Derivation of Formula for Total Surface Area of the Sphere by Integration

Tags: 
sphere
area by integration
surface area of the sphere
great circle
length of arc

The total surface area of the sphere is four times the area of great circle. To know more about great circle, see properties of a sphere. Given the radius r of the sphere, the total surface area is
 

$ A = 4\pi r^2 $

 

From the figure, the area of the strip is
$ dA = 2\pi \, x \, ds $
 

Where ds is the length of differential arc which is given by
$ ds = \sqrt{1 + \left( \dfrac{dy}{dx} \right)^2} \, dx = \sqrt{1 + \left( \dfrac{dx}{dy} \right)^2} \, dy $
 

Figure for the Derivation of Formula for Surface Area of the Sphere by Integration

 

See Length of Arc in Integral Calculus (link not yet active) for more information about ds.
 

The total area of the sphere is equal to twice the sum of the differential area dA from 0 to r.
$ \displaystyle A = 2 \left( \int_0^r 2\pi \, x \, ds \right) $

$ \displaystyle A = 4\pi \int_0^r x \sqrt{1 + \left( \dfrac{dy}{dx} \right)^2} \, dx $
 

From the figure,
$ x^2 + y^2 = r^2 $

$ y = \sqrt{r^2 - x^2} $

$ \dfrac{dy}{dx} = \dfrac{-2x}{2\sqrt{r^2 - x^2}} $

$ \dfrac{dy}{dx} = \dfrac{-x}{\sqrt{r^2 - x^2}} $

$ \left( \dfrac{dy}{dx} \right)^2 = \dfrac{x^2}{r^2 - x^2} $
 

Thus,
$ \displaystyle A = 4\pi \int_0^r x \sqrt{1 + \dfrac{x^2}{r^2 - x^2}} \, dx $

$ \displaystyle A = 4\pi \int_0^r x \sqrt{\dfrac{(r^2 - x^2) + x^2}{r^2 - x^2}} \, dx $

$ \displaystyle A = 4\pi \int_0^r x \sqrt{\dfrac{r^2}{r^2 - x^2}} \, dx $
 

Let
x = r sin θ
dx = r cos θ dθ

When x = 0, θ = 0
When x = r, θ = π/2
 

Thus,
$ \displaystyle A = 4\pi \int_0^{\pi/2} r \sin \theta \sqrt{\dfrac{r^2}{r^2 - r^2 \sin^2 \theta}} \, (r \cos \theta \, d\theta) $

$ \displaystyle A = 4\pi \int_0^{\pi/2} r^2 \sin \theta \cos \theta\sqrt{\dfrac{r^2}{r^2(1 - \sin^2 \theta)}} \, d\theta $

$ \displaystyle A = 4\pi r^2 \int_0^{\pi/2} \sin \theta \cos \theta\sqrt{\dfrac{1}{\cos^2 \theta}} \, d\theta $

$ \displaystyle A = 4\pi r^2 \int_0^{\pi/2} \sin \theta \cos \theta \left( \dfrac{1}{\cos \theta} \right) \, d\theta $

$ \displaystyle A = 4\pi r^2 \int_0^{\pi/2} \sin \theta \, d\theta $

$ A = 4\pi r^2 \bigg[-\cos \theta \bigg]_0^{\pi/2} $

$ A = 4\pi r^2 \bigg[-\cos \frac{1}{2}\pi + \cos 0 \bigg] $

$ A = 4\pi r^2 \bigg[ -0 + 1 \bigg] $

$ A = 4\pi r^2 $           okay!
 

»

Comments

Submitted by Guest (guest) on November 10, 2011 - 11:46am

you made mistakes: x=rcos(theta) not sin(theta)
cos(theta)= adj/hyp = x/r
so
x= r cos(theta)

you can't just ignore the + 1 you are left over with at the end...

Submitted by Romel Verterra on November 13, 2011 - 7:22pm

@Guest: Think of theta as angle from the vertical, and the +1 at the end is not an addition but a multiplication as indicated by the square brackets.

Submitted by Guest (guest) on February 22, 2012 - 4:02pm

There are many ways to solve for the area of a sphere. The above method is the hardest way I've seen yet.

Looking at the surface of 1/8 of a sphere (cut a sphere in half, then half again, then half again)

you have:

dA = (one side of a small element)*(the arc it makes)
dA = (R*sin(psi))*(R*d(theta), where psi, theta are angles
integrate 0 from pi/2 twice (the limits for each integral)

= (pi*R^2)/2

this answer represent 1/8 of the total surface area, so multiply it by 8.

= 4*pi*R^2

Submitted by Romel Verterra on February 25, 2012 - 7:15am

Thank you for sharing tim.

Submitted by Guest (guest) on August 24, 2012 - 11:29pm

In your first integral, the limits of integration of s should go from 0 to infinity rather than 0 to r.

Submitted by Guest (guest) on August 24, 2012 - 11:59pm

Correction: s goes from 0 to pi*R/2 in the first integral.

Submitted by Guest (guest) on October 28, 2012 - 8:06am

According to Miles Mathis, the above derivation is a "fake proof" that has been "fudged", "fails", and contains "disallowed substitutions".

The Mathis critique can be found at the following web site:
http://milesmathis.com/pascal2.pdf

Caution: Miles Mathis is a math and physics critic with a dubious reputation!

Submitted by Romel Verterra on October 28, 2012 - 10:55am

Thank you for sharing the link. I read his arguments and appreciate his keen observation, meticulous pursuit of truth, and passion on the subject. Not everyone can do what he is doing. I am not particular about the professional reputation of anyone as long as I understand what he is talking and he will not take my freedom to accept and reject ideas/facts, including my own.

With regards to the topic, I will not comment about it. I exist because I am targeting a different audience, a group that is more interested in application and those who want to improve their problem solving skill through examples.

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