
Archimedean Spiral
I have been given this question
The Archimedean spiral is given by the polar equation
r= C(theta) where C is a constant.
Find the area when theta is between 0 and 2pi?
Does this mean that i just use the area of a circle equation A= pi r^{2}
So the area will be pi(2pi^{2})pi(0)= pi(2pi^{2})

Re: Archimedean Spiral
Well, this has nothing to do with a circle so I find that reference very strange.
What is true is that if $\displaystyle r= f(\theta)$ and the graph does not overlap itself, the area is $\displaystyle \int_0^{2\pi} \int_0^{f(\theta)} r drd\theta= \frac{1}{2} \int_0^{2\pi} f^2(\theta)d\theta$.
Here $\displaystyle r= f(\theta)= C\theta$ so the area is $\displaystyle \frac{C^2}{2}\int_0^{2\pi} \theta^2 d\theta= \frac{C^2}{2}\frac{8\pi^3}{3}$.

1 Attachment(s)
Re: Archimedean Spiral
Hi,
No, you need to use the formula for the area of a polar curve:
Oops, I just noticed I said c=2, but the picture shows c=0.7
Attachment 30418

Re: Archimedean Spiral
Hi again,
HallsOfIvy in post 2 makes an important observation. The formula for polar areas $A={1\over2}\int_\alpha^\beta r(\theta)^2\,d\theta$ is applicable only when exactly one region is "swept out" for $\theta\in[\alpha,\beta]$. Here a region is swept out as follows: imagine a ray from the origin to a point $(r(\theta),\theta)$ on the curve. As $\theta$ varies from $\alpha$ to $\beta$, this ray "sweeps out" the region.
Example: You want the area of the spiral $r(\theta)=c\theta$ for $\theta\in[0,6\pi]$. As $\theta$ varies from 0 to $6\pi$, the region from 0 to $2\pi$ is swept out three times and that from $2\pi$ to $4\pi$ is swept out twice. So the desired area is not ${1\over2}\int_0^{6\pi}c^2\theta^2\,d\theta$, but is the integral ${1\over2}\int_{4\pi}^{6\pi}c^2\theta^2\,d\theta$
This is really why polar areas are tricky. I think it helps to see how a polar curve is drawn dynamically. A good calculator will allow you to watch this. For example, you can watch the spiral above as it is drawn at https://www.desmos.com/calculator