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Link to derivation: www.phys.uri.edu/~gerhard/PHY204/tsl36.pdf

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- Thread starter davezhan
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Link to derivation: www.phys.uri.edu/~gerhard/PHY204/tsl36.pdf

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berkeman

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Link to derivation: www.phys.uri.edu/~gerhard/PHY204/tsl36.pdf

How would you write the equation for the area of that ring section? What happens when you simplify what you've written?

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fluidistic

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HallsofIvy

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Here's another way to look at it: Imagine opening that strip up to a "rectangle". It's length is the circumference of the circle, [itex]2\pi a[/itex], and it's width is da. The area of that "rectangle" is "length times width", [itex]2\pi a da[/itex]. I have put "rectangle" in quotes because, of course, you cannot "open up" a circular strip into a rectangle. This is, again, only true in the

If you were to take da to be any finite length, [itex]2\pi a da[/itex] would give you an approximate area, not an exact area. But you can use "da" in an integral to get the exact area.

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Thank you for your help! The above post helped to clarify things tremendously.

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