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Question Number 177540 by yaslm last updated on 06/Oct/22

	Answered by aleks041103 last updated on 07/Oct/22

	$i t s e i t h e r c i r c l e i n s c r i b e d i n a n e l l i p s e o r$ $t h e o t h e r w a y a r o u n d .$ $\Rightarrow A n s . = ∣ π a b - π a^{2} ∣$ $\Rightarrow A n s . = π a ∣ b - a ∣$
	Commented by yaslm last updated on 07/Oct/22
	I need a solution by integration with draw it
	Answered by som(math1967) last updated on 07/Oct/22

	$A r e a o f c i r c l e r e g i o n$ $4 \int_{0}^{a} \sqrt{a^{2} - x^{2}} d x$ $4 {[\frac{x \sqrt{a^{2} - x^{2}}}{2} + \frac{a^{2}}{2} \sin^{- 1} (\frac{x}{a})]}_{0}^{a}$ $4 \times \frac{π}{4} \times a^{2} = π a^{2} s q u n i t$ $A r e a o f e l l i p s e r e g i o n$ $4 \int_{0}^{a} y d x$ $= 4 \times \frac{b}{a} \int_{0}^{a} \sqrt{a^{2} - x^{2}} d x$ $= 4 \times \frac{b}{a} {[\frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} \sin^{- 1} (\frac{x}{a})]}_{0}^{a}$ $= 4 \times \frac{b}{a} \times \frac{π}{4} \times a^{2} = π a b s q u n i t$ $a r e a o f g r e e n r i g i o n$ $π a^{2} - π a b = π a (a - b) s q u n i t f o r$ $g i v e n f i g u r e (a > b)$
	Commented by som(math1967) last updated on 07/Oct/22

	Commented by peter frank last updated on 07/Oct/22

	$thank you$
	Commented by yaslm last updated on 07/Oct/22
	thank you so much
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