Intuitive error in finding Volume of sphere using single integration.

Question

To calculate the volume of a sphere of radius $\mathbf R$, I considered a thin disc of radius $ r= \mathbf Rsin\theta$ ,where $\theta$ is the angle radius vector on the circumference makes with the axis of the disc. However I considered the volume d$\mathbf V$= $ \pi r^2 \mathbf R d\theta $ , $\mathbf R d\theta $ being the infinitesimal thickness of the disc, which is erroneous. The volume of the sphere comes out to be $\frac{3\pi^2 \mathbf R^3}{2}$. The correct thickness is $\mathbf R sin\theta d\theta $ which is found usually in other derivations gives the correct volume. However I do not understand what exactly is the issue in my approach and why the component of the curve length is considered.

Answer 1

Let the $x$-axis of ${\mathbb R}^3$ be the axis of the disc. The disc is then bounded by the sphere of radius $R$ and two planes $x=a$ and $x=b$. We now have to express $a$ and $b$ in terms of $\theta$ and $d\theta$. When $b=R\cos\theta$ then $a=R\cos(\theta+d\theta)<b$. The thickness of the disc is then given by $$b-a=R\bigl(\cos\theta-\cos(\theta+d\theta)\bigr)\approx R\bigl(-\cos'(\theta)\bigr)\>d\theta=R\sin\theta\>d\theta\ ,$$ and has nothing to do with the arc length $R\,d\theta$ along the meridian of $B$. We obtain $${\rm vol}(B)=\int_0^\pi \pi r^2(\theta)\>R\sin\theta\> d\theta={4\pi\over3}R^3\ .$$