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
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
Solution and Explanation
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
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