Question

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

• A. $\frac{a^2}{6}$
• B. $a^2$
• C. $\frac{a^2}{2}$
• D. $\frac{a^2}{4}$

Answer 1

The correct option is A $\frac{a^2}{6}$

$y=(\sqrt{a}-\sqrt{x}{)}^2$
$=a+x-2\sqrt{ax}$
Hence the required area
$={\int }_0^{a}y.dx$
$={\int }_0^{a}a+x-2\sqrt{ax}.dx$
$=[ax+\frac{x^2}{2}-\frac{4x\sqrt{ax}}{3}{]}_0^a$
$=a^2+\frac{a^2}{2}-\frac{4a^2}{3}$
$=\frac{3a^2}{2}-\frac{4a^2}{3}$
$=\frac{9a^2-8a^2}{6}$
$=\frac{a^2}{6}$. sq units.