The area bounded by the curve x+y=a and the coordinate axes is

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The area bounded by the curve $\sqrt{x} + \sqrt{y} = \sqrt{a}$ and the coordinate axes is

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$\frac{a^{2}}{4}$

$\frac{a^{2}}{8}$

$\frac{a^{2}}{6}$

$\frac{a}{6}$

answer is A.

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Detailed Solution

$a r e a = 4 \int_{0}^{a} (\sqrt{a} - \sqrt{x})^{2} d x$

$\begin{aligned} = \int_{0}^{a} a + x - 2 \sqrt{a} \sqrt{x} d x \\ = {[a x + \frac{x}{2} - 2 \sqrt{a} \cdot \frac{2}{3} x^{3 / 2}]}_{0}^{a} \\ = [\frac{3 a^{2}}{2} - \frac{4 a^{2}}{3}] \\ = a^{2} [\frac{3}{2} - \frac{4}{3}] \\ = a^{2} [\frac{9 - 8}{6}] = a^{2} [\frac{1}{6}] \\ = \frac{a^{2}}{6} \end{aligned}$