Question

use parametric equations of the semi ellipse to find the area that it encloses.
x=2cos(t), y=3sin(t), t=[0,pi]

Answer 1

\[\frac{ x ^{2} }{ 2^{2} }+\frac{ y ^{2} }{ 3^{2} }=1\]
solving for y
\[y=\sqrt{3^{2}(1-\frac{ x ^{2} }{ 2^{2} })}\]

Answer 2

\[y=3\sqrt{(1-\frac{ x ^{2} }{ 4 })}\]

Answer 3

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
\[=6\int\limits_{0}^{\pi/2}\frac{ (1+\cos 2t) }{ 2 }dt\]
=\[\frac{ 6 }{ 2 }\left[ t+\frac{ \sin 2t }{ 2 } \right]from 0 \to \frac{ \pi }{ 2 }\]
\[\frac{ 1 }{ 2 }A=3\left[ \frac{ \pi }{ 2 }+0-(0+0) \right]\]

\[A=3\pi. \]

Answer 4

thank you so much @mark_o.

Answer 5

YW good luck now :D ........ have fun solving :D