# Feynman's proof for Newton's shell theorem [closed]

I have two questions concerning this proof:

Firstly, what is the difference between the increments ds and dx? Are they not just the same thickness of the strip?

Secondly, why can the integral limits be the wrong way round without flipping the sign?

## closed as off-topic by GiorgioP, glS, HDE 226868, ZeroTheHero, mpvJun 26 at 12:49

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You can think of $$dx$$ as the projection of $$ds$$ onto the line $$OP$$.

The effect is most noticeable as $$dx$$ moves along line $$OP$$ towards the right.

There is an equation for $$dW$$ in terms of $$dx$$ and if you were to integrate that equation you would set the limits from $$x=-a$$ to $$x=+a$$.

With a change of variable to $$r$$ by using $$r^2 = y^2 +(R-x)^2$$ and $$y=0$$ the limits change to $$r=R+a$$ and $$r=R-a$$.

• Thank you, but why is the upper limit the smaller quantity (R-a) and not the lower limit? – Physics May 20 at 16:06
• @Physics That is what you get when changing the variable from $x$ to $r$ as stated in the last part of my answer. – Farcher May 20 at 16:09

1) If you study the image more closely, you see that ds is length along the sphere, while dx is the horizontal thickness. Because the incremental piece is inclined, ds and dx are different distances.