Why is the area of a sphere is coming out as $ \pi^2R^2 $ instead of $4\pi R^2$?

Question

I am trying to derive the formula for the area of a sphere using integration. It is coming out as $ \pi^2R^2 $ instead of $4\pi R^2$.

This is what I am doing :-

I am approximating the area of the sphere of radius R (kept at origin) using the Curved Surface Area of infinite infinitesimal cylinders along the X axis.

Now each infinitesimally small cylinder's Curved Surface Area is $ 2π f(x) dx $.

Therefore, Area of the sphere is :-

$$ \int_{-R}^R 2\pi f(x) dx \\ = 2\pi\int_{-R}^R \sqrt{R^2 - x^2} dx \\ \text{Put } x = R \sin \theta \text{, we get} \\ \text{Area} = 2\pi \int_{-\frac{\pi}{2}}^\frac{\pi}{2} R \cos \theta. R \cos \theta d\theta \\ = 2 \pi R^2 \int_{-\frac{\pi}{2}}^\frac{\pi}{2} \cos^2 \theta d \theta \\ = 2 \pi R^2 \int_{-\frac{\pi}{2}}^\frac{\pi}{2} \frac{1 + \cos2\theta}{2} d\theta \\ = \pi R^2 \left [\theta + \frac{\sin2\theta}{2} \right ]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \\ = \pi^2 R^2 $$

What am I doing wrong?

Answer 1

On the interval $(x_1,x_2)$ the better approximation is not the cone, which misses a lot of area of $f'(x)$ is large on the interval.

Rather, if we think of the section as approximated by a cone, then we have (with $\Delta x=x_2-x_1,$ and $\Delta f = f(x_2)-f(x_1))$ that the point of the cone must be $(c(x),0)$ where $$c(x)=x-f(x)\frac{\Delta x}{\Delta f}$$

Then the area of this region of the one is:

$$\left|\pi\cdot f(x_2)\sqrt{\left(\Delta x+f(x_1)\frac{\Delta x}{\Delta f}\right)^2+f(x_2)^2}-\pi\cdot f(x_1)\sqrt{\left(f(x_1)\frac{\Delta x}{\Delta f}\right)^2+f(x_1)^2}\right|$$

This can be written is:

$$\pi\frac{\Delta x}{\Delta f}\left| \left(f(x_2)^2- f(x_1)^2\right)\sqrt{1+\left(\frac{\Delta f}{\Delta x}\right)^2}\right|$$

Assuming $f'(x_1)\neq 0$ then as $x_2\to x_1$ you get:

$$2\pi \Delta x\left|f(x_1)\sqrt{1+f'(x_1)^2}\right|$$

But this is the value gives the cylinder when $f'(x_1)=0.$

So you get the integral:

$$2\pi \int_{-R}^R f(x)\sqrt{1+f'(x)^2}\,dx$$

Note that with $f(x)=\sqrt{R^2-x^2},$ you have $f'(x)=\frac{-x}{f(x)}$ and $f(x)$ is positive, you can rewrite this as:

$$2\pi \int_{-R}^R \sqrt{f(x)^2+x^2}\,dx=2R\pi\int_{-R}^{R}\,dx=4R^2\pi$$

Answer 2

Since $x^2+y^2=R^2$,$$xdx+ydy=0\implies\frac{dy}{dx}=-\frac{x}{y}\implies\frac{ds}{dx}=\sqrt{1+\left(\frac{dy}{dx}\right)^2}=\frac{R}{y}.$$The surface is the integral of $2\pi yds$, not $2\pi y dx$. For an intuitive understanding of why, see this video's comparison of the sphere's surface to a cylinder's. (It'll even help you compare them without ordinary calculus.) Since $yds=Rdx$, the final result is$$2\pi R\int_{x=-R}^{x=R}dx=4\pi R^2.$$

Answer 3

As pointed out by Nate in the comments, your approach for deriving the problem need a slight modification, as the sphere have a curvature in vertical direction unlike the cylinders. Take your infinitesimal surface as frustum, with slanted length $R d \theta$ (?) and circumference $ 2 \pi R \sin \theta$ (?) $$\implies dA = 2 \pi R^2 \sin \theta d\theta \implies A =2 \pi \int^{\pi}_0 R^2 \sin \theta d \theta$$