Polar coordinates: area by
integration
By Anonymous on Wednesday, March 14,
2001 - 02:15 pm :
Consider the curve r=acos(3q). It consists of three petals. If I
wanted to find the area of one of these, I could just integrate between
-p/6 to p/6 and get the required answer (using 1/2 r2 dq etc.)
Why can I not integrate between 0 and 2p/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 2p/3, you
will get 2 petals.
You get half of a petal from 0 to p/6, where r changes from a to 0.
Then, from p/6 to p/2, you actually get the petal on the other side,
as r < 0. From p/2 to 2p/3, you get another half of petal.
Thus, by integrating from 0 to 2p/3, you actually find twice the area of a
petal.
Kerwin
