Surface area of a sphere with integration of disks Why it is not correct to say that the surface area of a sphere is:
$$
2 \int_{0}^{R} 2\pi r \text{ } dr
$$
In my mind we are summing up the perimeters of disks from $r=0$ to $r=R$, so by 1 integration, we would have $\frac{1}{2}$ of the surface area of the sphere.
I know that it's not correct because that will give us $2\pi R^2$ that it's different from $4\pi R^2$, but why???
Thanks!

$$\text{Area will be}~~: 2 \int_{0}^{R} 2\pi x ~ ds$$
Where $ds$ is width of strip bounded by circles of radius $x$ and $x+dx$ situated at height $y$. Also $ds \neq dr$ it's tilted in $y$ direction too. Only horizontal projection of $ds$ is $dr$.What you have done is valid for Disk See image below.

$$(ds)^2=(dx)^2+(dy)^2$$
Therefore :
$$ds = \sqrt{1+\Bigg( \frac{dy}{dx}\Bigg)^2} \; dx$$
Now,
$$x^2 + y^2 = R^2$$
$$y = \sqrt{R^2 - x^2}$$
$$\dfrac{dy}{dx} = \dfrac{-2x}{2\sqrt{R^2 - x^2}}$$
$$\dfrac{dy}{dx} = \dfrac{-x}{\sqrt{R^2 - x^2}}$$
$$\left( \dfrac{dy}{dx} \right)^2 = \dfrac{x^2}{R^2 - x^2}$$
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
$$x = R \sin θ \implies
dx = R \cos θ dθ$$
When $x = 0, θ = 0$
When $x = R, θ = \pi/2$
Thus,
$$\displaystyle A = 4\pi \int_0^{\pi/2} R \sin \theta \sqrt{\dfrac{R^2}{R^2 - R^2 \sin^2 \theta}} \, (R \cos \theta \, d\theta)$$
$$\displaystyle A = 4\pi \int_0^{\pi/2} R^2 \sin \theta \cos \theta\sqrt{\dfrac{R^2}{R^2(1 - \sin^2 \theta)}} \, d\theta$$
$$\displaystyle A = 4\pi R^2 \int_0^{\pi/2} \sin \theta \cos \theta\sqrt{\dfrac{1}{\cos^2 \theta}} \, d\theta$$
$$\displaystyle A = 4\pi R^2 \int_0^{\pi/2} \sin \theta \cos \theta \left( \dfrac{1}{\cos \theta} \right) \, d\theta$$
$$\displaystyle A = 4\pi R^2 \int_0^{\pi/2} \sin \theta \, d\theta$$
$$A = 4\pi R^2 \bigg[-\cos \theta \bigg]_0^{\pi/2}$$
$$A = 4\pi R^2 \bigg[-\cos \frac{1}{2}\pi + \cos 0 \bigg]$$
$$A = 4\pi R^2 \bigg[ -0 + 1 \bigg]$$
$$A = 4\pi R^2$$
