# Area of an Archimedian spiral

1. Dec 18, 2012

### CAF123

1. The problem statement, all variables and given/known data

3. The attempt at a solution

I considered each shaded part separately. For the one between $2\pi$ and $4\pi$, I set up: $$\int_{2\pi}^{4\pi} ((4\pi +\frac{\pi}{2})- (2\pi + \frac{\pi}{2}) d\theta$$ I used a similar method for the other portions and for each portion, I end up with $4\pi^2$. This does not seem right because some of the shaded sections look larger than the others.

Where is the fault with my method?

Thanks!

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• ###### Arc.Spiral 002.jpg
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2. Dec 18, 2012

### Staff: Mentor

I would expect that your integration variable depends on θ in some way. You just integrate over 2 pi (why?).

3. Dec 18, 2012

### CAF123

In considering the first shaded section (between 2pi and 4pi), I say that the curves $r = \theta_1 = (4\pi + \pi/2)$ and $r = \theta_2 = (2\pi + \pi/2)$ bound this section.

4. Dec 18, 2012

### CAF123

I see what I have done wrong. $\theta$ is continually changing so indeed there should be a theta dependence. How do I get the curves that bound each section then?

5. Dec 18, 2012

### HallsofIvy

I think you you integral gives the area between two concentric circles of radii $2\pi+ \pi/4= 9\pi/4$ and $4\pi+ \pi/2= 9\pi/2$.

The two arms of the spiral, at the start where $\theta= 0$, a ray starts at $2\pi$ and ends at $4\pi[/b], but as [itex]\theta$ increases, so does the distance from the origin to those endpoints. On has $r= 2\pi+ \theta$

I confess that at first, I thought you were, in fact, using an integral that would give the distance between two concentric circles, but what you are doing is, basically, correct. At the beginning, $\theta= 0$, the x-axis cuts the first loop of the spiral at $2\pi$ and the second at $4\pi$ so its length is $4\pi- 2\pi= 2\pi$. As $\theta$ increases, a ray crosses the first loop at $2\pi+ \theta$ and the second at $4\pi+ \theta$ but, to my surprise, the difference is still $2\pi$- there is NO $\theta$ dependence. I don't see where your $2\pi$ terms come from. And I certainly cannot imagine why you are multiplying them!

Since we make one complete loop around the center, the integral does go from $\theta= 0$ to $2\pi$. Since area, in polar coordinates, is given by $\int rd\theta$, the area you want here is given by
$$\int_{\theta= 0}^{2\pi} 2\pi d\theta= 2\pi \int_{\theta= 0}^{2\pi} d\theta$$

6. Dec 18, 2012

### CAF123

Where am I multiplying them about?
Why do we consider theta from 0 to 2pi? this does not correspond to a shaded section?
Thanks.

7. Dec 18, 2012

### Staff: Mentor

The integral itself does not work. In polar coordinates, area is given by r dr dθ. You can perform the integration over r first (this is a good idea), but the result is not just the difference in radius, due to the additional factor of r inside. You will get expressions with the squared radius.

8. Dec 18, 2012

### CAF123

So we consider a double integral here? why? Would the limits of r and θ be the same?

9. Dec 18, 2012

### Staff: Mentor

You want to calculate an area - it is two-dimensional. In a cartesian coordinate system, many problems have a trivial integral in one coordinate - you can simplify the two-dimensional integral to one dimension without calculating anything.
In polar coordinates, this can work for θ, but it cannot work for r, as the area element depends on r.
The limits for r and θ are fine.

10. Dec 18, 2012

### LCKurtz

The polar area between the origin and $r = f(\theta)$ for $\theta$ between $\alpha$ and $\beta$ is$$A=\frac 1 2\int_\alpha^\beta f^2(\theta)\, d\theta$$The tricky thing about your problem is some of the area gets repeated as $\theta$ varies. If you let $\theta$ go from $4\pi$ to $6\pi$ you would get all the area that is shaded plus the white center area. Then it looks to me like you could subtract that white area by going from $2\pi$ to $4\pi$, unless I'm overlooking something.

11. Dec 18, 2012

### CAF123

Would you not subtract from $0$ to $4\pi$?

12. Dec 18, 2012

### CAF123

Could you explain a bit more what you mean here? I have done integrals that represent areas as single integrals, (I.e at high school level).

13. Dec 18, 2012

### LCKurtz

That counts some of the white area more than once, no? If you shade that inside area for $\theta$ sweeping from $0$ to $2\pi$, that same area is swept out again as $\theta$ goes from $2\pi$ to $4\pi$.

14. Dec 18, 2012

### CAF123

15. Dec 18, 2012

### SammyS

Staff Emeritus
For one thing, your reference to the shaded part "between $2\pi$ and $4\pi$" is at best ambiguous and at worst it makes no sense.

The spiral from θ = 2π to θ = 4π forms the "inner" boundary of the shaded region.

You can do this nicely with an iterated (double) integral in polar coordinates. (I assume we don't want that approach here.)

Consider a washer with inner radius, r, and outer radius, R. The area, ΔA, of this washer between θ and θ + Δθ is $\displaystyle \Delta A=(1/2)R^2\Delta\theta-(1/2)r^2\Delta\theta$

For the spiral in this problem, let r = θ and R = θ + 2π .

Integrate from θ = 2π to 4π .

16. Dec 18, 2012

### CAF123

Thanks for this method. Yes, I agree that what I wrote in the OP is nonsensical since θ is continually changing. I wouldn't mind seeing the double integral approach, I just don't see why it is necessary? Thanks!

17. Dec 18, 2012

### Staff: Mentor

As I said, one integral there was trivial.

If you want to calculate the area under the curve f(x) from a to b, you can express this as $$\int_a^b dx \int_0^{f(x)} 1 dy = \int_a^b dx (f(x)-0) dy= \int_a^b f(x) dx$$ which is the usual one-dimensional integral.
"1" comes from the cartesian coordinate system. It allows to perform the inner integral without writing it down, as it is simply the difference between the upper and lower bound.
In polar coordinates, you cannot do this with the integral over r.

$$\int_{2\pi}^{4\pi} d\theta \int_\theta^{\theta+2\pi} r dr = \int_{2\pi}^{4\pi} \frac{1}{2}\left((\theta+2\pi)^2 - \theta^2\right)d\theta = \int_{2\pi}^{4\pi} (2\pi\theta+2\pi^2)d\theta = 4\pi^3 + 2 \pi (16\pi^2 - 2\pi^2) = 32\pi^3$$

Hmm, looks quite big. Maybe I made an error somewhere.

18. Dec 18, 2012

### SammyS

Staff Emeritus
Of course, it's not necessary.

Do you really want to see it?

19. Dec 19, 2012

### CAF123

Hi SammyS,
What I meant by 'why is it necessary' was why the need for a double integral? Although mfb has cleared that up. Is the double integral approach what mfb showed? If not, yes, I would like to see it.
Thanks.

@mfb the answer is $16\pi^3$. But I do see your error. Thanks.

20. Dec 19, 2012

### Staff: Mentor

Oh, the last 16 should be 8, right.

Right.