Question

Find the area of the region${(x,y):y^2≤4x,4x^2+4y^2≤9}$

Answer 1

The correct answer is:$∫^{\frac{1}{2}}_02\sqrt{x}dx+∫^{\frac{3}{2}}_{\frac{1}{2}}\frac{1}{2}\sqrt{(3)^2-(2x)^2}dx$
The area bounded by the curves,${(x,y):y^2≤4x,4x^2+4y^2≤9}$,is represented as

The points of intersection of both the curves are$(\frac{1}{2},\sqrt{2})$and$(\frac{1}{2},-\sqrt{2}).$
The required area is given by OABCO.
It can be observed that area OABCO is symmetrical about $x$-axis.
$∴ Area\,\, OABCO=2\times Area\,\, OBC$
$Area\,\, OBCO=Area\,\, OMC+Area\,\, MBC$
$=∫^{\frac{1}{2}}_02\sqrt{x}dx+∫^{\frac{3}{2}}_{\frac{1}{2}}\frac{1}{2}\sqrt{9-4x^2}dx$
$=∫^{\frac{1}{2}}_02\sqrt{x}dx+∫^{\frac{3}{2}}_{\frac{1}{2}}\frac{1}{2}\sqrt{(3)^2-(2x)^2}dx$