Question-177451

Question Number 177451 by mr W last updated on 05/Oct/22

Commented by mr W last updated on 05/Oct/22

an elipse with a=10cm and b=6cm is moved by 5cm in x direction and by 3cm in y direction. find the shaded area.

$${an}\:{elipse}\:{with}\:{a}=\mathrm{10}{cm}\:{and}\:{b}=\mathrm{6}{cm}\:{is} \\ $$$${moved}\:{by}\:\mathrm{5}{cm}\:{in}\:{x}\:{direction}\:{and}\:{by} \\ $$$$\mathrm{3}{cm}\:{in}\:{y}\:{direction}.\:{find}\:{the}\:{shaded} \\ $$$${area}. \\ $$

Answered by mr W last updated on 05/Oct/22

Commented by mr W last updated on 05/Oct/22

red elipse: (x^2 /(10^2 ))+(y^2 /6^2 )=1 ⇒y=6(√(1−(x^2 /(100)))) blue elipse: (((x−5)^2 )/(10^2 ))+(((y−3)^2 )/6^2 )=1 line PQ: ((10x−25)/(10^2 ))+((6y−9)/6^2 )=0 ⇒y=3(1−(x/5)) point P and Q: (x^2 /(10^2 ))+(3^2 /6^2 )(1−(x/5))^2 =1 2x^2 −10x−75=0 x=((5±5(√7))/2) ⇒x_Q =((5(1+(√7)))/2), y_Q =((3(1−(√7)))/2) ⇒x_P =((5(1−(√7)))/2), y_P =((3(1+(√7)))/2) A_(segment) =∫_x_P ^x_Q [6(√(1−(x^2 /(100))))−3+((3x)/5)]dx+2∫_x_Q ^(10) 6(√(1−(x^2 /(100)))) dx =3∫_x_P ^x_Q [2(√(1−(x^2 /(100))))−1+(x/5)]dx+12∫_x_Q ^(10) (√(1−(x^2 /(100)))) dx =3[x(√(1−(x^2 /(100))))+10 sin^(−1) (x/(10))−x+(x^2 /(10))]_x_P ^x_Q +6[x(√(1−(x^2 /(100))))+10 sin^(−1) (x/(10))]_x_Q ^(10) =3{((5(1+(√7)))/2)(√(1−(((1+(√7))^2 )/(16))))−((5(1−(√7)))/2)(√(1−(((1−(√7))^2 )/(16))))+10( sin^(−1) ((1+(√7))/4)−sin^(−1) ((1−(√7))/4))−((5(1+(√7)−1+(√7)))/2)+((5[(1+(√7))^2 −(1−(√7))^2 ])/8)}+6{10(√(1−((100)/(100))))−((5(1+(√7)))/2)(√(1−(((1+(√7))^2 )/(16))))+10(sin^(−1) ((10)/(10))−sin^(−1) ((1+(√7))/4))} =((45)/2)+30( sin^(−1) ((1+(√7))/4)−sin^(−1) ((1−(√7))/4))−((15(√7))/2)−((45)/2)+30π−60 sin^(−1) ((1+(√7))/4) =30π−30( sin^(−1) ((1+(√7))/4)+ sin^(−1) ((1−(√7))/4))−((15(√7))/2) A_(elipse) =πab=10×6π=60π A_(shaded) =A_(elipse) −2A_(segment) =60π−60π+60(sin^(−1) ((1+(√7))/4)+sin^(−1) ((1−(√7))/4))+15(√7) =60(sin^(−1) ((1+(√7))/4)+ sin^(−1) ((1−(√7))/4))+15(√7) =60 cos^(−1) (3/4)+15(√7) ✓ ≈83.05

$${red}\:{elipse}: \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{10}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{\mathrm{6}^{\mathrm{2}} }=\mathrm{1}\:\Rightarrow{y}=\mathrm{6}\sqrt{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{100}}} \\ $$$${blue}\:{elipse}: \\ $$$$\frac{\left({x}−\mathrm{5}\right)^{\mathrm{2}} }{\mathrm{10}^{\mathrm{2}} }+\frac{\left({y}−\mathrm{3}\right)^{\mathrm{2}} }{\mathrm{6}^{\mathrm{2}} }=\mathrm{1} \\ $$$${line}\:{PQ}: \\ $$$$\frac{\mathrm{10}{x}−\mathrm{25}}{\mathrm{10}^{\mathrm{2}} }+\frac{\mathrm{6}{y}−\mathrm{9}}{\mathrm{6}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow{y}=\mathrm{3}\left(\mathrm{1}−\frac{{x}}{\mathrm{5}}\right) \\ $$$${point}\:{P}\:{and}\:{Q}: \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{10}^{\mathrm{2}} }+\frac{\mathrm{3}^{\mathrm{2}} }{\mathrm{6}^{\mathrm{2}} }\left(\mathrm{1}−\frac{{x}}{\mathrm{5}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{10}{x}−\mathrm{75}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{5}\pm\mathrm{5}\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$$$\Rightarrow{x}_{{Q}} =\frac{\mathrm{5}\left(\mathrm{1}+\sqrt{\mathrm{7}}\right)}{\mathrm{2}},\:{y}_{{Q}} =\frac{\mathrm{3}\left(\mathrm{1}−\sqrt{\mathrm{7}}\right)}{\mathrm{2}} \\ $$$$\Rightarrow{x}_{{P}} =\frac{\mathrm{5}\left(\mathrm{1}−\sqrt{\mathrm{7}}\right)}{\mathrm{2}},\:{y}_{{P}} =\frac{\mathrm{3}\left(\mathrm{1}+\sqrt{\mathrm{7}}\right)}{\mathrm{2}} \\ $$$${A}_{{segment}} =\int_{{x}_{{P}} } ^{{x}_{{Q}} } \left[\mathrm{6}\sqrt{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{100}}}−\mathrm{3}+\frac{\mathrm{3}{x}}{\mathrm{5}}\right]{dx}+\mathrm{2}\int_{{x}_{{Q}} } ^{\mathrm{10}} \mathrm{6}\sqrt{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{100}}}\:{dx} \\ $$$$\:\:=\mathrm{3}\int_{{x}_{{P}} } ^{{x}_{{Q}} } \left[\mathrm{2}\sqrt{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{100}}}−\mathrm{1}+\frac{{x}}{\mathrm{5}}\right]{dx}+\mathrm{12}\int_{{x}_{{Q}} } ^{\mathrm{10}} \sqrt{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{100}}}\:{dx} \\ $$$$\:\:=\mathrm{3}\left[{x}\sqrt{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{100}}}+\mathrm{10}\:\mathrm{sin}^{−\mathrm{1}} \frac{{x}}{\mathrm{10}}−{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{10}}\right]_{{x}_{{P}} } ^{{x}_{{Q}} } +\mathrm{6}\left[{x}\sqrt{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{100}}}+\mathrm{10}\:\mathrm{sin}^{−\mathrm{1}} \frac{{x}}{\mathrm{10}}\right]_{{x}_{{Q}} } ^{\mathrm{10}} \\ $$$$\:\:=\mathrm{3}\left\{\frac{\mathrm{5}\left(\mathrm{1}+\sqrt{\mathrm{7}}\right)}{\mathrm{2}}\sqrt{\mathrm{1}−\frac{\left(\mathrm{1}+\sqrt{\mathrm{7}}\right)^{\mathrm{2}} }{\mathrm{16}}}−\frac{\mathrm{5}\left(\mathrm{1}−\sqrt{\mathrm{7}}\right)}{\mathrm{2}}\sqrt{\mathrm{1}−\frac{\left(\mathrm{1}−\sqrt{\mathrm{7}}\right)^{\mathrm{2}} }{\mathrm{16}}}+\mathrm{10}\left(\:\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}+\sqrt{\mathrm{7}}}{\mathrm{4}}−\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}−\sqrt{\mathrm{7}}}{\mathrm{4}}\right)−\frac{\mathrm{5}\left(\mathrm{1}+\sqrt{\mathrm{7}}−\mathrm{1}+\sqrt{\mathrm{7}}\right)}{\mathrm{2}}+\frac{\mathrm{5}\left[\left(\mathrm{1}+\sqrt{\mathrm{7}}\right)^{\mathrm{2}} −\left(\mathrm{1}−\sqrt{\mathrm{7}}\right)^{\mathrm{2}} \right]}{\mathrm{8}}\right\}+\mathrm{6}\left\{\mathrm{10}\sqrt{\mathrm{1}−\frac{\mathrm{100}}{\mathrm{100}}}−\frac{\mathrm{5}\left(\mathrm{1}+\sqrt{\mathrm{7}}\right)}{\mathrm{2}}\sqrt{\mathrm{1}−\frac{\left(\mathrm{1}+\sqrt{\mathrm{7}}\right)^{\mathrm{2}} }{\mathrm{16}}}+\mathrm{10}\left(\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{10}}{\mathrm{10}}−\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}+\sqrt{\mathrm{7}}}{\mathrm{4}}\right)\right\} \\ $$$$\:\:=\frac{\mathrm{45}}{\mathrm{2}}+\mathrm{30}\left(\:\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}+\sqrt{\mathrm{7}}}{\mathrm{4}}−\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}−\sqrt{\mathrm{7}}}{\mathrm{4}}\right)−\frac{\mathrm{15}\sqrt{\mathrm{7}}}{\mathrm{2}}−\frac{\mathrm{45}}{\mathrm{2}}+\mathrm{30}\pi−\mathrm{60}\:\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}+\sqrt{\mathrm{7}}}{\mathrm{4}} \\ $$$$\:\:=\mathrm{30}\pi−\mathrm{30}\left(\:\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}+\sqrt{\mathrm{7}}}{\mathrm{4}}+\:\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}−\sqrt{\mathrm{7}}}{\mathrm{4}}\right)−\frac{\mathrm{15}\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$$${A}_{{elipse}} =\pi{ab}=\mathrm{10}×\mathrm{6}\pi=\mathrm{60}\pi \\ $$$${A}_{{shaded}} ={A}_{{elipse}} −\mathrm{2}{A}_{{segment}} \\ $$$$\:\:=\mathrm{60}\pi−\mathrm{60}\pi+\mathrm{60}\left(\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}+\sqrt{\mathrm{7}}}{\mathrm{4}}+\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}−\sqrt{\mathrm{7}}}{\mathrm{4}}\right)+\mathrm{15}\sqrt{\mathrm{7}} \\ $$$$\:\:=\mathrm{60}\left(\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}+\sqrt{\mathrm{7}}}{\mathrm{4}}+\:\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}−\sqrt{\mathrm{7}}}{\mathrm{4}}\right)+\mathrm{15}\sqrt{\mathrm{7}} \\ $$$$\:\:=\mathrm{60}\:\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{3}}{\mathrm{4}}+\mathrm{15}\sqrt{\mathrm{7}}\:\checkmark \\ $$$$\:\:\approx\mathrm{83}.\mathrm{05} \\ $$

Commented by Ar Brandon last updated on 05/Oct/22

Sir, why ...+2∫_x_Q ^(10) 6(√(1−(x^2 /(100))))dx atA_(segment) ? Why not just ...+∫_x_Q ^(10) 6(√(1−(x^2 /(100))))dx

$$\mathrm{Sir},\:\mathrm{why}\:…+\mathrm{2}\int_{{x}_{{Q}} } ^{\mathrm{10}} \mathrm{6}\sqrt{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{100}}}{dx}\:\mathrm{at}{A}_{\mathrm{segment}} ? \\ $$$$\mathrm{Why}\:\mathrm{not}\:\mathrm{just}\:…+\int_{{x}_{{Q}} } ^{\mathrm{10}} \mathrm{6}\sqrt{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{100}}}{dx} \\ $$

Commented by mr W last updated on 05/Oct/22

$${see}\:{diagram}: \\ $$

Commented by mr W last updated on 05/Oct/22

Commented by Ar Brandon last updated on 05/Oct/22

I see, Sir. But I don′t understand why you multiplied it by 2

$$\mathrm{I}\:\mathrm{see},\:\mathrm{Sir}.\:\mathrm{But}\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand}\:\mathrm{why}\: \\ $$$$\mathrm{you}\:\mathrm{multiplied}\:\mathrm{it}\:\mathrm{by}\:\mathrm{2} \\ $$

Commented by mr W last updated on 06/Oct/22

this is obviously wrong sir! you can count the area from following graph. each small square has an area of 4. we count approximately 21 small squares, so the area should be approximately ≈21×4=84

$${this}\:{is}\:{obviously}\:{wrong}\:{sir}! \\ $$$${you}\:{can}\:{count}\:{the}\:{area}\:{from}\:{following} \\ $$$${graph}.\:{each}\:{small}\:{square}\:{has}\:{an}\:{area} \\ $$$${of}\:\mathrm{4}. \\ $$$${we}\:{count}\:{approximately}\:\mathrm{21}\:{small} \\ $$$${squares},\:{so}\:{the}\:{area}\:{should}\:{be}\: \\ $$$${approximately}\:\approx\mathrm{21}×\mathrm{4}=\mathrm{84} \\ $$

Commented by mr W last updated on 05/Oct/22

Commented by mr W last updated on 05/Oct/22

∫_x_Q ^(10) ydx is only the area above the x axis. but we have also the area under the x axis, therefore 2∫_x_Q ^(10) ydx.

$$\int_{{x}_{{Q}} } ^{\mathrm{10}} {ydx}\:{is}\:{only}\:{the}\:{area}\:{above}\:{the}\: \\ $$$${x}\:{axis}.\:{but}\:{we}\:{have}\:{also}\:{the}\:{area} \\ $$$${under}\:{the}\:{x}\:{axis},\:{therefore}\:\mathrm{2}\int_{{x}_{{Q}} } ^{\mathrm{10}} {ydx}. \\ $$

Commented by Ar Brandon last updated on 05/Oct/22

OK got it now. Thanks!

Commented by a.lgnaoui last updated on 05/Oct/22

y_1 ^2 =(6^2 /(10^2 ))(100−x^2 ) (1) (y_2 +3)^2 =(6^2 /(10^2 ))[(100−(x+5)^2 ] (2) y_1 =y_2 (points communs (1)et(2) (1) y_1 =(6/(10))(√(100−x^2 )) (y+3)^2 =[3+(3/(10))(√(100−x^2 )) ]^2 =(9/(25))[(100−(x+5)^2 ] (9+((9/(25)))(100−x^2 )+((18)/5)(√(100−x^2 )) )=(9/(25))(75−x^2 −10x) (45−((9x^2 )/(25))+((18)/5)(√(100−x^2 )) )=(27−((9x^2 )/(25))−((18)/5)) 1+(1/5)(√(100−x^2 )) +(1/5)=0 x=9,92 ⇒intersection =(4,92 ; 9,92) y=(3/5)(√(100−24,20)) =5,22 P(4,92 ; 5,22) Q(9,92;−2,25) Aure=∫(6/5)+(1/5)(√(100−x^2 )) dx =(6/5)x+2∫(√(1−((x^2 /(10))))) dx sin t=(x/(10)) dx =10cos t dt=10(√(1−t^2 )) dt =10∫(1−t^2 ) dt=10t−10(t^3 /3) Aire=∫_0 ^5 =[((6x)/5) ]_0 ^5 +10[sin^(−1) ((x/(10)))]_0 ^5 −((10)/3)[(sin^(−1) ((x/(10))))^3 ]_0 ^5 =6+10sin^(−1) ((1/2))−((10)/3)[sin^(−1) ((1/2))]^3 6+2[10(π/6)−((10)/3)×(π^3 /(216))]=6+21,44−0,86 Aire=26,58 bj

$${y}_{\mathrm{1}} ^{\mathrm{2}} =\frac{\mathrm{6}^{\mathrm{2}} }{\mathrm{10}^{\mathrm{2}} }\left(\mathrm{100}−{x}^{\mathrm{2}} \right)\:\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\left({y}_{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} =\frac{\mathrm{6}^{\mathrm{2}} }{\mathrm{10}^{\mathrm{2}} }\left[\left(\mathrm{100}−\left({x}+\mathrm{5}\right)^{\mathrm{2}} \right]\:\:\left(\mathrm{2}\right)\right. \\ $$$${y}_{\mathrm{1}} ={y}_{\mathrm{2}} \left({points}\:{communs}\:\left(\mathrm{1}\right){et}\left(\mathrm{2}\right)\right. \\ $$$$\left(\mathrm{1}\right)\:\:\:{y}_{\mathrm{1}} =\frac{\mathrm{6}}{\mathrm{10}}\sqrt{\mathrm{100}−{x}^{\mathrm{2}} \:} \\ $$$$\left({y}+\mathrm{3}\right)^{\mathrm{2}} =\left[\mathrm{3}+\frac{\mathrm{3}}{\mathrm{10}}\sqrt{\mathrm{100}−{x}^{\mathrm{2}} }\:\right]^{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{25}}\left[\left(\mathrm{100}−\left({x}+\mathrm{5}\right)^{\mathrm{2}} \right]\right. \\ $$$$\left(\mathrm{9}+\left(\frac{\mathrm{9}}{\mathrm{25}}\right)\left(\mathrm{100}−{x}^{\mathrm{2}} \right)+\frac{\mathrm{18}}{\mathrm{5}}\sqrt{\mathrm{100}−{x}^{\mathrm{2}} \:}\:\right)=\frac{\mathrm{9}}{\mathrm{25}}\left(\mathrm{75}−{x}^{\mathrm{2}} −\mathrm{10}{x}\right) \\ $$$$\left(\mathrm{45}−\frac{\mathrm{9}{x}^{\mathrm{2}} }{\mathrm{25}}+\frac{\mathrm{18}}{\mathrm{5}}\sqrt{\mathrm{100}−{x}^{\mathrm{2}} \:}\:\right)=\left(\mathrm{27}−\frac{\mathrm{9}{x}^{\mathrm{2}} }{\mathrm{25}}−\frac{\mathrm{18}}{\mathrm{5}}\right) \\ $$$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{5}}\sqrt{\mathrm{100}−{x}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{5}}=\mathrm{0} \\ $$$${x}=\mathrm{9},\mathrm{92}\:\:\Rightarrow{intersection}\:=\left(\mathrm{4},\mathrm{92}\:;\:\mathrm{9},\mathrm{92}\right) \\ $$$${y}=\frac{\mathrm{3}}{\mathrm{5}}\sqrt{\mathrm{100}−\mathrm{24},\mathrm{20}}\:=\mathrm{5},\mathrm{22} \\ $$$$\mathrm{P}\left(\mathrm{4},\mathrm{92}\:;\:\mathrm{5},\mathrm{22}\right)\:\:\:\:\mathrm{Q}\left(\mathrm{9},\mathrm{92};−\mathrm{2},\mathrm{25}\right) \\ $$$${Aure}=\int\frac{\mathrm{6}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{5}}\sqrt{\mathrm{100}−{x}^{\mathrm{2}} }\:{dx} \\ $$$$=\frac{\mathrm{6}}{\mathrm{5}}{x}+\mathrm{2}\int\sqrt{\mathrm{1}−\left(\frac{{x}^{\mathrm{2}} }{\mathrm{10}}\right)}\:{dx} \\ $$$$\mathrm{sin}\:{t}=\frac{{x}}{\mathrm{10}}\:\:\:\:{dx}\:=\mathrm{10cos}\:{t}\:{dt}=\mathrm{10}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} \:}\:{dt} \\ $$$$=\mathrm{10}\int\left(\mathrm{1}−{t}^{\mathrm{2}} \right)\:{dt}=\mathrm{10}{t}−\mathrm{10}\frac{{t}^{\mathrm{3}} }{\mathrm{3}} \\ $$$${Aire}=\int_{\mathrm{0}} ^{\mathrm{5}} \:\:=\left[\frac{\mathrm{6}{x}}{\mathrm{5}}\:\:\right]_{\mathrm{0}} ^{\mathrm{5}} +\mathrm{10}\left[\mathrm{sin}^{−\mathrm{1}} \left(\frac{{x}}{\mathrm{10}}\right)\right]_{\mathrm{0}} ^{\mathrm{5}} −\frac{\mathrm{10}}{\mathrm{3}}\left[\left(\mathrm{sin}^{−\mathrm{1}} \left(\frac{{x}}{\mathrm{10}}\right)\right)^{\mathrm{3}} \right]_{\mathrm{0}} ^{\mathrm{5}} \\ $$$$=\mathrm{6}+\mathrm{10sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\frac{\mathrm{10}}{\mathrm{3}}\left[\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right]^{\mathrm{3}} \\ $$$$\mathrm{6}+\mathrm{2}\left[\mathrm{10}\frac{\pi}{\mathrm{6}}−\frac{\mathrm{10}}{\mathrm{3}}×\frac{\pi^{\mathrm{3}} }{\mathrm{216}}\right]=\mathrm{6}+\mathrm{21},\mathrm{44}−\mathrm{0},\mathrm{86} \\ $$$${Aire}=\mathrm{26},\mathrm{58} \\ $$$${bj} \\ $$

Commented by mr W last updated on 05/Oct/22