Polar coordinates: area by integration

By Anonymous on Wednesday, March 14, 2001 - 02:15 pm :

Consider the curve

r = a \cos (3 θ)

. It consists of three petals. If I wanted to find the area of one of these, I could just integrate between

- π / 6

π / 6

and get the required answer (using

1 / 2 r^{2} d θ

etc.) Why can I not integrate between 0 and

2 π / 3

? Surely that consists of a petal too?
Thanks for you help.

By Kerwin Hui (Kwkh2) on Wednesday, March 14, 2001 - 03:20 pm :

The reason is that if you integrate from 0 to

2 π / 3

, you will get 2 petals.

You get half of a petal from 0 to $π / 6$ , where $r$ changes from $a$ to 0. Then, from $π / 6$ to $π / 2$ , you actually get the petal on the other side, as $r < 0$ . From $π / 2$ to $2 π / 3$ , you get another half of petal.

Thus, by integrating from 0 to $2 π / 3$ , you actually find twice the area of a petal.

Kerwin

$- π / 6$ to $π / 6$ 0 to $2 π / 3$

0 to 2pi/3