Equality of Surface area of sphere and Cylinder

Question

I seen a video that shows How the surface area of sphere of radius R is equals to the surface area of cylinder of radius R and height 2R.enter link description here

It was an geometrical proof and I understand it but I want to proove it in calculas way by intergral.It mean I want to show why when you take area element in sphere or in cylinder they give the same inegtral result? How these integral area and area element related to each other ie. what kind of transformation from one variable to other take you from one perception to other? So Please Help me.

Answer 1

In spherical coordinates, the surface-area element of the sphere is $R^2\sin\theta d\theta d\phi$ and its area is integrated as

$$\int_0^{2\pi} \int_0^\pi R^2\sin\theta d\theta d\phi=4\pi R^2$$

In cylindrical coordinates, the surface-area element of the cylinder is $R dz d\phi$ and its area is integrated as

$$\int_0^{2\pi} \int_0^{2R} R dz d\phi=4\pi R^2$$

Answer 2

What's going on geometrically here is the amazing fact (which can be proved in the style of Euclid — with no calculus — as in the video) that the surface area of a zone of a sphere (i.e., the portion between two parallel planes intersecting the sphere) is $2\pi R$ times the distance $h$ between the parallel planes.

With single-variable calculus, we can find the surface area when we rotate $y=f(x)$ about the $x$-axis, $a\le x\le a+h$, by computing the integral $$2\pi\int_a^{a+h} f(x)\sqrt{1+f'(x)^2}\,dx.$$ (See any calculus book for surface area of a surface of revolution.) With $f(x)=\sqrt{R^2-x^2}$, we get $$2\pi\int_a^{a+h} \sqrt{R^2-x^2}\sqrt{\frac{R^2}{R^2-x^2}}dx = 2\pi\int_a^{a+h} R\,dx = 2\pi Rh,$$ as desired. By taking $h=2R$, you get the surface area of the entire sphere, of course.

If you're looking for a direct explanation with calculus of what the video proved, let's use polar coordinates to give the circle of radius $R$ in the $yz$-plane. The infinitesimal area of the band of the sphere at height $z$ is $dS=2\pi y\,ds$, where $ds$ is the element of arclength on the circle. With $y=R\cos\theta$ and $z=R\sin\theta$, this gives us $$dS = 2\pi y\,ds = 2\pi (R\cos\theta)(R\,d\theta) = 2\pi R\, d(R\sin\theta) = 2\pi R\,dz,$$ which is the element of surface area on the cylinder.