Question

The area of the region bounded by the curve $\sqrt{x}+\sqrt{y}=\sqrt{a}(x,y>0)$ and the coordinate axes is

• A. $a^2$
• B. $\frac{a^2}{2}$
• C. $\frac{a^2}{3}$
• D. $\frac{a^2}{6}$

Answer 1

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

$\sqrt{x}+\sqrt{y}=\sqrt{a}\Rightarrow \sqrt{y}=\sqrt{a}-\sqrt{x}\Rightarrow y=a+x-2\sqrt{a}\sqrt{x}$
If y = 0, then x = a
$\therefore$ Area = ${\int }_0^{a}ydx={\int }_0^a(a+x-2\sqrt{a}\sqrt{x})dx={[\frac{x^2}{3}+ax-2\sqrt{a}\left(\frac{2}{3}\right)x^{\frac{3}{2}}]}_0^a$
$=\frac{a^2}{6}$