FP2 Polar Integration

The integral runs between those values because it's where theta is defined. Remember you're integrating w.r.t theta, and the equation is defined for $0 \leq \theta \leq 2\pi$. I think with your method you mistakenly assumed that the graph is symmetric about the axes, although if this is wrong I'm sure someone else will point it out. Does this make anything any clearer?

I didn't mean the initial line, I meant equal areas between 0 and pi/2 and pi/2 and pi which I don't think is the case. I think what you've just said there is more a matter of convention. An angle clockwise through 3pi/2 is exactly the same as an angle anti-clockwise through pi/2 (i.e. -pi/2). Convention I think is that theta should lie between -pi and pi, although I could be wrong...

I think with your method you mistakenly assumed that the graph is symmetric about the axes, although if this is wrong I'm sure someone else will point it out.

I think you'll find it is. Note that the curve with polar equation $r=f(\theta)$ is symmetrical about the line $\theta=\phi$ if $f(2\phi - \theta) \equiv f(\theta) ; \forall \theta \in \mathbb{R}$. Taking $\phi = \dfrac{\pi}{2}$ and $\phi=0$ confirms that this symmetry holds.

Would you be able to explain why they have just integrated between 0 and 2pi and why this has given the correct answer?

Thanks

I think it's because r>0 (please note, I'm no expert on polar coordinates). My reasoning is simple, think about the integration of cartesian functions f(x), the only time when the areas may cancel is if f(x)<0. This would be the case here if r<0 but since r>0 for all theta, this problem never arises and thus we can integrate between any limits we like.

I think it's because r>0 (please note, I'm no expert on polar coordinates). My reasoning is simple, think about the integration of cartesian functions f(x), the only time when the areas may cancel is if f(x)<0. This would be the case here if r<0 but since r>0 for all theta, this problem never arises and thus we can integrate between any limits we like.

But, even though we don't take into account anything for r<0, doesn't the integration take it into account? I mean, it seems as if it worked out that time, but I have a feeling it wouldn't always if I just integrated between 0 and 2 pi.

Oh, if anyone can answer my question above (with the other attachment) I'd be grateful.

That integral will give you the magnitude of that bounded area because r>0 across the interval you're integrating over. If the example wasn't as simple as this one and there were regions where r<0 then you just have to find the intervals of theta over which r<0 and deal with them accordingly.

As for your other question, I think we have the problem of $(r,\theta)=\left( -a\sqrt{\dfrac{\sqrt{3}}{2}}, \dfrac{\pi}{6}\right)$ implying that r<0, which is problematic. Not completely sure though because looking at the graph, the other point works fine. Best wait until tomorrow so someone more knowledgable can comment.

Eughh yet another question.. this isn't going well. I understand that r = a represents a circle of radius a. But I've now come across the equation r = 4cos(theta), which also apparently describes a circle, can someone explain that to me and tell me the general form of a circle?

A circle with a centre of $(r,\theta) = (r_0, \theta_0)$ and radius a is given by:

$r^2-2rr_0\cos (\theta - \theta_0)+r_0^2=a^2$.

$r=a$ is a circle with a centre at the pole of radius a and $r=4\cos \theta$ is a circle with centre $(2,0)$ and radius 2 i.e. $r_0=2, \theta_0=0, a=2$. Sub them in and rearrange.