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Question Number 120864 by mnjuly1970 last updated on 03/Nov/20

Commented by math35 last updated on 03/Nov/20

please sir how was the diagram made

$${please}\:{sir}\:{how}\:{was}\:{the} \\ $$$${diagram}\:{made} \\ $$

Commented by mnjuly1970 last updated on 03/Nov/20

hello sir Tanmay: I took a photo of the chart and sent it.using the attach image option. have a nice time...

$${hello}\:{sir}\:{Tanmay}: \\ $$$${I}\:{took}\:{a}\:{photo}\:{of}\:{the}\:{chart}\: \\ $$$$\:{and}\:{sent}\:{it}.{using}\:{the}\:{attach}\:{image} \\ $$$${option}.\:{have}\:{a}\:{nice}\:{time}... \\ $$

Commented by TANMAY PANACEA last updated on 03/Nov/20

pls chk my answer...i do not remember formula

$${pls}\:{chk}\:{my}\:{answer}...{i}\:{do}\:{not}\:{remember}\:{formula} \\ $$

Answered by TANMAY PANACEA last updated on 03/Nov/20

D is origin(0,0) C(−4,0) eqn of big semi circle→x^2 +y^2 =4^2 mid point C and D (−2,0) eqn of small semi circle centre(−2,0) and radius 2 (x+2)^2 +y^2 =2^2 eqn another small semi circle x^2 +(y−2)^2 =2^2 solve x^2 +4x+4+y^2 =4 x^2 +y^2 −4y+4=4 −4x−4y=0→x=−y so x^2 +4x+4+x^2 =4→2x^2 +4x=0 2x(x+2)=0 so point B(−2,2) eqn AB is y=2 point A ...solve y=2 and x^2 +y^2 =4^2 x^2 =16−4=12→x=±2(√3) point A(−2(√3) ,2) required area {∫_(−4) ^(−2(√3)) (√(16−x^2 )) dx+∫_(−2(√3) ) ^(−2) 2dx}−{∫_(−4) ^(−2) (√(4−(x+2)^2 )) dx} pls chk ∫(√(a^2 −x^2 )) dx=(x/2)(√(a^2 −x^2 )) +(a^2 /2)sin^(−1) ((x/a)) I_1 =∣(x/2)(√(16−x^2 )) +((16)/2)sin^(−1) ((x/4))∣_(−4) ^(−2(√3) ) =(((−2(√3))/2)(√(16−12)) +8sin^(−1) (((−2(√3))/4)) =−(√3) ×2+8(−(π/3))=−2(√3) −((8π)/3) I_2 =2∣x∣_(−2(√3)) ^(−2) =2(−2+2(√3) )=4(√3) −4 ★∫_(−4) ^(−2) (√(4−(x+2)^2 )) dx (x+2)=p ∫_(−2) ^0 (√(2^2 −p^2 )) dp=∣(p/2)(√(2^2 −p^2 )) +(4/2)sin^(−1) ((p/2))∣_(−2) ^0 =(0)−{((−2)/2)(√(4−4)) +2sin^(−1) (((−2)/2))} =0−0−2sin^(−1) (−1)=−2×((−π)/2)=π rewuired answer I_1 +I_2 −I_3 =(−2(√3) −((8π)/3))+(4(√3) −4)−(π) =2(√3) −4−((11π)/3)

$${D}\:{is}\:{origin}\left(\mathrm{0},\mathrm{0}\right)\:\:{C}\left(−\mathrm{4},\mathrm{0}\right) \\ $$$${eqn}\:{of}\:{big}\:{semi}\:{circle}\rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{4}^{\mathrm{2}} \\ $$$${mid}\:{point}\:{C}\:{and}\:{D}\:\left(−\mathrm{2},\mathrm{0}\right) \\ $$$${eqn}\:{of}\:{small}\:{semi}\:{circle}\:{centre}\left(−\mathrm{2},\mathrm{0}\right)\:{and}\:{radius}\:\mathrm{2} \\ $$$$\left({x}+\mathrm{2}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{2}^{\mathrm{2}} \\ $$$${eqn}\:{another}\:{small}\:{semi}\:{circle}\:{x}^{\mathrm{2}} +\left({y}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{2}^{\mathrm{2}} \\ $$$${solve}\:\:{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{4}+{y}^{\mathrm{2}} =\mathrm{4} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{4}{y}+\mathrm{4}=\mathrm{4} \\ $$$$−\mathrm{4}{x}−\mathrm{4}{y}=\mathrm{0}\rightarrow{x}=−{y} \\ $$$${so}\:{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{4}+{x}^{\mathrm{2}} =\mathrm{4}\rightarrow\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}{x}=\mathrm{0}\:\:\:\mathrm{2}{x}\left({x}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{point}}\:\boldsymbol{{B}}\left(−\mathrm{2},\mathrm{2}\right) \\ $$$${eqn}\:\:{AB}\:{is}\:{y}=\mathrm{2} \\ $$$${point}\:{A}\:...{solve}\:{y}=\mathrm{2}\:{and}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{4}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} =\mathrm{16}−\mathrm{4}=\mathrm{12}\rightarrow{x}=\pm\mathrm{2}\sqrt{\mathrm{3}}\: \\ $$$$\boldsymbol{{point}}\:\boldsymbol{{A}}\left(−\mathrm{2}\sqrt{\mathrm{3}}\:,\mathrm{2}\right) \\ $$$${required}\:{area} \\ $$$$\left\{\int_{−\mathrm{4}} ^{−\mathrm{2}\sqrt{\mathrm{3}}} \sqrt{\mathrm{16}−{x}^{\mathrm{2}} }\:{dx}+\int_{−\mathrm{2}\sqrt{\mathrm{3}}\:} ^{−\mathrm{2}} \mathrm{2}{dx}\right\}−\left\{\int_{−\mathrm{4}} ^{−\mathrm{2}} \sqrt{\mathrm{4}−\left({x}+\mathrm{2}\right)^{\mathrm{2}} }\:{dx}\right\} \\ $$$${pls}\:{chk} \\ $$$$\int\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }\:{dx}=\frac{{x}}{\mathrm{2}}\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} \:}\:+\frac{{a}^{\mathrm{2}} }{\mathrm{2}}{sin}^{−\mathrm{1}} \left(\frac{{x}}{{a}}\right) \\ $$$${I}_{\mathrm{1}} =\mid\frac{{x}}{\mathrm{2}}\sqrt{\mathrm{16}−{x}^{\mathrm{2}} }\:+\frac{\mathrm{16}}{\mathrm{2}}{sin}^{−\mathrm{1}} \left(\frac{{x}}{\mathrm{4}}\right)\mid_{−\mathrm{4}} ^{−\mathrm{2}\sqrt{\mathrm{3}}\:} \\ $$$$=\left(\frac{−\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{2}}\sqrt{\mathrm{16}−\mathrm{12}}\:+\mathrm{8}{sin}^{−\mathrm{1}} \left(\frac{−\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{4}}\right)\right. \\ $$$$=−\sqrt{\mathrm{3}}\:×\mathrm{2}+\mathrm{8}\left(−\frac{\pi}{\mathrm{3}}\right)=−\mathrm{2}\sqrt{\mathrm{3}}\:−\frac{\mathrm{8}\pi}{\mathrm{3}} \\ $$$${I}_{\mathrm{2}} =\mathrm{2}\mid{x}\mid_{−\mathrm{2}\sqrt{\mathrm{3}}} ^{−\mathrm{2}} =\mathrm{2}\left(−\mathrm{2}+\mathrm{2}\sqrt{\mathrm{3}}\:\right)=\mathrm{4}\sqrt{\mathrm{3}}\:−\mathrm{4} \\ $$$$\bigstar\int_{−\mathrm{4}} ^{−\mathrm{2}} \sqrt{\mathrm{4}−\left({x}+\mathrm{2}\right)^{\mathrm{2}} }\:{dx} \\ $$$$\left({x}+\mathrm{2}\right)={p} \\ $$$$\int_{−\mathrm{2}} ^{\mathrm{0}} \sqrt{\mathrm{2}^{\mathrm{2}} −{p}^{\mathrm{2}} }\:{dp}=\mid\frac{{p}}{\mathrm{2}}\sqrt{\mathrm{2}^{\mathrm{2}} −{p}^{\mathrm{2}} }\:+\frac{\mathrm{4}}{\mathrm{2}}{sin}^{−\mathrm{1}} \left(\frac{{p}}{\mathrm{2}}\right)\mid_{−\mathrm{2}} ^{\mathrm{0}} \\ $$$$=\left(\mathrm{0}\right)−\left\{\frac{−\mathrm{2}}{\mathrm{2}}\sqrt{\mathrm{4}−\mathrm{4}}\:+\mathrm{2}{sin}^{−\mathrm{1}} \left(\frac{−\mathrm{2}}{\mathrm{2}}\right)\right\} \\ $$$$=\mathrm{0}−\mathrm{0}−\mathrm{2}{sin}^{−\mathrm{1}} \left(−\mathrm{1}\right)=−\mathrm{2}×\frac{−\pi}{\mathrm{2}}=\pi \\ $$$${rewuired}\:{answer} \\ $$$${I}_{\mathrm{1}} +{I}_{\mathrm{2}} −{I}_{\mathrm{3}} \\ $$$$=\left(−\mathrm{2}\sqrt{\mathrm{3}}\:−\frac{\mathrm{8}\pi}{\mathrm{3}}\right)+\left(\mathrm{4}\sqrt{\mathrm{3}}\:−\mathrm{4}\right)−\left(\pi\right) \\ $$$$=\mathrm{2}\sqrt{\mathrm{3}}\:−\mathrm{4}−\frac{\mathrm{11}\pi}{\mathrm{3}} \\ $$

Commented by TANMAY PANACEA last updated on 03/Nov/20

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