86 Chapter 6. SLICING TECHNIQUES—FURTHER APPLICATIONS

Total mass =

∫ 6

0

ρ(x)dx =

∫ 2

0

x2dx +

∫ 3

2

3 dx +

∫ 6

3

4

x2dx

=x3

3

∣

∣

∣

∣

∣

2

0

+3 − 4

x

∣

∣

∣

∣

∣

6

3

=19

3kg.

Moment about origin =

∫ 6

0

xρ(x)dx

=

∫ 2

0

x · x2 dx +

∫ 3

2

x · 3 dx +

∫ 6

3

x · 4

x2dx

=x4

4

∣

∣

∣

∣

∣

2

0

+3

2x2

∣

∣

∣

∣

∣

3

2

+ 4 ln x

∣

∣

∣

∣

∣

6

3

=

(

23

2+ ln 16

)

kg · m.

Centre of mass =232 + ln 16

(

193

) m =3

38(23 + 2 ln 16) m.

(6.5) Length and area in polar coordinatesSH

6: 10.5,10.9.4

7: 9.6,9.9.3

8: 9.5,9.8.3

In this section, we show how to calculate area and arc

r

r

r=f

r l∆∆θ

(θ)

θ∆θ

∆lengths using polar coordinates. Whereas in cartesian co-ordinates it was natural to use rectangles to define area, inpolar coordinates the natural shape to consider is a wedge.

The area of the pictured wedge is

∆θ

2π× (area of circle of radius r) = 1

2r2∆θ.

So to find the area “enclosed” by the curve r = f(θ) for θ1 ≤θ ≤ θ2 we calculate

12

∫ θ2

θ1

r2 dθ.

The diagram also suggests

(∆`)2 ' (∆r)2 + (r∆θ)2

=

(

(∆r

∆θ

)2

+ r2

)

(

∆θ)2

,

so that arc length should be given by

∫ θ2

θ1

√

r2 +(dr

dθ

)2

dθ.

(6.5) Length and area in polar coordinates 87

Note that all the arc length formulæ we have seen can be obtained by manipulatingthe formal expression

arc length =

∫

√

(dx)2 + (dy)2.

For example, the connection between cartesian and polar coordinates is given byx = r cos θ, y = r sin θ, so that

dx = (−r sin θ +dr

dθcos θ) dθ

dy = (r cos θ +dr

dθsin θ) dθ.

Then√

(dx)2 + (dy)2 =(

(−r sin θ +dr

dθcos θ)2

+ (r cos θ +dr

dθsin θ)2

)1/2dθ

=(

r2 sin2 t − 2r sin θ cos θdr

dθ+

(dr

dθ

)2

cos2 θ

+ r2 cos2 t + 2r sin θ cos θdr

dθ+

(dr

dθ

)2

sin2 θ)1/2

dθ

=

√

r2 +(dr

dθ

)2

dθ.

Examples

(a) For the cardioid r = 1 + sin θ

(α) Calculate the area enclosed.

(β) Find the length of the curve.

Solution

(α) A = 2

∫ π2

−π2

1

2r2 dθ

=

∫ π2

−π2

(1 + sin θ)2dθ

=

∫ π2

−π2

(1 + 2 sin θ + sin2 θ) dθ

= π + 0 +

∫ π2

0

(1 − cos 2θ) dθ =3π

2

(β) ` = 2

∫ π2

−π2

√

(1 + sin θ)2 + cos2 θ dθ

88 Chapter 6. SLICING TECHNIQUES—FURTHER APPLICATIONS

= 2

∫ π2

−π2

√2 + 2 sin θdθ

= 2√

2

∫ π2

−π2

√1 + sin θdθ

= 2√

2

∫ π2

−π2

(cosθ

2+ sin

θ

2) dθ

N.B. the integrand is positive on[

− π

2,π

2

]

= 2√

2 · 2[

sinθ

2

]π2

−π2

= 2√

2 · 2 · 2√2

= 8.

* (b) Show that the area enclosed by the smaller loop of the

limacon r = 1 + 2 cos θ is2π − 3

√3

2.

(Note that we allow r to be negative in this example).

A =1

2

∫ 4π/3

2π/3

(1 + 2 cos θ)2 dθ

=

∫ π

2π/3

(1 + 4 cos θ + 4 cos2 θ) dθ

=

∫ π

2π/3

(3 + 4 cos θ + 2 cos 2θ) dθ

=[

3θ + 4 sin θ + sin 2θ]π

2π/3

=3π

3− 4

√3

2+

√3

2=

2π − 3√

3

2.

(c) Show that the area inside the circle r = 5 sin θ and

outside the limacon r = 2 + sin θ is8π

3+

√3.

Solution r = 5 sin θ is x2 + (y − 52 )2 = 25

4 . Let θ bethe angle where they intersect. Then 2+ sin θ = 5 sin θ.

So sin θ = 12 ; θ =

π

6,

5π

6.

A = 2

∫ π/2

π/6

1

2(r2

1 − r22) dθ

=

∫ π/2

π/6

(5 sin θ)2 − (2 + sin θ)2dθ

=

∫ π/2

π/6

(25 sin2 θ − 4 − 4 sin θ − sin2 θ) dθ

(6.6) Appendix for MATH1241 : Centres of Mass in General 89

= 4

∫ π/2

π/6

(6 sin2 θ − sin θ − 1) dθ

= 4

∫ π/2

π/6

(2 − 3 cos 2θ − sin θ)dθ =8π

3+

√3.

(d) Find the length of the parabolic spiral r = aθ2 betweenθ = 0 and θ = π .

` =

∫ π

0

√

r2 + (dr

dθ)2 dθ

=

∫ π

0

√

a4θ4 + (2aθ)2 dθ

= a

∫ π

0

√

θ4 + 4θ2 dθ

= a

∫ π

0

θ√

θ2 + 4 dθ

= a

[

(θ2 + 4)

3

3/2]π

0

= a( (π2 + 4)3/2 − 8

3

)

' .52278a.

* (6.6) Appendix for MATH1241 : Centres of Mass in General

The slicing idea can be extended in the following way to two-and three-dimensionalobjects with variable density.

* (6.6.1) Plane Regions (or Laminas) with Variable Density

Suppose that a lamina corresponding to a region R has

a x b

Sx

a variable density ρ(x, y) mass/unit area. If we slice theregion at x, the slice Sx is a line. It will contribute

M(x) =

∫

Sx

ρ(x, y)dy

to the total mass. Therefore the total mass is given by

M =

∫ b

a

M(x)dx.

The coordinates of the center of mass of the lamina are given by

x =

∫ b

a

(

∫

Sxxρ(x, y)dy

)

dx

M

y =

∫ b

a

(

∫

Sxyρ(x, y)dy

)

dx

M.

90 Chapter 6. SLICING TECHNIQUES—FURTHER APPLICATIONS

Compare these formulae with the formula for the x-coordinate of the centre of mass of arod (see section 6.4). The expressions

∫

Sx

x ρ(x, y)dy and

∫

Sx

y ρ(x, y)dy

are called the moments of the lamina about the y−axis and the x−axis respectively. Ifthe density ρ(x, y) is constant throughout the lamina then the center of mass is called thecentroid of the lamina.

*Example

π

4

x

1

y (

π

4,

1√2

)

y = sin x

y = cos x

Sx

A lamina has the shape of the regionin the first quadrant that is bounded by thegraphs y = sin x, y = cos x, between x = 0and x = π

4 . The density of the lamina isgiven by ρ(x, y) = y . Find the coordinatesof the centre of mass.

The mass of the strip Sx is

M(x) =

∫

Sx

ρ(x, y) dy =

∫

Sx

y dy

=

∫ cos x

sin x

y dy =

[

y2

2

]cos x

sin x

=1

2

(

cos2 x − sin2 x)

,

so the total mass is

M =

∫ π4

0

M(x) dx =

∫ π4

0

1

2

(

cos2 x − sin2 x)

dx

=1

2

∫ π4

0

cos 2x dx =1

2·[

1

2sin 2x

]π4

0

=1

4.

The x−coordinate of the center of mass is given by

x =

∫ π4

0

(

∫

Sxx · y dy

)

dx

M

=

∫ π4

0

(∫ cos x

sin xxy dy

)

dx

1/4

= 4

∫ π4

0

[

xy2

2

]y = cos x

y = sin x

dx

= 2

∫ π4

0

x cos 2x dx =π − 2

4.

(6.6) Appendix for MATH1241 : Centres of Mass in General 91

Similarly,

y =

∫ π4

0

(

∫

Sxy · y dy

)

dx

M

= 4

∫ π4

0

(∫ cos x

sin x

y2dy

)

dx

= 4

∫ π4

0

[

y3

3

]y = cos x

y = sin x

dx

=4

3

∫ π4

0

(

cos3 x − sin3 x)

dx

=4

3

∫ π4

0

(

cos x(1 − sin2 x) − sin x(1 − cos2 x))

dx

=4

3

[

sinx − sin3 x

3+ cos x − cos3 x

3

]

π4

0

=10√

2 − 8

9.

Hence the center of mass has the coordinates

(

π − 2

4,

10√

2 − 8

9

)

.

* (6.6.2) Three-Dimensional Objects with Variable Density

Suppose that a closed and bounded three-dimensional

xb

a

y

S

S

x

xy

object has a variable density ρ(x, y, z) mass/unit volume.The slice through x will be a plane region Sx and we cancompute its contribution M(x) to the total mass by the samemethod as we used to calculate the mass of a lamina: theslice of Sx through y will be a line Sxy which contributes

m(x, y) =

∫

Sxy

ρ(x, y, z)dz

to M(x), so

M(x) =

∫

Sx

m(x, y)dy

and the total mass of the object is

M =

∫ b

a

M(x)dx.

The coordinates of the center of mass of the object are given by

x =

∫ b

a

(

∫

Sx

(

∫

Sxyx ρ(x, y, z)dz

)

dy)

dx

M,

y =

∫ b

a

(

∫

Sx

(

∫

Sxyy ρ(x, y, z)dz

)

dy)

dx

M,

z =

∫ b

a

(

∫

Sx

(

∫

Sxyz ρ(x, y, z)dz

)

dy)

dx

M,

92 Chapter 6. SLICING TECHNIQUES—FURTHER APPLICATIONS

where the multiple integrals on the numerator of x, y and z are moments of the objectabout the coordinate planes yz, xz and xy.

Note that the centre of mass is independent of the choice of origin. This means thata clever choice of origin can allow the symmetry of a body to be exploited so as to reducethe amount of calculation required to find the centre of mass.