# Surface area of a sphere dilemma

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.

• The error was the thickness of the ring is not "dx." – user156213 Apr 10 at 22:21
• Why its not dx? – Yash Mittal Apr 11 at 17:51
• Because the surface is curved. – user156213 Apr 12 at 16:38

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)

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

• I am asking, why the other method is incorrect? – Yash Mittal Apr 11 at 17:51
• @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. – user1952500 Apr 11 at 17:52
• @YashMittal I have added an addendum using $dx$ – user1952500 Apr 11 at 18:00
• But what is theta here? I have not taken any theta in first method. – Yash Mittal Apr 12 at 10:28
• @YashMittal what is the radius of the ring at distance $x$? – user1952500 Apr 12 at 17:04