Question Number 47006 by Tawa1 last updated on 03/Nov/18
Commented by Tawa1 last updated on 05/Nov/18
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by ajfour last updated on 03/Nov/18
![let A(0,0) eq. of arc with center A x^2 +y^2 = 1 when x=(1/2) , y = ((√3)/2) when y=(1/2) , x= ((√3)/2) reqd area=4∫_(1/2) ^( (√3)/2) ((√(1−x^2 ))−(1/2))dx = 4[(x/2)(√(1−x^2 ))+(1/2)sin^(−1) x−(x/2)]_(1/2) ^((√3)/2) =4((π/(12))−(((√3)−1)/4)) = (π/3)+1−(√3) .](Q47013.png)
$${let}\:{A}\left(\mathrm{0},\mathrm{0}\right) \\ $$$${eq}.\:{of}\:{arc}\:{with}\:{center}\:{A} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\:\mathrm{1} \\ $$$${when}\:{x}=\frac{\mathrm{1}}{\mathrm{2}}\:,\:{y}\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\: \\ $$$${when}\:{y}=\frac{\mathrm{1}}{\mathrm{2}}\:,\:\:{x}=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${reqd}\:{area}=\mathrm{4}\int_{\mathrm{1}/\mathrm{2}} ^{\:\:\sqrt{\mathrm{3}}/\mathrm{2}} \left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}}\right){dx} \\ $$$$\:=\:\mathrm{4}\left[\frac{{x}}{\mathrm{2}}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} {x}−\frac{{x}}{\mathrm{2}}\right]_{\mathrm{1}/\mathrm{2}} ^{\sqrt{\mathrm{3}}/\mathrm{2}} \\ $$$$\:=\mathrm{4}\left(\frac{\pi}{\mathrm{12}}−\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{4}}\right)\:=\:\frac{\pi}{\mathrm{3}}+\mathrm{1}−\sqrt{\mathrm{3}}\:. \\ $$
Commented by gunawan last updated on 05/Nov/18
$$\mathrm{Area}=\mathrm{1}^{\mathrm{2}} −\mathrm{4}\left(\frac{\mathrm{1}}{\mathrm{2}}×\left(\mathrm{1}^{\mathrm{2}} \right)\mathrm{sin}\:\mathrm{60}°−\frac{\mathrm{30}°}{\mathrm{360}°}×\pi×\left(\mathrm{1}^{\mathrm{2}} \right)\right. \\ $$$$\mathrm{Area}=\mathrm{1}^{\mathrm{2}} −\mathrm{4}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{12}}\pi\right) \\ $$$$=\mathrm{1}−\sqrt{\mathrm{3}}+\frac{\pi}{\mathrm{3}} \\ $$
Commented by Tawa1 last updated on 05/Nov/18
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$