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While reading this chapter of the Feynman Lectures I came across a statement I didn't know how to prove.

He mentions below Eq. 4.30 that when you take a surface and tilt it by some angle $\theta$, the area of the new surface is increased by a factor of $\frac{1}{cos\theta}$.

Of course, simply changing the orientation of a surface does not change its area. What he means is that you cut the cone at a different angle and that the new surface is still sufficiently close to the original so you can use the same E. (They may for example share a vertex.)

I think this rule only works if the original surface is a spherical one like the ones mentioned above Eq. 4.30 although it's not said explicitly. And that it only works for infinitesimal surface areas too.

So my question is, how do I prove this? Tips, explanations and the proof itself are all welcome.

Thanks for reading.

m009
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    Thank you for bringing to my attention CalTech's beautiful, free, online copy of the Feynman Lectures! I had no idea this existed and am definitely sharing it with students! – Dan Jun 13 '14 at 23:24

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It can be shown in a precise way but I will try to give the intuation behind this.

If you understand the reason for two dimensional object, it become more easy.

Now, $L_2$ is a line and $L_1$ is a projection of it to $x$ axes then you can say that,

$$|L_2|=|L_1|\dfrac{1}{cos(\theta)}$$

where $\theta$ the angle between the $L_2$ and $L_1$.

Now, instead of a line piece, let have a piece of plane $P_2$ and its projection to $x,y$ plane $P_1$.

It is a little bit hard to see but $$area(P_2)=area(P_1)\dfrac{1}{cos(\theta)} $$ where $\theta$ is the angles between the normal vectors of $P_1,P_2$.

The reason is simply this, only one dimensional property of $P_2$ changes. (When you look at your shadow it can be longer than you but width is same and $P_1$ is shodow of $P_2$ ) Thus, the ratio of areas are same the ratio of length.

Note : When you see that this equality for planes, then it become approximation for any other surfaces by tangent planes.

mesel
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  • Could you write down the precise proof? Your intuition is useful but there are some things I don't understand. Because the surfaces are cross-sections of a cone, the higher up they are the bigger their surface area. But you seem to be ignoring that. You also mentioned that only one dimensional property changes when a surface is tilted but that is not necessarily true. How would you prove it for the other cases? – m009 Jun 14 '14 at 07:37
  • By using double integral it can be done, Is it what you want

    $$area(P_2)=\int_{p_1} \sqrt{1+f_x^2+f_y^2} dxdy$$ ?

     – mesel Jun 14 '14 at 10:22
  • Never mind, that integral gave me enough clues to work it out myself. But I still have one concern, the one I mentioned in my previous comment. "Because the surfaces are cross-sections of a cone, the higher up they are the bigger their surface area." We merely worked out the case for something like a pillar, I do not know the technical term, instead of a cone. How would you prove it for that? – m009 Jun 14 '14 at 11:25
  • @m009: I am not sure whether I understood what you mean completly but, when we show this for planes then since every surface are union of small piece of planes we are done. What I mean we actually get $$dS=\dfrac{1}{cos(\theta)} dxdy$$ for any small piece of surface. – mesel Jun 14 '14 at 12:00