Question Number 44424 by peter frank last updated on 28/Sep/18
$${by}\:{considering}\:\:{a}\:{sermicircle}\:{from}\:−{r}\:{to}\:\:{r}\:{prove}\:{that}\:{area}\:{of}\:{circle}\:{is}\:\pi{r}^{\mathrm{2}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 29/Sep/18
$$\mathrm{2}\int_{−{r}} ^{{r}} \sqrt{{r}^{\mathrm{2}} −{x}^{\mathrm{2}} }\:\:{dx} \\ $$$$\mathrm{2}×\mid\frac{{x}\sqrt{{r}^{\mathrm{2}} −{x}^{\mathrm{2}} }\:}{\mathrm{2}}+\frac{{r}^{\mathrm{2}} }{\mathrm{2}}{sin}^{−\mathrm{1}} \left(\frac{{x}}{{r}}\right)\mid_{−{r}} ^{{r}} \\ $$$$=\mathrm{2}×\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left\{{sin}^{−\mathrm{1}} \left(\frac{{r}}{{r}}\right)−{sin}^{−\mathrm{1}} \left(\frac{−{r}}{{r}}\right)\right\} \\ $$$$={r}^{\mathrm{2}} \left\{\frac{\pi}{\mathrm{2}}−\left(−\frac{\pi}{\mathrm{2}}\right)\right\} \\ $$$$={r}^{\mathrm{2}} ×\pi \\ $$$${area}\:{of}\:{circle}\:{of}\:{radius}\:{r}\:\:=\pi{r}^{\mathrm{2}} \\ $$
Commented by peter frank last updated on 29/Sep/18
$${where}\:{this}\:{came}\:{from}\:\sqrt{{r}^{\mathrm{2}} −{x}^{\mathrm{2}} \:?} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 29/Sep/18
$${eqn}\:{of}\:{circle}\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${y}=\sqrt{{r}^{\mathrm{2}} −{x}^{\mathrm{2}} }\: \\ $$
Commented by peter frank last updated on 29/Sep/18
$${okay}\:{sir}\:{thank}\:{you} \\ $$