 1) → R be a harmonic function. Show that for any B(0, R) with0 < R


35/45


Tutorial 10 Solution

by Chan Chi Ming

September 3, 2013


1. By Green’s theorem, we have

Area(D) =

D

dxdy =

1

2

D

∂x

∂x −

∂ (−y)

∂y

dxdy =

1

2

C

xdy − ydx; D

∂x

∂x −

∂ (0)

∂y

dxdy =

C

xdy; D

∂ (0)

∂x −

∂ (−y)

∂y

dxdy = −

C

ydx.

2. (a) Let D be the region bounded by C . By Green’s theorem, C

x2ydx − y2xdy =

D

(−y2 − x2)dxdy =

π0

a0

(−r2) · rdrdθ = −1

4πa4.

(b) Let D be the region bounded by C . By Green’s theorem, C

ex sin(y)dx + ex cos(y)dy =

D

ex cos(y) − ex cos(y)

dxdy = 0.

(c) Let D be the region bounded by C . By Green’s theorem,

C −ydx + xdyx2 + y2 =

D

y

2

− x

2

x2 + y2 + x

2

− y

2

x2 + y2

dxdy = 0.

(Direct computation is possible because the region does NOT include the origin.)

3. Suppose f is harmonic in D. Then by Green’s theorem, for every simple closed curves C in D , with R denoting the region bounded by C , we have

C

∂f

∂xdy −

∂ f

∂ydx =

R

∂ 2f

∂x2 +

∂ 2f

∂y2

dxdy =

R

0 · dxdy = 0.

Suppose now for every simple closed curves C in D , with R denoting the region boundedby C , we have

C

∂f

∂xdy −

∂ f

∂ydx = 0.

If f were not harmonic, then there is some point p ∈ D such that ∆f ( p) = 0. Withoutloss of generality, we may assume ∆f ( p) > 0. By continuity, there is a ball B( p, r) in Dsuch that ∆f > 0 in B ( p, r). Then by Green’s theorem,

0 =

∂B( p,r)

∂f

∂xdy −

∂ f

∂ydx =

B( p,r)

∂ 2f

∂x2 +

∂ 2f

∂y2

dxdy > 0,

which is impossible. Therefore, f is harmonic in D .

1


36/45

4. (a) Suppose C is given by C (t) ≡

x(t), y(t)

for any t ∈ [a, b]. Then N(t) =

y(t),−x(t)

/C (t)and so by Green’s theorem

∂D

f · ∂g

∂nds =

ba

f ·

∂g

∂x, f ·

∂ g

∂y

,

y(t),−x(t)

C (t)

· C (t)dt

= ba−f · ∂ g

∂y · x(t) + f ·

∂g

∂x · y(t) dt

=

∂D

−f · ∂ g

∂ydx + f ·

∂g

∂xdy

=

D

∂

∂x

f ·

∂g

∂x

−

∂

∂y

−f ·

∂ g

∂y

dxdy

=

D

f · ∆g + ∇f, ∇g

dxdy.

(b) By Q.4(a), we have

D f · ∆g + ∇f, ∇g dxdy = ∂D f · ∂g

∂nds

D

g · ∆f + ∇g, ∇f

dxdy =

∂D

g · ∂ f

∂nds,

so D

(f · ∆g − g · ∆f )dxdy =

∂D

f ·

∂g

∂n − g ·

∂ f

∂n

ds.

5. (a) By Q.4(a), we have

0 =

B(0,R)

∆fdxdy =

∂B(0,R)

∂f

∂nds.

Note that

∂

∂r 2π

0f


dθ

= 2π0

∇f


,

cos(θ), sin(θ)

dθ

=

2π0

∂f

∂n


dθ

= 1

r

∂B(0,r)

∂f

∂nds

= 0.

Hence,

1

2πR

∂B(0,R)

f ds = 1

2π

2π0

f

R cos(θ), R sin(θ)

dθ =

1

2π

2π0

f (0, 0)dθ = f (0, 0).

(b) By Q.5(a), we have

∂B(0,r)

f ds = 2πr · f (0, 0) for every 0 < r


37/45


Surface Integral of Vector Fields and Divergence Theorem

(1) Evaluate

S

(x,y,z ) · ndA, where S is the surface of the solid cylinder x2 + y2 ≤

1, 0 ≤ z ≤ 1 with outward pointing normal.

(2) Evaluate

S

xdy ∧ dz + ydz ∧ dx + zdx ∧ dy

(x2 + y2 + z 2)3

2

, where S is a sphere centered at the

origin with outward pointing normal.

(3) Evaluate

S (0, y,−z ) · ndA, where S is the surface given by the paraboloid y =

x2 + z 2, 0 ≤ y ≤ 1 and the disk x2 + z 2 ≤ 1 at y = 1 with outward pointing normal.

(4) Evaluate

S

−

2

y3,−6xy2,

2z 3

x3

· ndA, where S is the graph of f (u, v) = uv3,

(u, v) ∈ [1, 2]× [1, 2] and the parametrization φ(u, v) = (u,v,uv3) is consistent with

the orientation.

(5) Find a function g(ρ) such that, if F(x,y,z ) = g(ρ)(x,y,z ), then ∇ · F = ρm, where

m ≥ 0 and ρ = (x2 + y2 + z 2)1

2 . Prove that

V

ρmdxdydz = 1

m + 3

∂V

ρm(x,y,z ) · ndA,

where V is a nice solid with boundary.

(6) Let F =

2x− y

r3 , x + 2y

r3 ,

2z

r3

be a vector field on R3 \ {0}.

(a) Find

S

F · ndA, where S = {(x,y,z ) : x2 + y2 + z 2 = 4}.

(b) Show that ∇ · F = 0.

(c) Find

∂V

F · ndA, where ∂V is the boundary of the following V:

(i) V = {(x,y,z ) : 1 ≤ z ≤ 7 − x2 − y4}

(ii) V = {(x,y,z ) : −1 ≤ z ≤ 7 − x2 − y4}

1


38/45



by Chan Chi Ming

September 3, 2013


1. Let S 1 ≡ {(x,y, 0) ∈ R3 : x2 + y2 = 1}, S 2 ≡ {(x,y, 1) ∈ R

3 : x2 + y2 = 1} andS 3 ≡ {(x,y ,z) ∈ R

3 : x2 + y2 = 1, 0 ≤ z ≤ 1}. Then S = S 1 ∪ S 2 ∪ S 3. Note that S 1

(x,y ,z), n dS =

S 1

(x,y, 0), (0, 0, −1) dS = 0,

S 2

(x,y ,z), n dS =

S 2

(x,y, 1), (0, 0, 1) dS = 2π.

In cylindrical coordinates, S 3 is parameterized to be {(1, u , v) : 0 ≤ u ≤ 2π, 0 ≤ v ≤ 1}.So

S 3

(x,y ,z), n dS =

1

0

2π

0

cos(u) sin(u) v

− sin(u) cos(u) 00 0 1

dudv = 2π.

Recall that a, b × c =

a1 a2 a3b1 b2 b3c1 c2 c3

.

Therefore,

S

(x,y ,z), n dS =

S 1

(x,y ,z), n dS +

S 2

(x,y ,z), n dS +

S 3

(x,y ,z), n dS = 3π.

2. Consider the sphere S given by {(x,y ,z) ∈ R

3

: x2

+ y2

+ z2

= R2

}. It is easy to seethat the unit outward normal n at a point (x,y ,z) on S is given by n = (x/R,y/R,z/R).Therefore,

S

xdy ∧ dz + ydz ∧ dx + zdx ∧ dy

(x2 + y2 + z2)3

2

=

S

xR3

, y

R3,

z

R3

, n

dS

=

S

xR3

, y

R3,

z

R3

, x

R, y

R, z

R

dS

=

S

1

R2dS

= 4π.

(Note that the value is independent of R.)

3. Let S 1 ≡ {(x, 1, z) ∈ R3 : x2 + z2 = 1} and S 2 = S \ S 1. Then S = S 1 ∪ S 2 and

S 2 =

u cos(v), u2, u sin(v)

: 0 ≤ u ≤ 1, 0 ≤ v ≤ 2π

.

Note that S 1

(0, y, −z), n dS =

S 1

(0, 1, −z), (0, 1, 0) dS =

S 1

dS = π

1


39/45

and S 2

(0, y, −z), n dS =

2π

0

1

0

0 u2 −u sin(v)

cos(v) 2u sin(v)−u sin(v) 0 u cos(v)

dudv

=

2π

0

1

0 − u3 cos2(v) − 3u3 sin2(v)

dudv

= −π.

Therefore, S

(0, y, −z), n dS =

S 1

(0, y, −z), n dS +

S 1

(0, y, −z), n dS = 0.

4. By the given parametrization, we have

S

−

2

y3, −6xy2,

2z3

x3

, n

dS =

2

1

2

1

− 2

v3 −6uv2 2v9

1 0 v3

0 1 3uv2

dudv

= 21

21

(2 + 18u2v4 + 2v9)dudv

= 467.

5. Note that

∇, F =

g(ρ) + xg(ρ) ·

x

ρ

+

g(ρ) + yg (ρ) ·

y

ρ

+

g(ρ) + zg (ρ) ·

z

ρ

= 3g(ρ)+ ρg(ρ).

Then it becomes an ordinary differential equation: ρm = 3g(ρ) + ρg(ρ), ρ > 0. Now

ρm+2 = 3ρ2g(ρ) + ρ3g(ρ) =

ρ3g(ρ)

.

So a choice of the function g is g(ρ) = ρm

/(m + 3) for any ρ > 0. In particular, bydivergence theorem, V

ρmdxdydz =

V

∇, F dxdydz =

∂V

F, n dS = 1

m + 3

∂V

ρm(x,y ,z), n dS.

6. (a) It is easy to see that the unit outward normal n at a point (x,y ,z) on S is given byn = (x/r, y/r, z/r). Then

S

F, n dS =

S

2x − y

r3 ,

x + 2y

r3 ,

2z

r3

,x

r, y

r, z

r

dS =

S

2

r2dS = 8π.

(b) ∇, F = . . . = 0.

(c) (i) Since F is smooth in V , by divergence theorem, ∂V

F, n dS =

V

∇, F dS = 0.

(ii) Let ε > 0 be (small enough) so that B ≡ B(0, ε) ⊂ V . Then F is smooth inV \ B and so by divergence theorem,

0 =

V \B

∇, F dV =

∂V

F, n dS −

∂B

F, n dS =

∂V

F, n dS −8π.

Therefore,

∂V

F, n dS = 8π.

2


40/45


(A) (Continued) Surface Integrals and Divergence Theorem

(1) Evaluate

S

−

2

y3,−6xy2,

2z 3

x3

·ndA, where S is the graph of f (u, v) = uv3,

(u, v) ∈ [1, 2] × [1, 2] and the parametrization φ(u, v) = (u,v,uv3) is consistent

with the orientation.

(2) Let S be the union of truncated paraboloids z = 4 − x2 − y2, 0 ≤ z ≤ 4 and

z = x2 + y2 − 4,−4 ≤ z ≤ 0 . Let n be its outward unit normal vector. Find

S

(x + y2 + sin z, x + y2 + cos z, cosx + sin y + z ) · ndS

(3) Let F = (2x−yr3 , x+2y

r3 , 2z

r3) be a vector field on R3 − 0.

(a) Find

S

F · ndA, where S = {(x,y,z ) : x2 + y2 + z 2 = 4}.

(b) Show that ∇ · F = 0.

(c) Find

∂V

F · ndA, where ∂V is the boundary of the following V :

(i) V = {(x,y,z ) : 1 ≤ z ≤ 7 − x

2

− y

4

}(ii) V = {(x,y,z ) : −1 ≤ z ≤ 7 − x2 − y4}

(B) Stokes’ Theorem

(1) Use Stokes’ Theorem to evaluate

C

F · dr where F = z 2i + y2 j + xk and C

is the triangle with vertices (1,0,0), (0,1,0) and (0,0,1) with counter-clockwise

rotation.

(2) Let C be the intersection of the graphs z = y3 and x2 + y2 = 3, oriented in

the clockwise direction when viewed from high up the positive z -axis. Find theline integral

C

(ex + z )dx + xydy + zeydz

(3) Using Stokes’ Theorem, or otherwise, evaluate

C

xzdx − ydy + x2ydz,

where C is the closed curve consisting of the three straight line segments con-

necting (1,0,0), (0,1,0) to (0,0,1) (in this order).

1


41/45



by Chan Chi Ming

September 3, 2013



2. Let V be the solid enclosed by S , V 1 ≡ {(x,y,z) ∈ R3, 0 ≤ z ≤ 4 − x2 − y2} andV 2 ≡ {(x,y,z) ∈ R3, x2 + y2 − 4 ≤ z ≤ 0}. Then by divergence theorem,

S

x + y2 + sin(z), x + y2 + cos(z), cos(x) + sin(y) + z

,n

dS

=

V

(2 + 2y)dV

= V 1

(2 + 2y)dV + V 2

(2 + 2y)dV

=

2π

0

2

0

4−r2

0

2 + 2r sin(θ)

rdzdrdθ +

2π

0

2

0

0

r2−4

2 + 2r sin(θ)

rdzdrdθ

= 32π.


4. Let S be the surface enclosed by C . Then S = {(u,v, 1−u−v) : 0 ≤ u, v ≤ 1} and the unitnormal (determined by the orientation of the curve) is given by n = (1/

√ 3, 1/

√ 3, 1/

√ 3).

By Stokes’ theorem,

C

(z2, y2, x), dr

=

S

∇ × (z2, y2, x),n dS =

1√ 3

S

(0, 2z − 1, 0), (1, 1, 1) dS

= 1√

3

1

0

1−v

0

(1 − 2u − 2v) ·√

3dudv

= −16.

5. Let S be the surface enclosed by C . Then S ≡ {(x,y,y3) ∈ R3 : x2 + y2 ≤ 3} and it canbe parameterized by the function Φ given by

Φ(u, v) ≡ (u,v,v3)

for any (u, v) ∈ D ≡ {(x, y) ∈ R2 : x2 + y2 ≤ 3}. Then

Φu × Φv =e1 e2 e31 0 0

0 1 3v2

= (0,−3v2, 1)

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Note that C is negatively oriented, by Stokes’ theorem,

C

(ex + z)dx + xydy + zeydz =

S

∇ × (ex + z,xy,zey),−n dS

= −

S

(zey, 1, y),n dS

= − D

(v − 3v2)dudv

= −

2π

0

√ 3

0

r sin(θ) − 3r2 sin2(θ) · rdrdθ

= 27

4 π.

6. Let S be the surface enclosed by C . Then S ≡ {(u,v, 1−u−v) : 0 ≤ u, v ≤ 1} and the unitnormal (determined by the orientation of the curve) is given by n = (1/

√ 3, 1/

√ 3, 1/

√ 3).

By Stokes’ theorem,

C

xzdx − ydy + x2ydz = S

∇ × (xz,−y, x2y),n dS =

1√ 3

S

(x2, x − 2xy, 0), (1, 1, 1) dS

= 1√

3

1

0

1−v

0

(u2 + u − 2uv) ·√

3dudv

= 1

6.

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43/45


Revision

(1) Let F (x,y,z ) = (x + z 2, 0,−z − 3), where (x,y,z ) ∈ R3. Is it possible to find a

C 2-function G : R3 → R3 such that curl G = F ? If it is possible, find such G.

(2) (a) Using Stokes’ Theorem, compute

Γ

3ydx − xzdy + yz 2dz , where Γ is the

curve defined by intersecting the plane {(x,y,z )|z = 2} with the paraboloid

{(x,y,z )|x2 + y2 = 2z }, where Γ rotates counterclockwise viewing from high

above.

(b) Compute the line integral in (a) directly by choosing a suitable parametrizationof Γ.

(3) Let f : R → R be a C 1 function, Γ be a piecewise C 1 curve in {(x, y)|y > 0} joining

the point (a, b) with the point (c, d).

Consider I =

Γ

1

y[1 + y2f (xy)]dx +

x

y2[y2f (xy) − 1]dy.

(a) Show that I is path-independent.

(b) Compute I , assuming that ab = cd.

(4) (a) Let F(x,y,z ) = (2xz, 6yz,h(x,y,z )) be a vector field defined on R3. Find

h(x,y,z ) satisfying h(0, 0, 0) = 0 such that F is irrotational.

(b) Find a potential φ = φ(x,y,z ) satisfying

∇φ = F.

(5) (a) Evaluate the surface integral

S

(x + y + z )2dS,

where S = {(x,y,z )|x2 + y2 + z 2 = 1}.

(b) Let S be a smooth closed surface without boundary enclosing a region R in R3.

Show that the volume enclosed is given by

1

3

S

r · ndS,

where r is the position vector of a point on S .

1


44/45



by Chan Chi Ming

September 3, 2013


Recall that a necessary condition for a vector field F to have a vector potential G is ∇, F =(∇,∇× G) = 0. So, one should always check whether ∇, F = 0 before “finding” a vectorpotential. Suppose now such a vector potential G = (G1, G2, G3) exists. Then observe that

G −∇

zz0

G3(x,y ,t)dt = ( G1, G2, 0) and ∇ × G −∇ zz0

G3(x,y ,t)dt

= ∇ × G,

we may assume G = (G1, G2, 0).

1. By direction computation, ∇, F = 0. Suppose G : R3 → R3 is a vector potential suchthat G = (G1, G2, 0) and ∇ × G = F. Then we have

−(G2)z = x + z2, (G1)z = 0, (G2)x − (G1)y = −z − 3.

Therefore, G1(x,y ,z) = g1(x, y), G2 = −xz − z3/3 + h1(x, y) and so

−z + (h1)x − (g1)y = (G2)x − (G1)y = −z − 3.

Hence, we choose g1(x, y) ≡ 0 and h1(x, y) = −3x in R2.

So G(x,y ,z) = (0,−3x − xz − z3/3, 0) in R3.

2. (a) Let S be the surface enclosed by Γ. Then S = {(x,y, 2) ∈ R3 : x2 + y2 = 4} and theunit normal (determined by the orientation of the curve) is given by n = (0, 0, 1).

Γ

3ydx − xzdy + yz2dz =

S

∇× (3y, −xz,yz2), n

dS

=

S

(x + z2, 0,−z − 3), (0, 0, 1)

dS

= −5

S

dS

= −20π.

(b) Parameterize Γ by the function γ given by

γ (t) ≡ 2 cos(t), 2sin(t), 2for any t ∈ [0, 2π]. Then Γ

3ydx−xzdy +yz2dz =

2π

0

6sin(t)·

−2sin(t)

−4 cos(t)·2cos(t)+0

dt = −20π.

3. (a) Since∂

∂x

x

y2

y2f (xy) − 1

− ∂

∂y

1

y

1 + y2f (xy)

= . . . = 0,

by Green’s theorem, I is path-independent.

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(b) Let η ≡ a/c = d/b and assume a ≤ c. Then choose Γ(t) ≡ (t, ab/t) for any t ∈ [a, c].Therefore,

I =

ca

t

ab

1 +

ab

t

2f (ab)

· 1 +

t3

a2b2

ab

t

2f (ab) − 1

· −ab

t2

dt =

c

d −

a

b.

4. (a) Recall that F is irrational if ∇ × F = 0. Hence,

hy − 6y = 0, −hx + 2x = 0.

Therefore, h(x,y ,z) = 3y2 + h1(x, z) and so

(h1)x = hx = 2x.

Hence, h1(x, z) = x2 + g2(z). Since h(0, 0, 0) = 0, g2(z) ≡ 0.

Therefore, h(x,y ,z) = x2 + 3y2 in R3.

(b) Suppose φ : R3 → R is a function such that F = ∇φ. Then we have

φx = 2xz, φy = 6yz, φz = x2 + 3y2.

Therefore, φ(x,y ,z) = x2z + g1(y, z) and so

(g1)y(y, z) = φy = 6yz.

Hence, g1(y, z) = 3y2z + g2(z) and so

x2 + 3y2 + g2

(z) = φz = x2 + 3y2.

Thus, g2(z) ≡ C and φ(x,y ,z) = x2z + 3y2z + C in R3.

5. (a) Parameterize S by the function Φ given by

Φ(θ, φ) ≡

cos(θ) sin(φ), sin(θ) sin(φ), cos(φ)

for any (θ, φ) ∈ D ≡ [0, 2π] × [0, π]. Then

Φθ × Φφ =

e1 e2 e3

− sin(θ)sin(φ) cos(θ) sin(φ) 0cos(θ)cos(φ) sin(θ)cos(φ) − sin(φ)

= − sin(φ) ·

cos(θ)sin(φ), sin(θ) sin(φ), cos(φ)

and so Φθ × Φφ = sin(φ) in D . Therefore, D

(x + y + z)2dS =

D

cos(θ)sin(φ)+sin(θ) sin(φ)+cos(φ)

2· sin(φ)dθdφ = 4π.

(b) Note that r(x,y ,z) = (x,y ,z). By divergence theorem,

1

3

S

r, n dS = 1

3

R

∇, r dV = 1

3

R

3dV = Vol(R).

math 2020 b solution

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