That's the domain of theta. The function is a function of 3 theta.
(a) Sketch the curve with polar equation r = a cos 3θ , a > 0, for 0 ≤ θ ≤ pi.
(b) Show that the total area enclosed by the curve r = a cos 3θ is (pi.a^2)
I am able to draw the curve although the answer drew it all the way from 0 to 2pi which I thought didn't make sense given the limits are 0<theta<n - for part b it also finds the area of all 3 of the leafs rather than half which I thought made sense given the limits,
This is the same problem as here : http://www.mathhelpforum.com/math-he...lar-curve.html
Though I didn't answer to the main problem yet, because I only have an informal idea of it...
Firstly remember for polar coordinates that the value can be negative, a negative value of r just means it is going 'backwards' so you should appreciate that for example is the same as . To get a rought idea of how r is varying with theta I suggest that you draw the curve form 0 to .
Follow the above carefully and you should be able produce a diagram similar to what moo posted on the other thread. then try and understand why the area is