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how to derive relation between solid angle and surface area and the radius of sphere ?

I know $s=r^2\Omega$ but how they got it using integral ?

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  • $\begingroup$ How do you define solid angle if not by this? The factor $r^2$ is needed to make the dimensions consistent. $\endgroup$ – Ross Millikan Oct 1 '13 at 22:34
  • $\begingroup$ @RossMillikan for example I can find in a circle that $l=r \theta $ using integral but here in my problem I dont know how $\endgroup$ – user79560 Oct 1 '13 at 22:36
  • $\begingroup$ You can do similar, using a small patch of the sphere as a differential of area and integrating. You use the known formula for the volume of a pyramid like you use the area of a triangle in the 2D case $\endgroup$ – Ross Millikan Oct 1 '13 at 22:44
  • $\begingroup$ @RossMillikan can you add your way to an answer with more explain ? $\endgroup$ – user79560 Oct 1 '13 at 22:45
  • $\begingroup$ Do you already know the volume of a sphere is $\frac 43 \pi r^3$? These are related $\endgroup$ – Ross Millikan Oct 1 '13 at 22:49
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One way is to start from the volume of a unit sphere. Using the disk method, it is $\int_{-1}^1 \pi (1-z^2)dz=\pi(z-\frac{z^3}3)|_{-1}^1=\frac 43\pi$ The volume of a sphere of radius $r$ is then $\frac 43 \pi r^3$. The surface area is then the derivative of this with respect to $r$-think of increasing $r$ a small amount $dr$ and the volume added. This gives the total solid angle is $\frac d{dr} \frac 43 \pi r^3=4\pi r^2$

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  • $\begingroup$ but how to prove that surface area is the derivative of the volume with respect to r ? $\endgroup$ – user79560 Oct 1 '13 at 23:05
  • $\begingroup$ If we put a shell around the sphere of thickness $dr$, the volume is $dr$ times the area of the sphere. $\endgroup$ – Ross Millikan Oct 1 '13 at 23:06
  • $\begingroup$ mmmmm , can you explain more please ? and its better for me when you using another way :) $\endgroup$ – user79560 Oct 1 '13 at 23:09
  • $\begingroup$ Maybe think of expanding a cube the same way. If you have a cube of side $a$, the volume is $a^3$. If you expand the side to $a+2da)$ (where I added the factor $2$ because the side of the cube is like the diameter of a sphere) the volume becomes $a^3+3 a^2(2da) + 3a(2da)^2+(2da)^3$ Each term of this can be identified with pieces of the larger cube. The $a^3$ is the original cube, the next are the extensions of the faces, the next are long skinny pieces along the edges, and the final is the corners. All but the faces are another order smaller and can be ignored. $\endgroup$ – Ross Millikan Oct 1 '13 at 23:49

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