anonymous
  • anonymous
use parametric equations of the semi ellipse to find the area that it encloses. x=2cos(t), y=3sin(t), t=[0,pi]
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\frac{ x ^{2} }{ 2^{2} }+\frac{ y ^{2} }{ 3^{2} }=1\] solving for y \[y=\sqrt{3^{2}(1-\frac{ x ^{2} }{ 2^{2} })}\]
anonymous
  • anonymous
\[y=3\sqrt{(1-\frac{ x ^{2} }{ 4 })}\]
anonymous
  • anonymous
1/2 area of semi ellipse A= \[=\int\limits_{0}^{2}3\sqrt{(1-\frac{ x ^{?} }{ 4 })} dx\] substitute sin t=x/2, then dx=2 cos t dt, also t= [0,pi/2] 1/2 A=\[\int\limits_{0}^{\frac{ \pi }{ 2 }}3\sqrt{(1-\sin ^{2}t)}(2\cos t dt)\] =\[\int\limits_{0}^{\frac{ 0 }{ 2 }}6\cos ^{2}tdt\] but cos^2 t=(1+cos 2t)/2 \[=6\int\limits_{0}^{\pi/2}\frac{ (1+\cos 2t) }{ 2 }dt\] =\[\frac{ 6 }{ 2 }\left[ t+\frac{ \sin 2t }{ 2 } \right]from 0 \to \frac{ \pi }{ 2 }\] \[\frac{ 1 }{ 2 }A=3\left[ \frac{ \pi }{ 2 }+0-(0+0) \right]\] \[A=3\pi. \]

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anonymous
  • anonymous
thank you so much @mark_o.
anonymous
  • anonymous
YW good luck now :D ........ have fun solving :D

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