Mathematics 1E2: Lecture 3
Sergei Fedotov
Sergei Fedotov
Mathematics 1E2: Lecture 3
1 / 15
More on area evaluation: In polar coordinates (r , θ)
r = f (θ)
r = f (θ)
2
δA = r2 dθ
A
dθ
θ1
θ0
We split the angular sector into equal size small angular sectors dθ.
The area of each small sector is δA = (r 2 /2)dθ
Z θ1 2
r
The same idea of Riemann sum leads to A =
dθ
θ0 2
Ex: Calculate the area trapped by the curve
r = θ between θ = 0 and θ = π (On board)
Sergei Fedotov
Mathematics 1E2: Lecture 3
2 / 15
More on area evaluation: a summary
Evaluate the area of a region bounded by
The x-axis,a curve y = f (x) for a ≤ x ≤ b
A=
b
Z
f (x) dx
a
The y -axis,a curve x = g (y ) for c ≤ y ≤ d
A=
Z
d
g (y ) dy
c
An angular sector θ0 ≤ θ ≤ θ1 and a curve r = f (θ)
A=
Sergei Fedotov
Z
θ1 2
r
θ0
2
Z
θ1
dθ =
θ0
(f (θ))2
dθ
2
Mathematics 1E2: Lecture 3
3 / 15
Arc length: how long is a curve?
y
ds
dy
dx
x
Pythagoras
570 BC - 495 BC
We want to find the length L of the blue curve.
We split the curve in equally sized straight elements ds.
Z s1
Letting the number of elements → ∞, we get L =
ds
s0
p
What is ds? Using Pythagoras, we get ds = dx 2 + dy 2
Sergei Fedotov
Mathematics 1E2: Lecture 3
4 / 15
Classical situations
Compute the length of a curve that is given by:
A function y = f (x)
s
ds =
p
dx 2
+
dy 2
=
1+
dy
dx
2
dx =
q
1 + (f 0 (x))2 dx
+ 1 dy =
q
1 + (g 0 (y ))2 dy
A function x = g (y )
p
ds = dx 2 + dy 2 =
s
dx
dy
2
A parametric curve x = g (t), y = f (t)
p
ds = dx 2 + dy 2 =
Sergei Fedotov
s
dx
dt
2
+
dy
dt
2
dt =
Mathematics 1E2: Lecture 3
q
(g 0 (t))2 + (f 0 (t))2 dt
5 / 15
Some practical examples (On the board)
Ex 1: Calculate the length of the curve given by y = cosh(x) from
x = −1 to x = 1.
Ex 2: Calculate the length of the parametric curve given by
x(t) = cos(t) and y (t) = sin(t) between t = 0 and t = π.
Sergei Fedotov
Mathematics 1E2: Lecture 3
6 / 15
The particular case of polar coordinates
What is the length of a curve defined by r = r (θ)?
Remember that in polar coordinates, we have x = r cos(θ) and
y = r sin(θ)
But in this case r = r (θ) so...
dx
dr
dy
dr
=
cos(θ) − r sin(θ) and
=
sin(θ) + r cos(θ)
dθ
dθ
dθ
dθ
q
This leads to ds =
r 2 + (r 0 (θ))2 dθ (On board)
and hence
L=
Sergei Fedotov
Z
θ1 q
r 2 + (r 0 (θ))2 dθ
θ0
Mathematics 1E2: Lecture 3
7 / 15
Volume of Revolution
y
y = f (x)
a
b
x
Consider a curve given by y = f (x) for a ≤ x ≤ b
rotate that curve about the x-axis to obtain a 3D object.
Questions
How big is the volume?
What is the lateral surface area?
Sergei Fedotov
Mathematics 1E2: Lecture 3
8 / 15
Calculation of the volume
y
y = f (x)
dx
f (x)
a
b
x
δV = π(f (x))2dx
Decompose the volume into cylinders of length dx and radius f (x)
The volume of each small element is δV = π(f (x))2 dx
Let the size of dx be smaller and smaller and add all the elements
together, we find that the total volume is given by
Z b
V=
π(f (x))2 dx
a
Sergei Fedotov
Mathematics 1E2: Lecture 3
9 / 15
Surface of a portion of a cone
l
r2
A
r1
Thales
624 BC - 546 BC
Consider a circular cone defined by its length L and end radius r ,
its lateral surface area S is given by S = πrL
So by taking the difference between two cones, the lateral surface A
of a portion of a cone defined by r1 , r2 and l is
A = S2 − S1 = π (r1 + r2 ) l
proof using the intercept (or Thales’) theorem (On board)
Sergei Fedotov
Mathematics 1E2: Lecture 3
10 / 15
Lateral surface area of volume of revolution
y
y = f (x)
ds
y + dy
2
a
y − dy
2
b
x
δA = 2πyds
Decompose the lateral surface into portion of cones of length l = ds
The two lateral radii are r1 = y − dy /2 and r2 = y + dy /2
So surface area of element is δA = 2πy ds
q
Remember that ds = 1 + (f 0 (x))2 dx
Finally the total surface area A is given by
Z b
q
A = 2π
f (x) 1 + (f 0 (x))2 dx
a
Sergei Fedotov
Mathematics 1E2: Lecture 3
11 / 15
Examples and work for next time...
Questions you should try
Small tutorial question “week 2”
Examples p44:
I
I
I
2(a), 2(b)
3(a), 3(b)
4(a), 4(b), 4(c)
Sergei Fedotov
Mathematics 1E2: Lecture 3
12 / 15