polar coordinates question

**Bryn** · Jun 15th 2008, 10:01 AM

(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

**sean.1986** · Jun 15th 2008, 10:04 AM

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

**Moo** · Jun 15th 2008, 10:05 AM

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...

**Bryn** · Jun 15th 2008, 10:23 AM

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

**bobak** · Jun 15th 2008, 12:02 PM

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

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

Bobak

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