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