Problem: Show that the area of the rose curve r = a sin(nθ) does not

depend on the value of n.

Solution:

Since the number of petals of the rose curve depends on the number n (that

being there r will be an n petaled rose curve if n is odd and a 2n petaled

rose curve if n is even (if the reader wants to confirm this for themselves they

should visit http://www.shodor.org/interactivate/activities/PolarCoordinates/

and test the number of petals for different values of n)) we must investigate the

area of the rose curve in two cases: the first for even n and the second for odd

n.

CASE 1: n is even. Therefore there are 2n petals spaced out over 2π

radians making one petal over each πn radians. Since r(0) = 0 the first petal is

mapped from θ = 0 to θ = πn . Finding the area of the petal produced by r from

θ = 0 to θ = πn produces the integral:

Area =1

2

∫ πn

0

r2dθ =1

2

∫ πn

0

a2 sin2(nθ)dθ =

a2

2

∫ πn

0

1

2(1 − cos(2nθ))dθ =

a2

4

(θ − sin(2nθ)

2n

)∣∣∣∣πn0

=a2

4

π

n

If this is the area of the first of 2n petals then the total area of the rose curve

is...

Area = 2n · a2

4

π

n=a2

2

So the area is half the area of circle of radius a. Let us now investigate the

case when n is odd.

CASE 2: n is odd. Therefore there are n petals spaced out over 2π radians

making one petal over each 2πn radians. Since r(0) = 0 the first petal is mapped

from θ = 0 to θ = 2πn . Finding the area of the petal produced by r from θ = 0

to θ = 2πn produces the integral:

Area =1

2

∫ 2πn

0

r2dθ =1

2

∫ 2πn

0

a2 sin2(nθ)dθ =

a2

2

∫ 2πn

0

1

2(1 − cos(2nθ))dθ =

a2

4

(θ − sin(2nθ)

2n

)∣∣∣∣πn0

=a2

4

2π

n

If this is the area of the first of n petals then the total area of the rose curve

is...

Area = n · a2

4

2π

n=a2

2

Again we get the same Area a2

2 . So the area of the rose curve r = a sin(nθ)

does not depend on the value of n.