# Fallacy in integration technique to find a sphere’s surface area?

Spherical Coordinates 3D Animation
Spherical Coordinates 3D Animation

The problem is to find the surface area of a sphere.

Consider the first approach as shown in the following picture (the approach is used in Wikipedia too):

In this first approach, the sphere is divided into infinitesimal rings where a ring has a radius of $R \sin\theta$ and a height of $R\,\text{d}\theta$. With this approach the integration is carried from $\theta = 0$ to $\theta = \frac\pi2$ obtaining the correct result as follows: $$\int_{0}^{\frac\pi2} \text{d}A = 2 \int_{0}^{\frac\pi2} 2\pi R^2\sin\theta\text{d}\theta$$ $$A = 4\pi R^2 \int_{0}^{\frac\pi2}\sin\theta\text{d}\theta$$ $$A = 4\pi R^2 \left(-\cos\theta\Big|_{0}^{\frac\pi2}\right)$$ $$A = 4\pi R^2 (-\cos(\pi / 2) + \cos(0))$$ $$A = 4\pi R^2 (0 + 1)$$ $$A = 4\pi R^2$$

Now consider the second approach as shown in the following picture:

In the second approach, the sphere is also divided into infinitesimal rings. But, a ring has a radius of $R$ and a height of $R\,\text{d}\theta$. With this approach the integration is also carried from $\theta = 0$ to $\theta = \frac\pi2$ but obtaining a result too big by $\frac{\pi}{2} – 1$:

$$\int_{0}^{\frac\pi2} \text{d}A = 2 \int_{0}^{\frac\pi2} 2\pi R^2\,\text{d}\theta$$ $$A = 4\pi R^2 \int_{0}^{\frac\pi2}\text{d}\theta$$ $$A = 4\pi R^2 \left(\theta\Big|_{0}^{\frac\pi2}\right)$$ $$A = 4\pi R^2 (\pi / 2 – 0)$$ $$A = 4\pi R^2 (\pi / 2)$$

What is wrong with the second approach? If the first approach is allowed to infinitesimally divide the sphere vertically, why should the second approach not be allowed to infinitesimally divide the sphere diagonally? Any sound mathematical argument to forbid the second approach?

As indicated in the comments, the fallacy lies in not seeing the overlap depicted in the following picture:

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