Thread: polar coordinates question

(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)
/4

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,

Thanks,

Bryn

That's the domain of theta. The function is a function of 3 theta.

Hi,

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...

Thanks, could you elaborate on what you mean by it being the domain, I'm not understanding its relevance in this polar coordinates.

thanks again

Firstly remember for polar coordinates that the $\displaystyle r$ value can be negative, a negative value of r just means it is going 'backwards' so you should appreciate that for example $\displaystyle (-r , \pi /2)$ is the same as $\displaystyle (r , 3 \pi /2)$. To get a rought idea of how r is varying with theta I suggest that you draw the curve $\displaystyle y = \cos 3x$ form 0 to $\displaystyle \pi$.

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 $\displaystyle 6 \times 1/2 \int_{0}^{\frac{\pi}{6}} r^2 d \theta$

Bobak