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I recently found that surface area of a sphere can't be found with the following method. What's the flaw in it? First, I have taken a very thin ring of thickness $dx$ at a distance of $x$ from the centre. Then, I integrated it, using substitution of $x=R\sin\theta$

But this is giving me answer $4\pi^2R^2$.

I also tried to solve many other problems related to area of sphere, like I found the magnetic field at the centre due to a revolving sphere carrying a charge $Q$, but this method is giving me wrong result.

The correct result is found by taking a thin ring subtending an angle $2\theta$ at centre, and thickness of the ring would be $Rd\theta$.

But, why my method is not giving the correct answer?

Sorry for formatting errors.

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  • $\begingroup$ The error was the thickness of the ring is not "dx." $\endgroup$ – user156213 Apr 10 at 22:21
  • $\begingroup$ Why its not dx? $\endgroup$ – Yash Mittal Apr 11 at 17:51
  • $\begingroup$ Because the surface is curved. $\endgroup$ – user156213 Apr 12 at 16:38
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For a ring at angle $\theta$, we have Radius = $R \sin\theta$

The area of this ring is the circumference times its thickness. The thickness is $R \cdot d\theta$ $$dA = 2\pi \cdot (R \sin\theta) \cdot R d\theta$$

$$\implies A = 2\pi R^2\int_{0}^{\pi}\sin\theta\ d\theta$$

$$ = 2\pi R^2(-\cos\theta)\big|^{\pi}_0$$ $$\implies \boxed{A = 4\pi R^2}$$


If you want to go by the $dx$ route, note that the thickness of the ring will be $$\frac{dx}{\cos{(90-\theta)}} = \frac{dx}{\sin\theta}$$

(How? Consider the tangent line from a point on the ring to the X-axis)

$$\implies dA = 2\pi\cdot(R\sin\theta)\cdot\left(\frac{dx}{\sin\theta}\right) = 2\pi R\ dx$$ $$A = \int_{-R}^{R}2\pi R\ dx = 2\pi R x\bigg|^R_{-R} = 4\pi R^2$$

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  • $\begingroup$ I am asking, why the other method is incorrect? $\endgroup$ – Yash Mittal Apr 11 at 17:51
  • $\begingroup$ @YashMittal that is because $dx$ is not the right thickness of the ring. If you post your entire derivation, I can take a look at the incorrect areas. There could be more than one but the thickness is one of them. $\endgroup$ – user1952500 Apr 11 at 17:52
  • $\begingroup$ @YashMittal I have added an addendum using $dx$ $\endgroup$ – user1952500 Apr 11 at 18:00
  • $\begingroup$ But what is theta here? I have not taken any theta in first method. $\endgroup$ – Yash Mittal Apr 12 at 10:28
  • $\begingroup$ @YashMittal what is the radius of the ring at distance $x$? $\endgroup$ – user1952500 Apr 12 at 17:04

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