# Deriving the Surface Area of Sphere

How do we come to 4.pi.r^2?

CompuChip
Homework Helper
In spherical coordinates,
$$\int_0^{2\pi} \int_0^\pi R^2 \sin\phi \, \mathrm{d}\theta \, \mathrm{d}\phi$$.
You can also calculate it from geometrical arguments: one way would be to divide the sphere into infinitesimal circles with radius r and thickness dz. Now express r in terms of z (at z = radius of the sphere R, r = 0 and at z = 0, r = R) and integrate $2 \pi r(z)$ over z.

CompuChip
Homework Helper
I thought about it some more, and I find this a particularly nice way.
First, we prove that the volume is $\frac43 \pi R^3$, as follows. Dissect the sphere into infinitesimal disks, with surface $\pi r^2$ (this can again be proven by straightforward integration, but I'll leave this to you if you wish, and focus on the main line) and height dz. From the Pythagorean theorem, we can relate the radius r to the height coordinate z, as shown in the attachment: $r^2 = R^2 - z^2$. Now the total volume of the sphere is found by adding the volumes $\pi (R^2 - z^2) \, \mathrm{d}z$ from z = -R to R, which indeed gives us
$$\int_{-R}^R \pi (R^2 - z^2) \, \mathrm{d}z = \frac43 \pi R^3$$.

Now, suppose we also have a sphere with a slightly smaller radius, R - h. If we substract the volumes, we get the volume of a spherical shell. You should probably convince yourself geometrically, that if we take h very small, we can get the volume of the spherical shell of radius R and thickness h; and by dividing out by the "thickness" we get the surface, which is hence
$$\frac{ \frac43 \pi R^3 - \frac43 \pi (R - h)^3 }{ h }$$.
Now we see that taking the limit for h to zero, will give us the surface area, but this is actually the definition of
$$\frac{d}{dR} \left( \frac43 \pi R^3 \right)$$
which equals $4 \pi R^3$.

I realize the weak point is the step where I claim that the surface is the derivative of the volume, so you should think a bit more about this and convince yourself that it is actually allowed (I don't know the exact details, but I believe that for compact sets in Euclidean space, it generally is).

#### Attachments

• sphereDimensions.jpg
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HallsofIvy
Homework Helper
I'm not sure what you mean by "the surface is the derivative of the volume". If you mean the surface area of a figure is equal to the derivative of the volume function, that's true only for spheres. For example, the volume of a cube of side length x is x3 which has derivative 3x2 while the surface area is 6x2. You can however, show that the surface area formula is a constant times the derivative of the volume formula.

Now what do we do to calculate the surface area of only a certain section of the sphere cap which has an effective radius(say, r) less than the actual radius of the sphere(R)?Also, how do we derive the area of a segment of a circle which has a certain length(d) joining the two curved ends and a perpendicular height(h) from the midpoint?Can we write the radius of the circle from which the segment has been formed as a function of 'h' and 'd'?

the attachment isn't really posted..could u attach it again?

HallsofIvy
Homework Helper
"Effective radius"? Do you mean the radius of the circle forming the bottom of the cap?

In that case, set up your coordinate system so that the center of the cape is at (0,0,R) on the z-axis. If the radius of the base of the cap is r, then that base circle is at
$$sin(\phi)= \frac{r}{R}$$
In the integral for surface area, take $\phi$ ranging from 0 to
[tex] arcsin(\frac{r}{R})[/itex]

In your next question, you are talking about what the Babylonians called a "bow and arrow". h is the length of the "string" of that bow and d is the length of the arrow. If you continue the "arrow" to the center of the circle as well as drawing the two radii at the ends of the bow you have two congruent right triangles. They have base equal to R-d, height h/2 and hypotenuse R. By the Pythagorean theorem, $R^2= (R-d)^2+ (h/2)^2$. Multiplying that out, $R^2= R^2- 2dR+ d^2+ h^2/4$ or
$2dR= d^2+ h^2/4$.

[tex]R= \frac{1}{4}\sqrt{4d^2+ h^2}[/itex]

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Wow...is that so I had actually thought of this in grade 6.Anyway, I have an attachment which has a derivation of the segment. I found the area of the triangle and subtracted that from the sector formed by that portion.