Area of an Archimedian spiral

SammyS

Of course, it's not necessary.

Do you really want to see it?

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.

mfb

Oh, the last 16 should be 8, right.

Right.

CAF123

##\pi \theta^2 + 2\pi^2 \theta## evaluated between ##2\pi## and ##4\pi## is $$16\pi^3 + 8\pi^3 - 4\pi^3 - 4\pi^3$$

SammyS

That's what I thought you meant ... and the answer is that this can be done without a double integral, so using a double integral is not necessary.

Since you want to see the double integral solution, here it is:

[itex]\displaystyle \int_{2\pi}^{4\pi}\int_{\theta}^{\theta+2\pi}r\,dr\,d\theta[/itex]

mfb

You just do the first integration step ("area, ΔA, of this washer") before writing down the integral over θ.

CAF123

Actually, could you explain how you get those limits: I think the θ limits are from 2π to 4π because that is the 'inner' curve that bounds the shaded section. Correct? Why is r from θ to θ + 2π ?

mfb

θ=r is the inner part of your spiral (starting at 2pi and increasing to 4pi), and the outer part is the next winding 2 pi away, so θ=r+2pi (starting at 4pi and increasing to 6pi).

You can change the limits, if you like, for example like this:
$$\int_0^{2\pi} d\theta \int_{2\pi+\theta}^{4\pi+\theta} r dr$$

SammyS

The θ limits are indeed from 2π to 4π. That's the outer integral.

As for

"Why is r from θ to θ + 2π ?"

Look at your figure. r goes from the inner boundary to the outer boundary. The inner boundary is at r = θ . To get to the outer boundary, r = θ + 2π .