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