[Math] how to derive relation between solid angle and surface area and the radius of sphere using definite integral

definite integralseuclidean-geometrygeometrysolid angle

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 ?

Best Answer

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 $\frac d{dr} \frac 43 \pi r^3=4\pi r^2$

Related Solutions

[Math] Solid angle and projections

When a plane shape of area $A$ is tilted by the angle $\theta>0$ versus the horizontal plane and then projected orthogonally onto this plane, then the area of the "shadow" is exactly $\cos\theta\>A$.

Now your "infinitesimal" surface element $\Delta A$ centered at the point $p\in S$ can be considered as such a plane shape. This shape is then projected orthogonally on a sphere of radius $r$ passing through $p$ and having normal $n$ there. The shadow $\Delta A_n$ covers such a tiny part of the sphere that it can as well be considered as plane. Thereforet we can apply the formula found above for the "linear model" of this situation, and we obtain $|\Delta A_n|=\cos\theta\>|\Delta A|$.

[Math] How to derive the volume and surface area formula for truncated cylinder without calculus

Let $C(A, B)$ denote a truncated cylinder whose left edge has length $A$ and whose right edge has length $B$. (I'm assuming that it's placed so that its longest and shortest edges are to the left and right).

The volume of $C(A, 0)$ is $\frac{1}{2} \pi r^2 A$. Reason: Stack up $C(A, 0)$ atop $C(0, A)$ to get a cylinder of height $A$. The volume of $C(A, 0)$ is clearly half the volume of this cylinder.
Now for a cylinder C(A, B), where $A > B$, consider the stack

C(0, Q)
C(A, B)
C(B, A)
C(Q, 0)

where $Q = A - B$. THat's a cylinder of height $A + B + (A-B) = 2A$. If we subtract from it the volumes of the top and bottom pieces (which we computed in part 1), we get $$ \pi r^2 (2A) - \pi r^2 (Q) = \pi r^2 (2A - (A-B)) = \pi r^2 (A + B) $$ But the volume we've just computed is twice the volume of the sliced cylinder $C(A,B)$ that we're interested in. So the volume of $C(A,B)$ isjust $$ \pi r^2 \frac{A + B}{2} $$

A rather similar proof works for surface area.

Note that the proof given above works for a cylinder cut by ANY two planes, as long as the longest and shortest "line" of the truncated cylinder are diametrically opposite each other.

Best Answer

Related Solutions

[Math] Solid angle and projections

[Math] How to derive the volume and surface area formula for truncated cylinder without calculus

Related Question