Can the Surface Area of a Sphere be Derived by Integrating Circumferences?

[armolinasf]

Homework Statement

So I wanted to try and derive the surface area of a sphere with radius r. My plan was to basically integrate the circumferences of disks from 0 to r and then multiply it by 2.

The Attempt at a Solution

I got this:

4$$\pi$$r$$^{2}$$ $$\int^{\pi/2}_{0}$$cos$$^{2}$$$$\theta$$

evaluating gives $$\pi^{2}$$r$$^{2}$$ which i obviously not the SA of a sphere.

So I went on to wikipedia and read about archimedes and the proof about how you can derive the SA by the fact that its the derivative of the volume. So my question is why can't it be proven by integrating circumferences or in some similar manner?

 

[the_kool_guy]

do this way
at some height above centre,let radius be x.
its width is Rd(theta)
integrate 2*pie*x*r*d(theta)
but x=Rsin(theta)
integrate theta 0 to pie

 

[HallsofIvy]

Homework Statement

So I wanted to try and derive the surface area of a sphere with radius r. My plan was to basically integrate the circumferences of disks from 0 to r and then multiply it by 2.

The Attempt at a Solution

I got this:

4$$\pi$$r$$^{2}$$ $$\int^{\pi/2}_{0}$$cos$$^{2}$$$$\theta$$

evaluating gives $$\pi^{2}$$r$$^{2}$$ which i obviously not the SA of a sphere.

So I went on to wikipedia and read about archimedes and the proof about how you can derive the SA by the fact that its the derivative of the volume. So my question is why can't it be proven by integrating circumferences or in some similar manner?

Essentially, for the same reason that you cannot, for example, find the length of the straight line from (0, 0) to (1, 1) by approximating it as n horizontal segments, n vertical segments, summing the lengths and then taking the limit as n goes to infinity. (Each of the n horizontal segments will have length 1/n so their sum is 1 for all n. Each of the n vertical segments will have length 1/n so their sum is 1 for all n. The two sums will be 2 for all n.)

To get the surface area of each such disk, you have to multiply the circumerence of the disk by the differential of arc length for the circle, not just the height of the disk.

At each $\phi$ in polar coordinates, the radius of the disk will be $r= R\sin(\phi)$, where R is the radius of the sphere. The circumference of such a disk will be $2\pi R \sin(\phi)$.

Now, we need to find the differential of arc-length for the circle where, say, $\theta= 0$. Parametric equations for a sphere of radius R are $x= R\cos(\theta)\sin(\phi)$, $y= R\sin(\theta)\sin(\phi)$, and $z= R\cos(\phi)$. The circle at $\theta= 0$ is given by $x= R sin(\phi)$, $y= 0$, and [itex]z= R\cos(\phi)[/tex]. The differential of arclength is given by <br /> $$ds= \sqrt{\left(\frac{dx}{d\phi}\right)^2+ \left(\frac{dy}{d\phi}\right)^2+ \left(\frac{dz}{d\phi}\right)^2}d\phi$$<br /> $$= \sqrt{R^2 cos^2(\phi)+ R^2 sin^2(\phi)}d\phi= R d\phi$$<br /> <br /> The surface area of the sphere of radius R is given by <br /> [tex]\int_{\phi= 0}^\pi (2\pi R sin(\phi))(R d\phi)[/tex<br /> <br /> If you use your "r", the height above the xy-plane, instead of my "$\phi$" The raidus of each disk is given by $\sqrt{R^2- r^2}$ rather than $R cos(\phi)$ so the circumference of a disk is $2\pi \sqrt{R^2- r^2}$<br /> <br /> Now for the arclength, your "r" is equivalent to "y" in $x^2+ y^2= R^2$ from which we have $x= \sqrt{R^2- y^2}$,<br /> $$\frac{dx}{dy}= \frac{y}{\sqrt{R^2- y^2}}$$<br /> so that <br /> $$ds= \sqrt{1+ \frac{y^2}{R^2- y^2}}dy= \frac{R}{\sqrt{R^2- y^2}}dy$$<br /> or, using your "r",<br /> $$ds= \frac{R}{\sqrt{R^2- r^2}}dr$$<br /> so the integral for surface area is again very easy.[/tex][/itex]

 

[armolinasf]

Wow, thanks, that was very helpful. I'm wondering if there's something I could read about to help with writing equations with latec. Thanks again for the explanation

 

[HallsofIvy]

There is a LaTex tutorial on this website at
https://www.physicsforums.com/showthread.php?t=8997