Question

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

• A. $a^2/6$
• B. $a^2/2$
• C. $a^2/3$
• D. $a^2$

Answer 1

The correct option is A $a^2/6$

$A={\int }_0^{a}f\left(x\right)dx\sqrt{x}+\sqrt{y}=\sqrt{a}\sqrt{y}=\sqrt{a}-\sqrt{x}y=(\sqrt{a}-\sqrt{x}{)}^{2}y=a+x-2\sqrt{a}\sqrt{x}A=in{t}_0^a(a+x-2\sqrt{a}\sqrt{x})dx={\int }_0^{a}dx+{\int }_0^{a}xdx-2\sqrt{a}{\int }_0^a{x}^{1/2}dx=a[x{]}_0^a+[{]}_0^a+[\frac{x^2}{2}{]}_0^a-2\sqrt{a}[\frac{2}{3}{x}^{3/2}{]}_0^a=a[a-0]+[\frac{a^2}{2}-\frac{0^2}{2}]-2\sqrt{a}[\frac{2}{3}{a}^{3/2}]=a^2+\frac{a^2}{2}-\frac{4}{3}{a}^{3/2+1/2}=\frac{3}{2}{a}^2-\frac{4}{3}{a}^2=\frac{1}{6}{a}^2$

Co-ordinates axes$=\frac{a^2}{6}$