Question Number 177540 by yaslm last updated on 06/Oct/22
Answered by aleks041103 last updated on 07/Oct/22
$${its}\:{either}\:{circle}\:{inscribed}\:{in}\:{an}\:{ellipse}\:{or} \\ $$$${the}\:{other}\:{way}\:{around}. \\ $$$$\Rightarrow{Ans}.\:=\:\mid\pi{ab}−\pi{a}^{\mathrm{2}} \mid \\ $$$$\Rightarrow{Ans}.\:=\:\pi{a}\mid{b}−{a}\mid \\ $$
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
![Area of circle region 4∫_0 ^a (√(a^2 −x^2 ))dx 4[((x(√(a^2 −x^2 )))/2) +(a^2 /2)sin^(−1) ((x/a))]_0 ^a 4×(π/4)×a^2 =πa^2 sq unit Area of ellipse region 4∫_0 ^a ydx =4×(b/a)∫_0 ^a (√(a^2 −x^2 ))dx =4×(b/a)[(x/2)(√(a^2 −x^2 )) +(a^2 /2)sin^(−1) ((x/a))]_0 ^a =4×(b/a)×(π/4)×a^2 =πab sq unit area of green rigion πa^2 −πab=πa(a−b)sq unit for given figure (a>b)](https://www.tinkutara.com/question/Q177561.png)
$$\:{Area}\:{of}\:{circle}\:{region} \\ $$$$\:\mathrm{4}\int_{\mathrm{0}} ^{{a}} \sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }{dx} \\ $$$$\mathrm{4}\left[\frac{{x}\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }}{\mathrm{2}}\:+\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \left(\frac{{x}}{{a}}\right)\right]_{\mathrm{0}} ^{{a}} \\ $$$$\:\mathrm{4}×\frac{\pi}{\mathrm{4}}×{a}^{\mathrm{2}} =\pi{a}^{\mathrm{2}} {sq}\:{unit} \\ $$$${Area}\:{of}\:{ellipse}\:{region} \\ $$$$\:\mathrm{4}\int_{\mathrm{0}} ^{{a}} {ydx} \\ $$$$=\mathrm{4}×\frac{{b}}{{a}}\int_{\mathrm{0}} ^{{a}} \sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }{dx} \\ $$$$=\mathrm{4}×\frac{{b}}{{a}}\left[\frac{{x}}{\mathrm{2}}\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }\:+\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \left(\frac{{x}}{{a}}\right)\right]_{\mathrm{0}} ^{{a}} \\ $$$$=\mathrm{4}×\frac{{b}}{{a}}×\frac{\pi}{\mathrm{4}}×{a}^{\mathrm{2}} =\pi{ab}\:{sq}\:{unit} \\ $$$${area}\:{of}\:{green}\:{rigion} \\ $$$$\:\pi{a}^{\mathrm{2}} −\pi{ab}=\pi{a}\left({a}−{b}\right){sq}\:{unit}\:{for} \\ $$$${given}\:{figure}\:\left({a}>{b}\right) \\ $$$$ \\ $$
Commented by som(math1967) last updated on 07/Oct/22
Commented by peter frank last updated on 07/Oct/22
$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by yaslm last updated on 07/Oct/22
thank you so much