Deriving the formula for the volume of a sphere A circle $x^2 +y^2 =a^2$ is rotated about the $y$-axis to form a solid sphere of radius $a$. 
How do you express this motion mathematically in such a way that it allows me to arrive at the formula for the volume of the sphere?
 A: 
$$dV=(dr)(rd\theta)(rcos\theta d\phi)=[(r^{2}cos\theta dr) d\theta] d\phi$$
$$\rightarrow V=\int \int  \int [(r^{2}cos\theta dr) d\theta] d\phi$$
So I guess if you integrate w.r.t dr,d$\theta$ you get a disk.Now to 'rotate' this disk integrate the result w.r.t d$\phi$
Note: Using this method one can find that there are 6 ways to do this integration to find out the volume.(each way is unique and interesting)
A: Each point $(\rho,z)$ in the meridian half-plane $M:=\{(\rho,z)\>|\>\rho\geq0,\ z\in{\mathbb R}\}$ produces under rotation around the $z$-axis a circle
$$\phi\mapsto(x,y,z):=(\rho\cos\phi,\rho\sin\phi, z)\ ,$$
and an area element ${\rm d}A={\rm d}(\rho,z)$ in $M$ produces a ring-shaped body of volume
$${\rm d}V=2\pi\rho\ {\rm d}A=2\pi \rho\ {\rm d}(\rho,z)\ .$$
In order to obtain a solid sphere $B_a$ of radius $a>0$ we have to rotate the half circle $$H:=\{(\rho,z)\>|\> 0\leq\rho\leq a,\ |z|\leq\sqrt{a^2-\rho^2}\}$$
around the $z$-axis. In this way we get
$${\rm vol}(B_a)=\int_{B_a}{\rm d}V=2\pi\int\nolimits_{H}\rho\ {\rm d}(\rho,z)=2\pi\int_0^a \rho\ \int_{-\sqrt{a^2-\rho^2}}^\sqrt{a^2-\rho^2}dz\ d\rho=\ldots={4\pi\over3}a^3\ .$$ 
A: Here is a neat way to do it using the divergence theorem, see exercise II-60,chapter-1, Div, Curl , Grad and all that problem II-26. We have the formula:
$$ \frac13 \int_{S} \hat{n} \cdot \vec{r}dS = V$$
Where $\vec{r}$ is a position vector to a point on the surface, $\hat{n}$ is surface normal
The above equation can be verified via divergence theorem.
Calculating the volume:
For our calculations , we will consider spherical coordinates with the $\hat{r}$ unit vector and center our origin at the Center of sphere of radius R. This leads to:
$$ V = \frac13 \int_S \hat{r} \cdot R \hat{r} dS= \frac{R}{3} \int dS= \frac43 \pi R^3 $$
QED
A: Every point of the half-circle follows a trajectory of length $2\pi x$. If you integrate over that half-circle,
$$\int_{-r}^r\int_0^{\sqrt{r^2-y^2}}2\pi x\,dx\,dy=\int_{-r}^r\pi(r^2-y^2)\,dy=2\pi\left(r^3-\dfrac{r^3}3\right).$$
