22,22=1
and
.7.(Or) Find the area between the ellipses
Answers
Answer:
The area of the smaller region by the ellipse \dfrac {x^2}{9}+\dfrac {y^2}{4}=1
9
x
2
+
4
y
2
=1 and the line \dfrac {x}{3}+\dfrac {y}{2}=1
3
x
+
2
y
=1 represented by the shaded region BCABBCAB
\therefore Area\ BCAB = Area\ (OBCAO) - Area\ (OBAO)∴Area BCAB=Area (OBCAO)−Area (OBAO)
=\displaystyle \int_{0}^{3}2\sqrt {1-\dfrac {x^2}{9}}dx-\int_{0}^{3}2\left (1-\dfrac {x}{3}\right )dx=∫
0
3
2
1−
9
x
2
dx−∫
0
3
2(1−
3
x
)dx
=\dfrac {2}{3}\left [\displaystyle \int_{0}^{3}\sqrt {9-x^2}dx\right ]-\dfrac {2}{3}\displaystyle \int_{0}^{3}(3-x)dx=
3
2
[∫
0
3
9−x
2
dx]−
3
2
∫
0
3
(3−x)dx
=\dfrac {2}{3}\left [\dfrac {x}{2}\sqrt {9-x^2}+\dfrac {9}{2}\sin^{-1}\dfrac {x}{3}\right ]_{0}^{3}-\dfrac {2}{3}\left [3x-\dfrac {x^2}{2}\right ]_{0}^{3}=
3
2
[
2
x
9−x
2
+
2
9
sin
−1
3
x
]
0
3
−
3
2
[3x−
2
x
2
]
0
3
=\dfrac {2}{3}\left [\dfrac {9}{2}\left (\dfrac {\pi}{2}\right )\right ]-\dfrac {2}{3}\left [9-\dfrac {9}{2}\right ]=
3
2
[
2
9
(
2
π
)]−
3
2
[9−
2
9
]
=\dfrac {2}{3}\left [\dfrac {9\pi}{4}-\dfrac {9}{2}\right ]=
3
2
[
4
9π
−
2
9
]
=\dfrac {2}{3}\times \dfrac {9}{4}(\pi-2)=
3
2
×
4
9
(π−2)
=\dfrac {3}{2}(\pi-2)=
2
3
(π−2) sq. units