Question-208896

Question Number 208896 by efronzo1 last updated on 26/Jun/24

Answered by MM42 last updated on 27/Jun/24

s_1 =32−∫_0 ^4 (√(64−x^2 ))dx ; x=8sinθ ⇒=32−64∫_0 ^(π/6) cos^2 θdθ ⇒s_1 =32−8(√3)−((16π)/3)= y ∫_4 ^8 (√(64−x^2 )) dx=((32π)/3)−8(√3) ∫_4 ^8 (√(16−(x−4)^2 ))dx ; x−4=u =∫_0 ^4 (√(16−u^2 )) du=4π ⇒s_2 =((20π)/3)−8(√3) =x ⇒Ans=((32−8(√3)−((16π)/3))/(((20π)/3)−8(√3))) ✓

$${s}_{\mathrm{1}} =\mathrm{32}−\int_{\mathrm{0}} ^{\mathrm{4}} \sqrt{\mathrm{64}−{x}^{\mathrm{2}} }{dx}\:\:\:\:\:;\:\:{x}=\mathrm{8}{sin}\theta \\ $$$$\Rightarrow=\mathrm{32}−\mathrm{64}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:{cos}^{\mathrm{2}} \theta{d}\theta \\ $$$$\Rightarrow{s}_{\mathrm{1}} =\mathrm{32}−\mathrm{8}\sqrt{\mathrm{3}}−\frac{\mathrm{16}\pi}{\mathrm{3}}=\:{y} \\ $$$$\int_{\mathrm{4}} ^{\mathrm{8}} \sqrt{\mathrm{64}−{x}^{\mathrm{2}} }\:{dx}=\frac{\mathrm{32}\pi}{\mathrm{3}}−\mathrm{8}\sqrt{\mathrm{3}} \\ $$$$\int_{\mathrm{4}} ^{\mathrm{8}} \sqrt{\mathrm{16}−\left({x}−\mathrm{4}\right)^{\mathrm{2}} }{dx}\:\:\:\:;\:\:{x}−\mathrm{4}={u} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{4}} \sqrt{\mathrm{16}−{u}^{\mathrm{2}} }\:{du}=\mathrm{4}\pi \\ $$$$\Rightarrow{s}_{\mathrm{2}} =\frac{\mathrm{20}\pi}{\mathrm{3}}−\mathrm{8}\sqrt{\mathrm{3}}\:={x} \\ $$$$\Rightarrow{Ans}=\frac{\mathrm{32}−\mathrm{8}\sqrt{\mathrm{3}}−\frac{\mathrm{16}\pi}{\mathrm{3}}}{\frac{\mathrm{20}\pi}{\mathrm{3}}−\mathrm{8}\sqrt{\mathrm{3}}}\:\:\:\checkmark \\ $$$$ \\ $$

Commented by A5T last updated on 26/Jun/24

4=8sinθ⇒sinθ=(1/2)⇒θ=(π/6) Shouldn′t the second line be: 32−64∫_0 ^(π/6) cos^2 θdθ?

$$\mathrm{4}=\mathrm{8}{sin}\theta\Rightarrow{sin}\theta=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\theta=\frac{\pi}{\mathrm{6}} \\ $$$${Shouldn}'{t}\:{the}\:{second}\:{line}\:{be}:\:\mathrm{32}−\mathrm{64}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} {cos}^{\mathrm{2}} \theta{d}\theta? \\ $$

Commented by A5T last updated on 26/Jun/24

Also, shouldn′t s_2 =x=∫_4 ^8 (√(64−x^2 ))dx−((16π)/4) =((20π)/3)−8(√3) ?

$${Also},\:{shouldn}'{t}\:{s}_{\mathrm{2}} ={x}=\int_{\mathrm{4}} ^{\mathrm{8}} \sqrt{\mathrm{64}−{x}^{\mathrm{2}} }{dx}−\frac{\mathrm{16}\pi}{\mathrm{4}} \\ $$$$=\frac{\mathrm{20}\pi}{\mathrm{3}}−\mathrm{8}\sqrt{\mathrm{3}}\:? \\ $$

Answered by mr W last updated on 26/Jun/24

big circle: y=(√(8^2 −x^2 )) small circle: y=(√(4^2 −(x−4)^2 ))=(√(8x−x^2 )) Y=∫_0 ^4 (8−(√(8^2 −x^2 )))dx=32−8(√3)−((16π)/3) X=∫_4 ^8 ((√(8^2 −x^2 ))−(√(8x−x^2 )))dx=((20π)/3)−8(√3) (Y/X)=((32−8(√3)−((16π)/3))/(((20π)/3)−8(√3)))≈0.196 or X=4×8−(8×8−((8^2 π)/4)−Y)−((4^2 π)/4) =−32+12π+32−8(√3)−((16π)/3) =((20π)/3)−8(√3)

$${big}\:{circle}: \\ $$$${y}=\sqrt{\mathrm{8}^{\mathrm{2}} −{x}^{\mathrm{2}} } \\ $$$${small}\:{circle}: \\ $$$${y}=\sqrt{\mathrm{4}^{\mathrm{2}} −\left({x}−\mathrm{4}\right)^{\mathrm{2}} }=\sqrt{\mathrm{8}{x}−{x}^{\mathrm{2}} } \\ $$$${Y}=\int_{\mathrm{0}} ^{\mathrm{4}} \left(\mathrm{8}−\sqrt{\mathrm{8}^{\mathrm{2}} −{x}^{\mathrm{2}} }\right){dx}=\mathrm{32}−\mathrm{8}\sqrt{\mathrm{3}}−\frac{\mathrm{16}\pi}{\mathrm{3}} \\ $$$${X}=\int_{\mathrm{4}} ^{\mathrm{8}} \left(\sqrt{\mathrm{8}^{\mathrm{2}} −{x}^{\mathrm{2}} }−\sqrt{\mathrm{8}{x}−{x}^{\mathrm{2}} }\right){dx}=\frac{\mathrm{20}\pi}{\mathrm{3}}−\mathrm{8}\sqrt{\mathrm{3}} \\ $$$$\frac{{Y}}{{X}}=\frac{\mathrm{32}−\mathrm{8}\sqrt{\mathrm{3}}−\frac{\mathrm{16}\pi}{\mathrm{3}}}{\frac{\mathrm{20}\pi}{\mathrm{3}}−\mathrm{8}\sqrt{\mathrm{3}}}\approx\mathrm{0}.\mathrm{196} \\ $$$${or} \\ $$$${X}=\mathrm{4}×\mathrm{8}−\left(\mathrm{8}×\mathrm{8}−\frac{\mathrm{8}^{\mathrm{2}} \pi}{\mathrm{4}}−{Y}\right)−\frac{\mathrm{4}^{\mathrm{2}} \pi}{\mathrm{4}} \\ $$$$\:\:\:\:=−\mathrm{32}+\mathrm{12}\pi+\mathrm{32}−\mathrm{8}\sqrt{\mathrm{3}}−\frac{\mathrm{16}\pi}{\mathrm{3}} \\ $$$$\:\:\:\:=\frac{\mathrm{20}\pi}{\mathrm{3}}−\mathrm{8}\sqrt{\mathrm{3}} \\ $$

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