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

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

$ds = \sqrt{1 + \left( \dfrac{dy}{dx} \right)^2} \, dx = \sqrt{1 + \left( \dfrac{dx}{dy} \right)^2} \, dy$

See Length of Arc in Integral Calculus 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!*