I want to know if the way I derived the surface area of a sphere by integration is correct?

Question

I am using the alias of Sillysack Buttowski and this is my first question. I searched on other links on stack exchange regarding "how to find the surface area of a sphere by integration". They seemed overly complex to me and not what I was searching for. Hence, I decided to post my own question.

Okay, so this is what I came up with:

Now, in the image I wrote that the limit of $dr$, as $dr$ approaches $0$ is equal to the circumference of the circle or $2\pi r$. I imagine that the sphere is made up of thin slices of circles where $dr$ is their thickness. Therefore, I integrated the integrand from $0$ to $2r$ and got the correct answer of $4\pi r^2$. Now is this correct? I doubt this is. It's too simple to be right. calculus is a difficult subject. Therefore, I can only infer that my answer may be correct, however, the way I got to answer is wrong, probably.

Now, I want any criticism you have and to tell me the holes and errors of my thought processes. By the way, I am not an expert in calculus as you can see. Thanks for your help in advance!

EDIT: Okay, so the $lim$ of $dr$ as $dr$ approaches $0$ is equal to $0$. The Circumference and $2\pi r$ are not related in anyway to it. So just cut off that part over there.

Answer 1

Since you need the surface area, it would be more neat to take the area element spherically rather than linearly. It may also lead to errors : see Why can't I use the disk method to compute surface area?

Consider a sphere of radius $R$, and also consider an axis passing vertically through the centre of the sphere.

Now consider an Area element $\rm dS$ (pink ribbon) on the sphere inclined at $\theta$ angle from the axis, and subtending $\rm d\theta$ on the centre of sphere. Let the thickness of this area element be $\rm dl$.

When you open up this ribbon, we get a rectangle with width $\rm dl$, and length $= \rm2\pi\times BC$. The area of this rectangle is same as area of the elementary slice on sphere, ie, $\rm dS$

$$\rm dS=2\pi \text{BC}\times dl$$

• Using the definition of angle in radians, we can calculate $\rm dl$ is: $$\rm \theta = \dfrac{\text{arc length}}{radius} \\ d\theta = \dfrac{dl}{dR} \\ \therefore\rm dl = Rd\theta$$

• For calculating $\rm BC$, consider the right triangle $\rm ABC$. $$\rm \sin\theta = \dfrac{BC}{AC} = \dfrac{BC}{R}\\ \therefore BC = R\sin\theta$$

$$\rm\therefore dS=2\pi R\sin\theta\times Rd\theta$$

Then we integrate this element a complete angle of $\pi$ radians.

$$\rm S = \int_{0}^{\pi}2\pi R^2\sin\theta\times d\theta$$

$$\rm =2\pi R^2(1-\cos\pi)$$

$$\rm S=4\pi R^2$$

Answer 2

First of all you should use a clearer notation. For example: you cannot use $r$ to denote both the radius of the sphere and the integration variable. If I understand what you are trying to do, you want to calculate the surface area of a sphere by summing up the areas of the tiny strips comprised between two planes, perpendicular to a diameter of the sphere. There is nothing wrong with that and it is the very method Archimedes used a long time ago.

Let's call $x$ (and not $r$) the distance of the first plane from one end of the diameter, and $dx$ the distance between planes. The idea now is to compute the area of the strip to first order in $dx$. The result, however, is not $2\pi x\,dx$ (your result), but it is instead $2\pi r\, dx$.

To see why, look at the picture below: the area of the strip (to first order in $dx$) is $dA=2\pi\, l\cdot h$, where $h$ is different from $dx$. But by similarity of yellow and green triangles one has: $$ h:dx=r:l,\quad\hbox{that is:}\quad l\cdot h= r\cdot dx, $$ so that $dA=2\pi\,r\cdot dx$. Integrate now from $x=0$ to $x=2r$ and you are done.