by considering a sermicircle from −r to r prove that area of circle is πr^2


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 44424 by peter frank last updated on 28/Sep/18

$b y c o n s i d e r i n g a s e r m i c i r c l e f r o m - r t o r p r o v e t h a t a r e a o f c i r c l e i s π r^{2}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 29/Sep/18
	$2∫_(−r) ^r (√(r^2 −x^2 )) dx 2×∣((x(√(r^2 −x^2 )) )/2)+(r^2 /2)sin^(−1) ((x/r))∣_(−r) ^r =2×(r^2 /2){sin^(−1) ((r/r))−sin^(−1) (((−r)/r))} =r^2 {(π/2)−(−(π/2))} =r^2 ×π area of circle of radius r =πr^2$
	$2 \int_{- r}^{r} \sqrt{r^{2} - x^{2}} d x$ $2 \times ∣ \frac{x \sqrt{r^{2} - x^{2}}}{2} + \frac{r^{2}}{2} {s i n}^{- 1} (\frac{x}{r}) ∣_{- r}^{r}$ $= 2 \times \frac{r^{2}}{2} {{s i n}^{- 1} (\frac{r}{r}) - {s i n}^{- 1} (\frac{- r}{r})}$ $= r^{2} {\frac{π}{2} - (- \frac{π}{2})}$ $= r^{2} \times π$ $a r e a o f c i r c l e o f r a d i u s r = π r^{2}$
	Commented by peter frank last updated on 29/Sep/18

	$w h e r e t h i s c a m e f r o m \sqrt{r^{2} - x^{2} ?}$
	Commented by tanmay.chaudhury50@gmail.com last updated on 29/Sep/18

	$e q n o f c i r c l e x^{2} + y^{2} = r^{2}$ $y = \sqrt{r^{2} - x^{2}}$
	Commented by peter frank last updated on 29/Sep/18

	$o k a y s i r t h a n k y o u$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum