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Question Number 226003 by ajfour last updated on 17/Nov/25
If r^2 +r((√3)−(1/( (√3))))sin θ=(2/3) find A=∫_(π/6) ^( π/2) ((r^2 /2))dθ
$${If}\:\:{r}^{\mathrm{2}} +{r}\left(\sqrt{\mathrm{3}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\mathrm{sin}\:\theta=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${find}\:{A}=\int_{\pi/\mathrm{6}} ^{\:\pi/\mathrm{2}} \left(\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\right){d}\theta \\ $$$$\: \\ $$
Commented by mr W last updated on 18/Nov/25
what do you get for Q225980, sir?
$${what}\:{do}\:{you}\:{get}\:{for}\:{Q}\mathrm{225980},\:{sir}? \\ $$
Answered by mr W last updated on 18/Nov/25
r^2 +r((√3)−(1/( (√3)))) sin θ=(2/3) r^2 +((1/( (√3))))^2 −2r((1/( (√3)))) cos ((π/2)+θ)=1^2 acc. to law of cosines, this is a geometrical figure as shown in the diagram. A=∫_(π/6) ^(π/2) (r^2 /2)dθ is the hatched sector area for the range θ from (π/6) to (π/2). at θ=(π/6): ∠OCP=φ ((sin ((π/2)+θ))/1)=((sin (π−(π/2)−θ−φ))/(1/( (√3)))) cos θ=(√3) cos (θ+φ) cos (π/6)=(√3) cos ((π/6)+φ) ((√3)/2)=(√3) cos ((π/6)+φ) cos ((π/6)+φ)=(1/2) ⇒(π/6)+φ=(π/3) ⇒φ=(π/6) A=((1^2 ×φ)/2)−(1/2)×(1/( (√3)))×sin φ =(1^2 /2)×(π/6)−(1/2)×(1/( (√3)))×sin (π/6)=((π−(√3))/(12))
$${r}^{\mathrm{2}} +{r}\left(\sqrt{\mathrm{3}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\:\mathrm{sin}\:\theta=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${r}^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} −\mathrm{2}{r}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}+\theta\right)=\mathrm{1}^{\mathrm{2}} \\ $$$${acc}.\:{to}\:{law}\:{of}\:{cosines},\:{this}\:{is}\:{a} \\ $$$${geometrical}\:{figure}\:{as}\:{shown}\:{in} \\ $$$${the}\:{diagram}. \\ $$$${A}=\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{2}}} \frac{{r}^{\mathrm{2}} }{\mathrm{2}}{d}\theta\:{is}\:{the}\:{hatched}\:{sector}\: \\ $$$${area}\:{for}\:{the}\:{range}\:\theta\:{from}\:\frac{\pi}{\mathrm{6}}\:{to}\:\frac{\pi}{\mathrm{2}}. \\ $$$${at}\:\theta=\frac{\pi}{\mathrm{6}}:\: \\ $$$$\angle{OCP}=\phi \\ $$$$\frac{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}+\theta\right)}{\mathrm{1}}=\frac{\mathrm{sin}\:\left(\pi−\frac{\pi}{\mathrm{2}}−\theta−\phi\right)}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} \\ $$$$\mathrm{cos}\:\theta=\sqrt{\mathrm{3}}\:\mathrm{cos}\:\left(\theta+\phi\right) \\ $$$$\mathrm{cos}\:\frac{\pi}{\mathrm{6}}=\sqrt{\mathrm{3}}\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{6}}+\phi\right) \\ $$$$\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\sqrt{\mathrm{3}}\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{6}}+\phi\right) \\ $$$$\mathrm{cos}\:\left(\frac{\pi}{\mathrm{6}}+\phi\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\frac{\pi}{\mathrm{6}}+\phi=\frac{\pi}{\mathrm{3}}\:\Rightarrow\phi=\frac{\pi}{\mathrm{6}} \\ $$$${A}=\frac{\mathrm{1}^{\mathrm{2}} ×\phi}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}×\mathrm{sin}\:\phi \\ $$$$\:\:\:\:=\frac{\mathrm{1}^{\mathrm{2}} }{\mathrm{2}}×\frac{\pi}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}×\mathrm{sin}\:\frac{\pi}{\mathrm{6}}=\frac{\pi−\sqrt{\mathrm{3}}}{\mathrm{12}} \\ $$
Commented by mr W last updated on 18/Nov/25
Commented by ajfour last updated on 18/Nov/25
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Commented by ajfour last updated on 18/Nov/25
Commented by ajfour last updated on 18/Nov/25
yes! The common area of intersection of the circles =A=6(((π−(√3))/(12)))=(π/2)−((√3)/2)
$${yes}!\:{The}\:{common}\:{area}\:{of}\: \\ $$$${intersection}\:{of}\:{the}\:{circles} \\ $$$$={A}=\mathrm{6}\left(\frac{\pi−\sqrt{\mathrm{3}}}{\mathrm{12}}\right)=\frac{\pi}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$
Commented by mr W last updated on 18/Nov/25
or A=3∗ sector of 60°− 2∗ equilateral of sides 1 =3×((1^2 ×π)/(2×3))−2×(((√3)×1^2 )/4) =((π−(√3))/2)
$${or}\: \\ $$$${A}=\mathrm{3}\ast\:{sector}\:{of}\:\mathrm{60}°−\:\mathrm{2}\ast\:{equilateral}\:{of}\:{sides}\:\mathrm{1} \\ $$$$\:\:\:=\mathrm{3}×\frac{\mathrm{1}^{\mathrm{2}} ×\pi}{\mathrm{2}×\mathrm{3}}−\mathrm{2}×\frac{\sqrt{\mathrm{3}}×\mathrm{1}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:\:=\frac{\pi−\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$

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