- #1

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It turns out that the correct formula is actually

[itex]\int^{β}_{α}\sqrt{r^{2}+(\frac{dr}{dθ})^{2}} \ dθ [/itex]

I know how the correct formula is derived, I just can't figure out why the reasoning for the first formula is incorrect.

BiP

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- Thread starter Bipolarity
- Start date

- #1

- 775

- 1

It turns out that the correct formula is actually

[itex]\int^{β}_{α}\sqrt{r^{2}+(\frac{dr}{dθ})^{2}} \ dθ [/itex]

I know how the correct formula is derived, I just can't figure out why the reasoning for the first formula is incorrect.

BiP

- #2

- 36,175

- 6,795

By Pythagoras, ds^{2} = (rdθ)^{2}+dr^{2}. You can't go ignoring the dr term in general.

- #3

- 775

- 1

By Pythagoras, ds^{2}= (rdθ)^{2}+dr^{2}. You can't go ignoring the dr term in general.

My question was, why can't you use the arc length of the sector?

BiP

- #4

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Because in general the radial movement is of comparable magnitude to the tangential movement. In the extreme case, part of the curve might move directly away from the origin, so there dθ would be zero and you would only have dr.My question was, why can't you use the arc length of the sector?

BiP

- #5

arildno

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