Question

Find the area of the smaller region bounded by the ellipse $\frac{x^2}{9}$+$\frac{y^2}{4}$=1 and the line

$\frac{x}{3}$+$\frac{y}{2}$=1

Answer 1

The area of the smaller region bounded by the ellipse,x2/9+y2/4=1,and the line,

$\frac{x}{3}$+$\frac{y}{2}$=1,is represented by the shaded region BCAB as

∴Area BCAB=Area(OBCAO)–Area(OBAO)

=

\int_{0}^{3} 2\sqrt{1-\frac{x^2}{9}} \,dx$$$$-\int_{0}^{3} 2(1- \frac {x}{3}) \,dx

=$\frac 23$[

$\int_{0}^{3} \sqrt{9-1^2} \,dx$

]-

$\int_{0}^{3} (3-x) \,dx$

=$\frac 23$[\frac{x}{2}$$\sqrt{9-x^2}+$\frac{9}{2}$sin-1$\frac{x}{3}$]30-$\frac 23$[3x-$\frac{x^2}{2}$]30

=$\frac 23$[$\frac{9}{2}$($\frac{π}{2}$)]-$\frac 23$[9-$\frac{9}{2}$]

=$\frac 23$[$\frac{9π}{4}$-$\frac{9}{2}$]

=$\frac 23$×$\frac{9}{4}$(π-2)

=$\frac 32$(π-2)units