The volume of a sphere is given by:
[tex]V= \frac{4}{3} \pi r^{3}[/tex]
So, we need to deduct this equation. We will walk through Calculus on the concept of a solid of revolution that is a solid figure that is obtained by rotating a plane curve around some straight line (the axis of revolution) that lies on the same plane. We know from calculus that:
[tex]V=\pi \int_{a}^{b}[f(x)]^{2}dx[/tex]
Then, according to the concept of solid of revolution we are going to rotate a circumference shown in the figure, then:
[tex] x^{2}+y^{2}=r^{2}[/tex]
Isolationg y:
[tex]y= \sqrt{r^{2}-x^{2}}[/tex]
So,
[tex]f(x)=y=\sqrt{r^{2}-x^{2}}[/tex]
[tex]V=\pi \int_{a}^{b}[\sqrt{r^{2}-x^{2}}]^{2}dx[/tex]
[tex]V=\pi \int_{a}^{b}(r^{2}-x^{2})dx[/tex]
being -r and r the limits of this integral.
[tex]V=\pi \int_{-r}^{r}(r^{2}-x^{2})dx[/tex]
Solving:
[tex]V=\pi[r^{2}x-\frac{x^{3}}{3}]\right|_{-r}^{r}[/tex]
Finally:
[tex]V=\pi(r^{3}-\frac{r^{3}}{3})-\pi(-r^{3}+\frac{r^{3}}{3})= \frac{4}{3} \pi r^{3}[/tex]
