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?
[Math] Deriving the formula for the volume of a sphere
calculuseuclidean-geometrygeometryvolume
Related Solutions
What you need to look at is the second theorem of Pappus. When you rotate things in a circle, you have to account for the difference in the distance traveled by the point furthest from the axis (the bottom corner of our triangle) which goes the full $2\pi$ around. However, the vertical side doesn't actually move, so it doesn't get the full $2\pi$.
Pappus' theorem says you can use the radius of the centroid as your revolution radius, and it just so happens the the centroid of this triangle is a distance $r/3$ from the axis.
Another good read is this.
You can compute the volume of a sphere using the area of a circle, but you need a different trick. Take a sphere of radius r with centre the origin, and slice it into circles along a diameter. Well, not quite circles. The slice from $x$ to $x+dx$ with $|x|<r$ is a cylinder of volume $\pi[\rho(x)]^2dx$, where by the Pythagorean theorem $[\rho(x)]^2=r^2-x^2$. The sphere's volume is therefore $$\int^{r}_{-r}\pi[r^2-x^2]dx=\pi[r^2 x-\tfrac{x^3}{3}]^{r}_{-r}=\tfrac{4\pi}{3}r^3.$$ You can use this method to obtain a recurrence relation expressing the $d$-volume of a radius-$r$ $d$-ball (say $V_d r^d$) in terms of a radius-$r$ $(d-1)$-ball, viz. $$V_d r^d = \int^{r}_{-r} V_{d-1}[r^2-x^2]^{\tfrac{d-1}{2}}dx.$$ Hence $$\frac{V_d}{V_{d-1}}=2\int^1_0 [1-x^2]^{\tfrac{d-1}{2}}dx=\text{B}(\frac{1}{2},\,\frac{d+1}{2}).$$ Interestingly, comparing $V_d$ to $V_{d-2}$ gives an especially neat result: $$\frac{V_d}{V_{d-2}}=\frac{\sqrt{\pi} \Gamma(\frac{d}{2})\sqrt{\pi}\Gamma(\frac{d+1}{2})}{\Gamma(\frac{d+1}{2})\Gamma(\frac{d+2}{2})}=\frac{2\pi }{d}.$$ This explains why the series $$V_1 = 2, V_2 = \pi, V_3 = \frac{4\pi}{3}, V_4 = \frac{\pi^2}{2}, V_5 = \frac{8\pi^2}{15}$$ has integer powers of $\pi$, gaining a power at even $d$. Considering the even-$d$ case allows one to conjecture the correct general result, which is $$V_d=\frac{\pi^{d/2}}{\Gamma(\frac{d}{2}+1)}$$ (as induction easily verifies).
Best Answer
$$dV=(dr)(rd\theta)(rcos\theta d\phi)=[(r^{2}cos\theta dr) d\theta] d\phi$$
$$\rightarrow V=\int \int \int [(r^{2}cos\theta dr) d\theta] d\phi$$
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)