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This is a paragraph and a figure from The Feynman Lectures, He's trying to prove the Shell theorem for gravity: enter image description here

If we rearrange $dm=2\pi y \mu ds$ we obtain $\dfrac{dm}{2\pi yds}=\mu$. In this arrangement this equation is more intuitive to me. It says that the mass distribution $\mu$ of the ring equals its mass $dm$ divided by its area $2\pi yds$, Where $y$ is its radius and $ds$ is its thickness.(this is similar to the famous differential equation: $dA=2\pi r dr$, Where the infinitesimal area of a ring, equals the Circumference of the circle with radius $r$ multiplied by its thickness $dr$).

My question is in the paragraph, Feynman says that it's $dx$ not $ds$ that is the thickness of the ring, but the equation: $dm=2\pi y \mu ds$ implies that $ds$ is the thickness of the ring.Moreover, he says that $ds=\dfrac{dx}{sin\theta}$(compare the second and third term in the paragraph).

So what exactly are $ds$ and $dx$?

And Where does this relation between $ds$ and $dx$ come from?

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  • $\begingroup$ @diracpaul My problem is this, assuming that the angle between the radius $a$ and $ds$(look at Floris') is 90, is tantamount to stating that $ds$ is tangent on the circle. Tangent is defined as intersecting the circle at one point,But since $ds$ is a arc(very small) of a circle then it intersects with the circle at infinitely many points. $\endgroup$
    – Omar Nagib
    Jul 18 '15 at 11:02
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This answer has images only (An image is equivalent to 1000 words)

enter image description here

If you have glasses red-left / blue-right see image below in black & white 3D.

enter image description here

and moreover a colored 3D.

enter image description here

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  • $\begingroup$ I had one software in my previous laptop; I couldn't remember the name, yes "inkscape"! It was good but intricate. Thanks for commenting. Keep up doing this awesome job of helping others with the art that speaks more than 1000 words! $\endgroup$
    – user36790
    Jul 18 '15 at 13:11
  • $\begingroup$ @diracpaul - thanks; but somehow I was not getting through to Omar... maybe looking at it from a different angle will have helped. Meanwhile I learnt about a new tool - something to complement my powerpoint drawings. $\endgroup$
    – Floris
    Jul 20 '15 at 6:36
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$dx$ is measured along the $OP$ axis and $ds$ is measured along the surface. The relationship between them follows from simple geometry - $ds$ is at an angle $\pi/2-\theta$ to $dx$.

enter image description here

From this diagram, $\frac{dx}{ds}=\sin\theta$. The result follows from rearranging.

$ds$ is the width of a strip of the shell of constant surface density: $dx$ is the thickness of a shell measured along the $OP$ direction.

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