# Thread: Proving the formula for the surface area of a sphere

1. ## Proving the formula for the surface area of a sphere

Okay. (I've not posted on here before so forgive me if I'm not typing things properly or whatever).

In one of my assignment questions (I'm a 1st year Maths degree student, by the way) I have to prove the formula for the surface area of a sphere using the surface of revolution formula. So basically, my answer should be 4*Pi*r². I'm having difficulty with it, though. So far within the integral I have this horrible looking thing...

(I thought posting a picture would be easier than trying to type it all out). Those should be x²s on the ends of the fraction by the way - the bracket kinda cuts that off. That has to simplify a little, otherwise, I've done something wrong in my calculations prior to arriving at that.

Any ideas?

Cheers.

2. Originally Posted by chella182
Okay. (I've not posted on here before so forgive me if I'm not typing things properly or whatever).

In one of my assignment questions (I'm a 1st year Maths degree student, by the way) I have to prove the formula for the surface area of a sphere using the surface of revolution formula. So basically, my answer should be 4*Pi*r². I'm having difficulty with it, though. So far within the integral I have this horrible looking thing...

(I thought posting a picture would be easier than trying to type it all out). Those should be x²s on the ends of the fraction by the way - the bracket kinda cuts that off. That has to simplify a little, otherwise, I've done something wrong in my calculations prior to arriving at that.

Any ideas?

Cheers.
Using the formula for rotating about the x-axis (and using some symmetry) you have

$\displaystyle S = 4 \pi \int_{0}^{R} \sqrt{R^2 - x^2} \sqrt{1 + \frac{x^2}{R^2 - x^2}} \, dx$

$\displaystyle = 4 \pi \int_{0}^{R} \sqrt{R^2 - x^2} \, \sqrt{\frac{R^2}{R^2 - x^2}} \, dx$

$\displaystyle = 4 \pi \int_{0}^{R} R \, dx = 4 \pi R \int_{0}^{R} dx$.

3. That's not the formula I use for the surface of revolution
I use $\displaystyle 2\pi\int_{-R}^{R}y\sqrt{1+y'^2},dx$

4. Originally Posted by chella182

How have you gone from your second step to the last one?
Algebra.

Originally Posted by chella182

And also, doesn't the formula for 'S' have $\displaystyle 2\pi$ not $\displaystyle 4\pi$?

Oh, and the limits in the integral in my question are $\displaystyle -R$ to $\displaystyle R$ I think.
mr fantastic claims for symmetry. Have you ever heard about even functions?

(I see you edited your post. Glad you find it by yourself.)

5. Sorry, I edited again. You've used a different formula to me for the surface of revolution.

6. He just applied the formula for $\displaystyle y=\sqrt{R^2-x^2}$ an used symmetry.

7. Ah okay, I think I'm getting it now.
So when you have $\displaystyle 2\pi\int_{-R}^{R}\sqrt{R^2-x^2}\sqrt{\frac{R^2}{R^2-x^2}}dx$ the $\displaystyle \sqrt{R^2-x^2}$ terms just cancel to leave $\displaystyle \sqrt{R^2}$ in the integral, yes?

8. Originally Posted by chella182
Ah okay, I think I'm getting it now.
So when you have $\displaystyle 2\pi\int_{-R}^{R}\sqrt{R^2-x^2}\sqrt{\frac{R^2}{R^2-x^2}}dx$ the $\displaystyle \sqrt{R^2-x^2}$ terms just cancel to leave $\displaystyle \sqrt{R^2}$ in the integral, yes?
Yes.

And $\displaystyle 2\pi\int_{-R}^{R}\sqrt{R^2-x^2}\sqrt{\frac{R^2}{R^2-x^2}}dx = 4\pi\int_{0}^{R}\sqrt{R^2-x^2}\sqrt{\frac{R^2}{R^2-x^2}}dx$ by symmetry (since the integrand is clearly an even function).

Note: Questions at this level will assume an ability to perform basic algebraic manipulation and to recognise the application symmetry.