<s>[INST] Is the solution to the following correct? Graph the area that is bounded by $\sqrt{x} + \sqrt{y} = \sqrt{a}$
$y = +(\sqrt{a} - \sqrt{x})^2 $
First we find the zeros:
$$\sqrt{a} - \sqrt{x} = 0 \iff x = a$$
Also $a$ cannot be negative so we take integral bounds of lower bound of 0 and upper bound of a.
$$\frac{S}{2} = \int_{0}^{a} (\sqrt{a} - \sqrt{x})^2dx = \int_{0}^a (a + x - 2 \sqrt{ax}) = a\cdot a + a^2/2 -2\sqrt{a}\cdot a^{3/2}\frac{2}{3}) = a^2(1 + 1/2 - 2\cdot a\cdot\frac{2}{3}) = a^2(3/2 - a\cdot\frac{4}{3})
$$
Mistakes:
$$\frac{S}{2} \text{would have to be only } S$$
$$ \text{The second last line should be:}a\cdot a + a^2/2 -2\sqrt{a}\cdot a^{3/2}\frac{2}{3} = a^2(3/2-4/3)$$
 [/INST] No, note that the graph of the function can only lie in the first quadrant. Hence what you marked $\frac S2$ would actually be just $S$.
Also note the calculation error in the second last line, $a^{\frac 12}.a^{\frac 32}=a^2$ and not $a^3$.
</s>