Differential Calculus - Bartholomew Andrews

First Year Vector Calculus Problem Sheet 1 2011Differential Calculus

Numbers in italics refer to the Exercise number in Riley, Hobson and Bence “Mathematical Methods forPhysics and Engineering.” Answers and hints to odd-numbered questions are at end of each chapter ofthe book and overleaf. This sheet is perhaps a bit long. Start with (2), (4), and (6).

1. [2.1(c)] Obtain from first principles the first derivative of sin x.

2. [2.10(a)] Use the chain rule to show that the first derivative of y(x) = (1 + xm)n is nmxm−1(1 + xm)n−1

[Let u = 1 + xm]

3. By applying the Mean Value Theorem to f (x) =√

x in the interval 1 < x < 2 show that2216

<√

2 <

2316

.

4. [5.1(a),(b)] Using all the appropriate properties of ordinary (and partial) derivatives, perform thefollowing: (a) Find all the first partial derivatives of the following functions f (x, y): (i) x2y, (ii) x2+y2+4,

(iii) sin(x/y), (iv) tan−1(y/x), (v) r(x, y, z) =(x2 + y2 + z2

)1/2. For (i), (ii) and (v), find ∂2f/∂x2, ∂2f/∂y2

and ∂2f/∂x∂y.

5. [5.2] Determine which of the following are exact differentials:

(a) (3x + 2)y dx + x(x + 1) dy

(b) y tan x dx + x tan y dy

(c) y2(ln x + 1) dx + 2xy ln x dy

(d) y2(ln x + 1) dy + 2xy ln x dx

(e)[x/(x2 + y2)

]dy −

[y/(x2 + y2)

]dx

6. [5.9] The function f (x, y) satisfies the differential equation

y∂f∂x

+ x∂f∂y

= 0

By changing to new variables u = x2 − y2 and v = 2xy, show that f is, in fact, a function of x2 − y2

only.

7. [5.10] If x = eu cos θ and y = eu sin θ, show that

∂2φ

∂u2 +∂2φ

∂θ2 =(x2 + y2

) ( ∂2f∂x2 +

∂2f∂y2

)where f (x, y) = φ(u, θ).

Please turn over ...

1

Vector Calculus: Problem Set 1 Differential Calculus 2

Answers

1. [2.1(c)] cos x

2. [2.10(a)] –

3. –

4. [5.1(a),(b)] (a)(i) 2xy, x2; (ii) 2x, 2y; (iii) y−1 cos(x/y), (−x/y2) cos(x/y); (iv) −y/(x2 + y2), x/(x2 + y2);(v) x/r , y/r , z/r (b) (i) 2y, 0, 2x; (ii) 2, 2, 0; (v) (y2 + z2)r−3, (x2 + z2)r−3, −xyr−3

5. [5.2] (c) and (e) [Be careful in reading the question]

6. [5.9] The transformed equation is 2(x2 + y2)∂f/∂v = 0 hence f does not depend on v.

7. [5.10] –

Summary of Section 1: Differential CalculusBelow are various concepts and results from this section. Don’t quote a formula unless you know what

it means and are confident that it is applicable. Refer to the book(s) and/or your lecture notes.

Key concepts and results

Ordinary derivativedfdx

= lim∆x→0

f (x + ∆x) − f (x)∆x

Chain ruledf (u(x))

dx=

dfdu

dudx

Mean Value Theoremdfdx

(x = b) =f (c) − f (a)

c − afor some b in a < b < c

Partial derivative∂f (x, y)∂x

= lim∆x→0

f (x + ∆x, y) − f (x, y)∆x

∂f (x, y)∂y

= lim∆y→0

f (x, y + ∆y) − f (x, y)∆y

Total differential df =∂f∂x

dx +∂f∂y

dy

Total derivativedf (x, y(x))

dx=∂f∂x

+∂f∂y

dydx

Exact differential A dx + B dy ≡∂f∂x

dx +∂f∂y

dy if and only if∂A∂y

=∂B∂x

Chain rule (1 parameter)df (x(u), y(u))

du=∂f∂x

dxdu

+∂f∂y

dydu

Change of variables (x, y) → (u, v)s.t. x = x(u, v), y = y(u, v)

∂f∂u

=∂f∂x

∂x∂u

+∂f∂y

∂y∂u

∂f∂v

=∂f∂x

∂x∂v

+∂f∂y

∂y∂v

Steve Schwartz6 January 2011

First Year Vector Calculus Problem Sheet 2 2011Integration

Numbers in italics refer to the Exercise number in Riley, Hobson and Bence “Mathematical Methods forPhysics and Engineering.” Answers and hints to odd-numbered questions are at end of each chapter ofthe book and overleaf.

1. Let f (x) = x2 in the interval 0 ≤ x ≤ 1. Define the Nth Riemann Sum

SN =N∑

i=1

f (xi) (ξi − ξi−1)

with ξi = i/N and xi = (i−0.5)/N. Make a sketch of f (x) and SN for two or three values of N. EvaluateS1, S2, ... (easiest on a spreadsheet or via a simple computer program) and hence find the value ofN that gives

∫ 10 f (x) dx correct to 3 significant figures.

2. [2.37(a)] By making the substitution x = a cos2 θ + b sin2 θ, evaluate the definite integral

J =∫

[(x − a)(b − x)]−1/2 dx

between the limits x = a and x = b > a.

3. [2.40] Show, by changing variables to x + 1 = u2, that the indefinite integral of f (x) = x3/(x + 1)1/2 is

J =235

(5x3 − 6x2 + 8x − 16

)(x + 1)1/2

[This can also be reached after repeated integration by parts if you are brave enough and patientenough.]

4. Find the area enclosed by the curves y2 = 4x and y = x2/4.

5. [6.10(b)] A thin uniform circular disc lying in the x − y plane has mass M and radius a, so the massper unit area is σ = M/πa2. The moment of inertia about the diameter passing through the centreof the disc (which is at the origin) along the y-axis is defined by I =

!R x2 dM =

!R x2σ dA . By

converting this integral to plane polar coordinates (r , θ), show that I = Ma2/4. You might also try thisin Cartesian (x, y).

6. [6.16] Sketch the two families of curves y2 = 4u(u − x) and y2 = 4v(v + x) where u and v are

parameters. By transforming to the uv-plane, evaluate the integral of y/(x2 + y2

)1/2over the part of

the quadrant x > 0, y > 0 that is bounded by the lines x = 0, y = 0 and the curve y2 = 4a(a − x).

7. [6.18] Sketch the domain of integration for the integral

I =∫ 1

0

∫ 1/y

x=y

y3

xexp

[y2(x2 + x−2)

]dx dy

and characterise its boundaries in terms of new variables u = xy and v = y/x. Show that theJacobian for the change from (x, y) to (u, v) is equal to (2v)−1 and hence evaluate I.


1

Vector Calculus: Problem Set 2 Integration 2

Answers and Hints

1. N = 11

2. [2.37(a)] π [Note that dx = 2(b − a) cos θ sin θ dθ]

3. [2.40] –

4. 16/3

5. [6.10(b)] –

6. [6.16] a2. This is a bit challenging. Find the bounding curves x = 0, y = 0, and y2 = 4a(a − x) in theuv-plane. Show that x(u, v) = u − v and y(u, v) = 2

√uv and then transform the integral. Don’t forget

to calculate and include the Jacobian.

7. [6.18]18

(e − 1)2 The region in the uv-plane is a square bounded by the coordinate axes and the

lines u = 1 and v = 1. You should reach I ∝"

Ruv exp(u2 + v2) du dv.

Summary of Section 2: IntegrationBelow are various concepts and results from this section. Don’t quote a formula unless you know what

it means and are confident that it is applicable. Refer to the book(s) and/or your lecture notes.


Riemann Sum Sn =n∑

i=1

f (xi)(ξi − ξi−1)

Integral F(b) − F(a) =∫ b

af (x) dx = lim

n→∞Sn

Fundamental Theorem∫ b

a f (x) dx = F(b) − F(a) if f (x) integrable and f (x) = dF/dx

Change of variables F(b) − F(a) =∫ b(u)

a(u)f (x(u))

dxdu

du

Double Integral limit of SN where SN =N∑

sub−regions p=1

f (xp , yp)∆Ap

I =!

R f (x, y) dA =!

R f (x, y) dx dy

Jacobian J ≡∂(x, y)∂(u, v)

=

∣∣∣∣∣∣∣∣∣∣∣∣∂x∂u

∂y∂u

∂x∂v

∂y∂v

∣∣∣∣∣∣∣∣∣∣∣∣Change of variables I =

!R f (x, y) dx dy =

!R f (x(u, v), y(u, v)) |J| du dv

Plane polar coordinates x = r cos θ, y = r sin θ, |J| = r , dA = r dθ dr

Steve Schwartz27 Jan 2011

First Year Vector Calculus Problem Sheet 3 2011Triple Integration

Numbers in italics refer to the Exercise number in Riley, Hobson and Bence “Mathematical Methods forPhysics and Engineering.” Answers and hints to odd-numbered questions are at end of each chapter of thebook and overleaf.

1. [6.1] Identify [sketch] the curved wedge bounded by the surfaces y2 = 4ax, x + z = a and z = 0 andhence calculate its volume V .

2. [6.3] Find the volume integral of x2y over the tetrahedral volume bounded by the planes x = 0, y = 0,z = 0 and x + y + z = 1.

3. [6.11] The special moment of interia, I, about the origin of a uniform solid body of density ρ is given by

I = ρ$

V

(x2 + y2 + z2

)dV

Show (e.g., by using cylindrical polar coordinates) that the moment of inertia of a right circular cylinderof radius a, length 2b and mass M about its centre is

M(a2

2+

b2

3

)4. [after 6.13] Show that in spherical polar coordinates, the element of volume for a body that is symmet-

rical about the polar axis is dV = 2πr2 sin θ dr dθ while its surface area is 2πr sin θ[(dr)2 + r2(dθ)2

]1/2.

[This is just a volume of revolution expressed in spherical polar coordinates.] A particular surface isdefined by r = 2a cos θ where a is constant and 0 ≤ θ ≤ π/2. Find its total surface area and the volumeit encloses, and hence identify the surface.

5. Evaluate!

R F · dS where F = (xy,−x2, x + z) and R is the portion of the plane 2x + 2y + z = 6 in thefirst octant.

6. [after 11.26(a)] A vector field F is defined in cylindrical polar coordinates r , θ, z by

F = Fo

(x cos λza

i +y cos λz

aj + sin λz k

)≡

Fora

cos λz r + Fo sin λz k

where r = (x/r) i + (y/r) j is the unit vector in the radial direction. (a) Calculate, as a surface integral,the flux of F through the portion of the cylinder r = a between the planes z = 0 and z = aπ/2. Alsocalculate the flux through the circle r = a located at z = aπ/2.

7. [10.11] (a bit long; see hints) Parameterising the hyperboloid

x2

a2 +y2

b2 −z2

c2

by x = a cos θ sec φ, y = b sin θ sec φ, z = c tan φ, show that an area element on its surface is

dS = sec2 φ[c2 sec2 φ

(b2 cos2 θ + a2 sin2 θ

)+ a2b2 tan2 φ

]1/2dθ dφ

Use this formula to show that the area of the curved surface x2 + y2−z2 = a2 between the planes z = 0and z = 2a is

πa2(6 +

1√

2sinh−1 2

√2)

= πa2[6 +

1√

2ln

(2√

2 + 3)]

.


1

Vector Calculus: Problem Set 3 Triple Integration 2

Answers and Hints

1. 8a3/15

2. 1/360

3. –

4. Symmetric about the polar axis⇒ no φ-dependence. Also, consider increments dr and dθ in, e.g., theplane φ = 0 to express the arc length ds. Ans: V = 4πa3/3; S = 4πa2; surface is a sphere.

5. Write dS = n dS and find n from the standard form of a plane (e.g., r · n = d). Note that the angle α

between n and k is cosα = 1/3 so dS = dx dy/(1/3). Alternatively, write r = x i + y j + z(x, y) k and take

dS = N dx dy directly with N =∂r∂x×∂r∂y

. Ans: 27/4

6. Show, e.g., from a sketch of the differential surface, that on the cylinder dS = r dφ dz r. Ans:Fo(2πa/λ) sin(λaπ/2); πa2Fo sin(λaπ/2).

7. This is a bit long. Take the point on the surface as r = x(θ, φ) i + y(θ, φ) j + z(θ, φ) k and then |dS| =∣∣∣∣∣∂r∂θ×∂r∂φ

∣∣∣∣∣ dθ dφ gives the result surprisingly cleanly. Here θ turns out to be an azimuthal angle (not to

be confused with the θ in spherical polar coordinates) with 0 ≤ θ ≤ 2π and φ should just be considereda parameter. Show 0 ≤ φ ≤ tan−1 2. Finally, the substitution X =

√2 tan φ yields an integral of the form∫ (

1 + X2)1/2

=12

[X√

1 + X2 + ln(X +√

1 + X2)]

.

Summary of Section 3: Triple IntegrationBelow are various concepts and results from this section. Don’t quote a formula unless you know what it

means and are confident that it is applicable. Refer to the book(s) and/or your lecture notes.


Integral over volume$

Rf (x, y, z) dV =

$R

f dx dy dz

Change of variables dV =∣∣∣∣∣ ∂(x, y, z)∂(u, v, w)

∣∣∣∣∣ du dv dw

Jacobian J =∂(x, y, z)∂(u, v, w)

=

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

∂x∂u

∂y∂u

∂z∂u

∂x∂v

∂y∂v

∂z∂v

∂x∂w

∂y∂w

∂z∂w

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣Cylindrical polar coordinates x = r cos φ, y = r sin φ, z = z

use ρ =√

x2 + y2 instead if danger of confusion with r

J = r so dV = r dr dφ dz

Spherical polar coordinates x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ

J = r2 sin θ so J dV = r2 sin θ dθ dφ dr

Surface integrals weighted sum over dS

... continued

Vector Calculus: Problem Set 3 Triple Integration 3


Surface element If r(u, v) is on surface S then dS =(∂r∂u×∂r∂v

)du dv = N du dv =

n dS

Projected area element dS = dA/ cosα where α is angle begween dS on surface and itsprojection dA onto a coordinate plane.

Arc length ds =√

(dy)2 + (dx)2 so s =∫ √√√dy

dx

2

+ 1 dx

Volume of revolution Cast so z-axis is axis of revolution & use cylindrical polarcoordinates

V =$

dV =$

Rr dr dφ dz = π

∫r2(z) dz

Surface of revolution Use above properties of surface integrals

S = 2π∫

r(z)

√(drdz

)2

+ 1 dz


First Year Vector Calculus Problem Sheet 4 2011Line Integrals


1. [Boas 6.8.1] Evaluate the line integral∫ (

x2 − y2)

dx − 2xy dy along each of the following paths from(0, 0) to (1, 2)

(a) y = 2x2

(c) y = 0 from x = 0 to x = 2; then along the straight line joining (2, 0) to (1, 2).

2. [Boas 6.8.2] Evaluate the line integral∮

(x + 2y) dx − 2x dy along each of the following closed paths,taken counterclockwise

(a) the circle x2 + y2 = 1

(b) the square with corners (1, 1), (−1, 1), (−1,−1), (1,−1)

3. [11.3(c)] F is a vector field xy2 i + 2 j + x k and L is a path parameterised by x = ct , y = c/t , z = d forthe range 1 ≤ t ≤ 2. Evaluate (c)

∫L F · dr.

4. Show that the line integral∫

C F · dr of the force F = y sin 2x i + sin2 x j is independent of the path C. Forthis force

(a) Find the function φ(x, y, z) such that F =∂φ

∂xi +

∂φ

∂yj +

∂φ

∂zk

(b) Evaluate explicitly∫

C F · dr where C is the straight line from (0, 0) to (π/2, 1)

(c) Verify that your answer is φ(π/2, 1) − φ(0, 0)


1

Vector Calculus: Problem Set 4 Line Integrals 2

Answers and Hints

1. Ans: (a) -376/3 (c) -11/3

2. (a) Parameterise in terms of the circular polar angle θ. Ans: −4π (b) Split into 4 line segments. Be surethat the direction of the integration (i.e., the limits corresponding to each of the 4 segments) go aroundthe path in the counterclockwise direction. Ans: -16

3. Ans: c4(ln 2) − c

4. Recall sin 2x = 2 sin x cos x and cos 2x = cos2 x−sin2 x = 1−2 sin2 x (a) Ans: φ = y(1−cos 2x)/2+C =y sin2 x + C (b) Ans: 1

Summary of Section 4: Line IntegralsBelow are various concepts and results from this section. Don’t quote a formula unless you know what it



Work done by force F dW = F · dr

Vector line element dr = dx i + dy j + dz k

Line integrals∫

C φdr,∫

C F · dr,∫

C B × dr, etc.

Parameterisation of C∫

CF · dr = Fx

dxdu

du + Fydydu

du + Fzdzdu

du

Path independence F · dr = Fx dx + Fy dy + Fz dz is an exact differential

F · dr =∂φ

∂xdx +

∂φ

∂ydy +

∂φ

∂zdz = dφ for some φ(x, y, z)∫ B

AF · dr = φ(B) − φ(A ) independent of path A → B∮

C F · dr = 0 for all closed paths C.


First Year Vector Calculus Problem Sheet 5 2011Vector Fields


1. [Boas 6.4.5] The position of a particle at time t is given by r = i cos t + j sin t + k t . Show that the particlespeed [|v|] and magnitude of the acceleration are constant. Describe the motion.

2. [Boas 6.6.1] Find the gradient of w = x2y3z at (1, 2,−1).

3. [after Boas 6.6.3] Find the derivative of xy2 + yz at (1, 2, 3) in the direction of the vector a = 2 i− j + 2 k.

4. [Boas 6.6.6] Find a vector normal to the surface x2 +y2−z = 0 at the point (3, 4, 25). Find the equationsof the tangent plane and normal line to the surface at that point.

5. In Section 3.5 of the course we studied the surface of revolution z2 = (x2 + y2)1/2. We found that thesurface element dS = N dx dy with

N =(−

x2ρ−3/2 i −

y2ρ−3/2 j + k

)with ρ =

√x2 + y2.

(a) Normalise N to find the unit normal n to this surface

(b) Consider the surface φ = z2 − (x2 + y2)1/2 = constant. Find ∇φ and hence calculate n = ∇φ/|∇φ|.Verify that for the surface φ = 0 your result agrees with that in (a).

6. BONUS Consider a function V (r , θ, φ) expressed in spherical polar coordinates. This question derivesan expression for ∇V in spherical polar coordinates,

(a) Write down the differential dV in terms of dr , dθ, and dφ from the general expression (the “mother”equation) for differentials

(b) Write down the vector path element dr in spherical polar coordinates

(c) Define ∇V by dV = ∇V · dr

(d) Let ∇V = A r + B θθθ + C φφφ and dot this with dr noting that spherical polar coordinates form anorthogonal system, i.e., r · r = 1, r · θθθ = 0, etc.

(e) Since dr , dθ, and dφ are independent, their coefficients in dV and in ∇V · dr must match. Hencefind an expression for ∇V in spherical polar coordinates.


1

Vector Calculus: Problem Set 5 Vector Fields 2

Answers and Hints

1. [Boas 6.4.5] Motion is a helix (sketch this—)

2. [Boas 6.6.1] −16 i − 12 j + 8 k

3. [Boas 6.6.3] Don’t forget to normalise a. 5/3

4. [Boas 6.6.6] 6x + 8y − z = 25; (x − 3)/6 = (y − 4)/8 = (z − 25)/(−1)

5. -

6. (b) dr = dr r + r dθ θθθ + r sin θ dφ φφφ (e) ∇V =∂V∂r

r +1r∂V∂θ

θθθ +1

r sin θ∂V∂φ

φφφ.

Summary of Section 5: Vector FieldsBelow are various concepts and results from this section. Don’t quote a formula unless you know what it



Differentiation of vector e.g., r(t) = x(t) i + y(t) j + z(t) k

drdt

=dxdt

i +dydt

j +dzdt

k

Vector differential dF =dFdu

du

Gradient ∇φ =∂φ

∂xi +

∂φ

∂yj +

∂φ

∂zk

Differential dφ = ∇φ · dr

Directional derivativedφds

= ∇φ · a

Normal to surface φ = const n = ∇φ/|∇φ|

Steve Schwartz10 Feb 2011

First Year Vector Calculus Problem Sheet 6 Corrected 2011Divergence


1. [Boas 6.7.7] Compute the divergence of the vector field V = x sin y i + cos y j + xy k

2. [Boas 6.7.19] For r = x i + y j + z k evaluate ∇ ·(

r|r|

). Do this in cartesian coordinates and then check

your answer by doing it more trivially in spherical polar coordinates, in which r = r r and so |r| = r .

3. We can expand

∇ ·

(r|r|

)=

1r∇ · r −

rr2 · ∇r

by using the fact that ∇ obeys the quotient rule for differentiation, and by being careful that the dotproduct is retained. Evaluate the right hand side and verify that you obtain the same answer as youfound in the preceding question.

4. [Boas 6.10.2] Given V = x2 i + y2 j + z2 k, integrate V · n dS ≡ V · dS over the whole surface of the cubeof side 1 with four of its vertices at (0, 0, 0), (0, 0, 1), (0, 1, 0), (1, 0, 0). [i.e., in the corner of the firstoctant of the cartesian system]. Evaluate the same integral by means of the divergence theorem.

5. By applying the divergence theorem to the vector field φc where c is an arbitrary constant vector,establish the result

#V ∇φ =

�S φdS.

6. [11.26] A vector field F is defined in cylindrical polar coordinates r , θ, z by

F =Fo ra

cos λz r + Fo sin λz k

(a) Calculate, as a surface integral, the flux of F through the closed surface bounded by the cylindersr = a and r = 2a and the planes z = ±aπ/2.

(b) Evaluate the same integral using the divergence theorem.

7. [11.19] Evaluate the surface integral!

r ·dS where r is the position vector, over that part of the surfacez = a2 − x2 − y2 for which z ≥ 0, by each of the following methods.

(a) Parameterise the surface as x = a sin θ cos φ, y = a sin θ sin φ, z = a2 cos2 θ and show that

r · dS = a4(2 sin3 θ cos θ + cos3 θ sin θ

)dθ dφ

(b) Apply the divergence theorem to the volume bounded by the surface and the plane z = 0.


1

Vector Calculus: Problem Set 6 Divergence 2

Answers and Hints

1. [Boas 6.7.7] 0

2. [Boas 6.7.19] 2/r

3. Do this in spherical polars. Recall from Problem Sheet 5 that ∇V =∂V∂r

r +1r∂V∂θ

θθθ +1

r sin θ∂V∂φ

φφφ in

spherical polars.

4. [Boas 6.10.2] 3

5. Be careful to keep the dot between c and ∇. You will find this done in many books, but try it firstyourself.

6. [11.26] Note dS points out of the closed surface. For the cylinder at r = a this is toward the z-axis, i.e.,in the direction of −r . See the handout on Objects of Revolution for a diagram of cylindrical polars fromwhich you can see that on a cylinder dS = ±r r dφ dz.

7. [11.19] Note the surface in question is an open surface. In (b) the divergence theorem will provideyou with a relationship between the surface integral in (a), the surface integral over the bottom z = 0surface [what is r · dS here?], and a simple volume integral (e.g., treating it as a volume of revolution).(a) dS = (2a3 cos θ sin2 θ cos φ i + 2a3 cos θ sin2 θ sin φ j + a2 cos θ sin θ k) dθ dφ. (b) ∇ · r = 3; over theplane z = 0, r · dS = 0. Ans to both (a) and (b) is 3πa4/2

Summary of Section 6: DivergenceBelow are various concepts and results from this section. Don’t quote a formula unless you know what it



Divergence ∇ · B =∂Bx

∂x+∂By

∂y+∂Bz

∂z

Laplacian ∇2φ =∂2φ

∂x2 +∂2φ

∂y2 +∂2φ

∂z2

Divergence Theorem

SB · dS =

$V∇ · B dV

Spherical Polar Coordinates ∇ · B =1r2

∂(r2Br )∂r

+1

r sin θ∂(sin θBθ)

∂θ+

1r sin θ

∂Bφ

∂φ

Cylindrical Polar Coordinates ∇ · B =1r∂(r Br )∂r

+1r∂Bφ

∂φ+∂Bz

∂z

Some reminders from Section 3:

Surface element If r(u, v) is on surface S then dS =(∂r∂u×∂r∂v

)du dv = N du dv =

n dS

Volume of revolution V =$

dV =$

Rr dr dφ dz = π

∫r2(z) dz

Steve Schwartz15 Feb 2011; 28 Feb 2011

First Year Vector Calculus Problem Sheet 7 2011Curl


1. [11.1] The vector field F is defined by F = 2xz i + 2yz2 j + (x2 + 2y2z − 1) k. Calculate ∇× F and deducethat F can be written F = ∇φ. Determine the form of φ.

2. [Boas 6.12.29] Evaluate

V · dr around the boundary of the square with vertices (1, 0), (0, 1), (−1, 0),(0,−1) if V = x2 i + 5x j. [Do this in the simplest way possible, e.g., either directly or by use ofGreen’s/Stoke’s Theorem. If you want, you might then attempt to do it the “harder” way.]

3. Verify the following vector operator identities, e.g., by expanding out each side in cartesian coordinates:

(a) ∇ × (φV) = φ (∇ × V) − V × ∇φ

(b) ∇ · (U × V) = V · (∇ × U) − U · (∇ × V)

(c) BONUS: ∇ × (∇ × E) = ∇ (∇ · E) − ∇2E. [see hint]

4. [Boas 6.11.3] Use Stoke’s Theorem to evaluate!

S ∇ × V · dS where V = x2 i + z2 j − y2 k and S is thepart of the surface z = 4 − x2 − y2 above the x − y plane.

5. [Boas 6.12.27] For the vector field V = (2−y) i+xz j+xyz k evaluate!

S ∇ × V ·dS over the open surfacebased on the cube of side 2 in the first octant that has as faces the x − z and y − z planes and threefurther faces parallel to the coordinate axes. (This is simply the cube in the corner of the first octant,with diagonal (0, 0, 0) → (2, 2, 2) but without the face in the x − y plane. It looks like an upside-downbox.) Again decide the “easy” way to do this. You might then also attempt the hard way.

6. A vector vield B given in cylindrical polar coordinates by B = Kr4 φφφ where K is a constant. Evaluate

explicitly both sides of Stoke’s Theorem�

CB · dr =

"S∇ × B · dS where S is a circle of radius a in

the plane z = 0 with centre at r = 0.

In questions where the orientation of the surface element dS isn’t given, choose dS to have a positivek component. This isn’t a problem for closed surfaces, where the convention is to choose dS outward, butfor open surfaces there isn’t, and can’t be, any consistent convention. Really the problem should specifyit. Evaluating both sides of Stoke’s Theorem will always give the same answer, but of course reversing theorientation of dS will introduce a minus sign into the surface integral, and traverse the bounding curve C inthe opposite sense to keep that traversal right-handed.


1

Vector Calculus: Problem Set 7 Curl 2

Answers and Hints

1. [11.1] Show that ∇ × F = 0. φ(r) = x2z + y2z2 − z (plus a constant!).

2. [Boas 6.12.29] 10

3. Brute force expansion of all these will produce the required results, being careful to keep track of boththe order of vector cross products and also what is, and is not, being differentiated. Note that theLaplacian of a vector is the vector each component of which is the Laplacian of the correspondingcomponent of the original vector, i.e., ∇2A = ∇2Ax i +∇2Ay j +∇2Az k. The BONUS question isn’t reallyharder than the others, just a bit easier to get lost.

4. [Boas 6.11.3] 0

5. [Boas 6.12.27] 4

6. In cylindrical polars, r = r r (φ) + z k, so that dr = dr r + r d r + dz k = dr r + r (d r /dφ) dφ + dz k =dr r + r dφ φφφ + z k (since r = cos φ i + sin φ j, it is easy to see that d r /dφ = φφφ). This is also what youwould see in a diagram if you increment r by dr , φ by dφ and z by dz. Be careful here, as the r incylindrical polar coordinates is the distance from the z-axis, NOT the distance from the origin (andNOT |r|). [As an aside, you might find it amusing to calculate dr in spherical polar coordinates this wayby writing r = r r (θ, φ) where in this case r IS |r|.] Answer: 2πKa5.

Summary of Section 7: CurlBelow are various concepts and results from this section. Don’t quote a formula unless you know what it



Curl ∇ × A =

∣∣∣∣∣∣∣∣∣∣∣i j k∂

∂x∂

∂y∂

∂zAx Ay Az

∣∣∣∣∣∣∣∣∣∣∣=

(∂Az

∂y−∂Ay

∂z

)i +

(∂Ax

∂z−∂Az

∂x

)j +

(∂Ay

∂x−∂Ax

∂y

)k

Green’s Theorem"

R

(∂Q∂x−∂P∂y

)dx dy =

�C

P dx + Q dy

Stoke’s Theorem�

CA · dr =

"S∇ × A · dS

Cylindrical Polar form ∇×A =

1r

∂Az

∂φ−∂Aφ

∂z

r +

∂Ar

∂z−∂Az

∂r

φφφ+

1r

∂

∂r

(rAφ

)−

1r

∂Ar

∂φ

k

Spherical Polar form ∇ × A =

1r sin θ

∂

∂θ

(Aφ sin θ

)−

1r sin θ

∂Aθ

∂φ

r +

+

1r sin θ

∂Ar

∂φ−

1r

∂

∂r

(rAφ

) θθθ +

1r

∂

∂r(rAθ) −

1r

∂Ar

∂θ

φφφ

Steve Schwartz22 Feb 2011

Differential Calculus - Bartholomew Andrews

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