The surface area of a sphere

Question

My idea was like this. There is a circle which is represented by the equation $x^2 + y^2 = R^2 $, or if rearranged for $x > 0, \ y > 0$, $y = \sqrt{R^2 - x^2}$.

$$A = 2 \int_0^R 2 \pi y \ dx = 4\pi \int_0^R \sqrt{R^2 - x^2} \ dx$$

But if I solve that I obtain ${\pi}^2 R^2$, which is not $4 \pi R^2$.

What is wrong?

Answer 1

You need to use the correct formula for the surface area of the revolution which is $$A = 2 \int_0^R 2 \pi y \ ds $$ where $$ds = \sqrt {1+(dy/{dx})^2} dx$$

With $$y=\sqrt {R^2-x^2}$$ you find $$ds = \frac {R}{\sqrt {R^2-x^2}}dx $$

You will find the correct answer with $$A = 2 \int_0^R 2 \pi y \ ds=2\int_0^R 2\pi R dx = 4\pi R^2$$

Answer 2

The equation for the sphere is

$$x^2+y^2+z^2=R^2$$

then it is convevient use spherical coordinates

• $x=R\sin \phi \cos \theta$

• $y=R\sin \phi \sin \theta$

• $z=R\cos \phi $

and the set up for the surface area becomes

$$S=\int_0^\pi \:d\phi \int_0^{2\pi}R^2\sin \phi \:d\theta$$