Mentor

## Area of an Archimedian spiral

 Quote by 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!
Of course, it's not necessary.

Do you really want to see it?
 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.

Mentor
Oh, the last 16 should be 8, right.

 Is the double integral approach what mfb showed?
Right.

 Quote by mfb Where did you find the error?
##\pi \theta^2 + 2\pi^2 \theta## evaluated between ##2\pi## and ##4\pi## is $$16\pi^3 + 8\pi^3 - 4\pi^3 - 4\pi^3$$

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 Quote by 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.
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:

$\displaystyle \int_{2\pi}^{4\pi}\int_{\theta}^{\theta+2\pi}r\,dr\,d\theta$

Mentor
 Quote by 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.
You just do the first integration step ("area, ΔA, of this washer") before writing down the integral over θ.
 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 ?
 Mentor 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: $$\int_0^{2\pi} d\theta \int_{2\pi+\theta}^{4\pi+\theta} r dr$$

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 Quote by 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 ?
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 .