The area of the portion enclosed by the curve $$\sqrt x + \sqrt y = \sqrt a $$    and the axes of reference is :

Solution :
$$\eqalign{ & y = {\left( {\sqrt a - \sqrt x } \right)^2} \cr & = a + x - 2\sqrt {ax} \cr & {\text{Hence the required area}} \cr & = \int_0^a {y.dx} \cr & = \int_0^a {a + x - 2\sqrt {ax} .dx} \cr & = \left[ {ax + \frac{{{x^2}}}{2} - \frac{{4x\sqrt {ax} }}{3}} \right]_0^a \cr & = {a^2} + \frac{{{a^2}}}{2} - \frac{{4{a^2}}}{3} \cr & = \frac{{3{a^2}}}{2} - \frac{{4{a^2}}}{3} \cr & = \frac{{9{a^2} - 8{a^2}}}{6} \cr & = \frac{{{a^2}}}{6}\,\,{\text{sq}}{\text{.}}\,{\text{units}} \cr} $$