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- Thread starter Periapsis
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Office_Shredder

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So to integrate anything, you need a theoretical higher exponent type dimension? Then why can you take a two dimensional surface area and integrate it to find the volume? V = ∫ f(r) 4πr

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Office_Shredder

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theoretical higher exponent type dimension?

I have no idea what these words in succession mean.

Then why can you take a two dimensional surface area and integrate it to find the volume? V = ∫ f(r) 4πr^{2}dr right? Integrating that would result in a cubed radius?

You can take a two dimensional surface area and integrate it to find the volume, in the sense that you can integrate the surface area of a sphere to get the volume of a ball. I don't know what your function f(r) is supposed to be here - if it's completely arbitrary, then you're just integrating an arbitrary function.

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Office_Shredder

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If you have a different definition then we can work through how one is a derivative of another.

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Office_Shredder

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s'(t) = v(t)

s''(t) = v'(t) = a(t)

s'''(t) = v''(t) = a'(t) = j(t)

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pwsnafu

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Okay, now I understand that, thanks. But now what about using that same principal to solve for anything? What is circumference a function of?

C=2πr. So the only candidate is radius.

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