# Deriving Formulae Related to Circles using Integration

## Key Questions

• By using polar coordinates, the area of a circle centered at the origin with radius $R$ can be expressed:
$A = {\int}_{0}^{2 \pi} {\int}_{0}^{R} r \mathrm{dr} d \theta = \pi {R}^{2}$

Let us evaluate the integral,
$A = {\int}_{0}^{2 \pi} {\int}_{0}^{R} r \mathrm{dr} d \theta$
by evaluating the inner integral,
$= {\int}_{0}^{2 \pi} {\left[\frac{{r}^{2}}{2}\right]}_{0}^{R} d \theta = {\int}_{0}^{2 \pi} {R}^{2} / 2 d \theta$
by kicking the constant ${R}^{2} / 2$ out of the integral,
${R}^{2} / 2 {\int}_{0}^{2 \pi} d \theta = {R}^{2} / 2 {\left[\theta\right]}_{0}^{2 \pi} = {R}^{2} / 2 \cdot 2 \pi = \pi {R}^{2}$

• Since a sphere with radius $r$ can be obtained by rotating the region bounded by the semicircle $y = \sqrt{{r}^{2} - {x}^{2}}$ and the x-axis about the x-axis, the volume $V$ of the solid can be found by Disk Method.

$V = \pi {\int}_{- r}^{r} {\left(\sqrt{{r}^{2} - {x}^{2}}\right)}^{2} \mathrm{dx}$

by the symmetry about the y-axis,

$= 2 \pi {\int}_{0}^{r} \left({r}^{2} - {x}^{2}\right) \mathrm{dx}$

$= 2 \pi {\left[{r}^{2} x - {x}^{3} / 3\right]}_{0}^{r}$

$= 2 \pi \left({r}^{3} - {r}^{3} / 3\right)$

$= \frac{4}{3} \pi {r}^{3}$