by-considering-a-sermicircle-from-r-to-r-prove-that-area-of-circle-is-pir-2-

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 πr^2

$${by}\:{considering}\:\:{a}\:{sermicircle}\:{from}\:−{r}\:{to}\:\:{r}\:{prove}\:{that}\:{area}\:{of}\:{circle}\:{is}\:\pi{r}^{\mathrm{2}} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 29/Sep/18

2∫_(−r) ^r (√(r^2 −x^2 )) dx 2×∣((x(√(r^2 −x^2 )) )/2)+(r^2 /2)sin^(−1) ((x/r))∣_(−r) ^r =2×(r^2 /2){sin^(−1) ((r/r))−sin^(−1) (((−r)/r))} =r^2 {(π/2)−(−(π/2))} =r^2 ×π area of circle of radius r =πr^2

$$\mathrm{2}\int_{−{r}} ^{{r}} \sqrt{{r}^{\mathrm{2}} −{x}^{\mathrm{2}} }\:\:{dx} \\ $$$$\mathrm{2}×\mid\frac{{x}\sqrt{{r}^{\mathrm{2}} −{x}^{\mathrm{2}} }\:}{\mathrm{2}}+\frac{{r}^{\mathrm{2}} }{\mathrm{2}}{sin}^{−\mathrm{1}} \left(\frac{{x}}{{r}}\right)\mid_{−{r}} ^{{r}} \\ $$$$=\mathrm{2}×\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left\{{sin}^{−\mathrm{1}} \left(\frac{{r}}{{r}}\right)−{sin}^{−\mathrm{1}} \left(\frac{−{r}}{{r}}\right)\right\} \\ $$$$={r}^{\mathrm{2}} \left\{\frac{\pi}{\mathrm{2}}−\left(−\frac{\pi}{\mathrm{2}}\right)\right\} \\ $$$$={r}^{\mathrm{2}} ×\pi \\ $$$${area}\:{of}\:{circle}\:{of}\:{radius}\:{r}\:\:=\pi{r}^{\mathrm{2}} \\ $$

Commented by peter frank last updated on 29/Sep/18