Calculate the area bounded by the curve sqrt(x) + sqrt(y)

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

We will evaluate the area by expressing $y$ explicitly in terms of $x$ and integrating.

Step 1: Understand the curve

The equation is:

\[\sqrt{x} + \sqrt{y} = \sqrt{a}\]

where $a > 0$, $x \geq 0$, and $y \geq 0$ (since we need square roots to be real).

Step 2: Find intercepts

$x$-intercept: Set $y = 0$:

\[\sqrt{x} + 0 = \sqrt{a} \implies \sqrt{x} = \sqrt{a} \implies x = a\]

$y$-intercept: Set $x = 0$:

\[0 + \sqrt{y} = \sqrt{a} \implies \sqrt{y} = \sqrt{a} \implies y = a\]

So the curve passes through $(a, 0)$ and $(0, a)$.

Step 3: Express $y$ as a function of $x$

From $\sqrt{x} + \sqrt{y} = \sqrt{a}$:

\[\sqrt{y} = \sqrt{a} - \sqrt{x}\]

Squaring both sides:

\[y = (\sqrt{a} - \sqrt{x})^2 = a - 2\sqrt{a}\sqrt{x} + x\]

\[y = a - 2\sqrt{ax} + x\]

Step 4: Set up the integral

The area bounded by the curve and the coordinate axes is:

\[A = \int_0^a y \, dx = \int_0^a (a - 2\sqrt{ax} + x) \, dx\]

\[= \int_0^a a \, dx - \int_0^a 2\sqrt{ax} \, dx + \int_0^a x \, dx\]

Step 5: Evaluate each integral

First integral:

\[\int_0^a a \, dx = [ax]_0^a = a^2 - 0 = a^2\]

Second integral:

\[\int_0^a 2\sqrt{ax} \, dx = 2\sqrt{a} \int_0^a \sqrt{x} \, dx = 2\sqrt{a} \int_0^a x^{1/2} \, dx\]

\[= 2\sqrt{a} \left[\frac{x^{3/2}}{3/2}\right]_0^a = 2\sqrt{a} \cdot \frac{2}{3}[x^{3/2}]_0^a\]

\[= \frac{4\sqrt{a}}{3}[a^{3/2} - 0] = \frac{4\sqrt{a}}{3} \cdot a\sqrt{a} = \frac{4a^2}{3}\]

Third integral:

\[\int_0^a x \, dx = \left[\frac{x^2}{2}\right]_0^a = \frac{a^2}{2} - 0 = \frac{a^2}{2}\]

Step 6: Combine the results

\[A = a^2 - \frac{4a^2}{3} + \frac{a^2}{2}\]

Finding common denominator (6):

\[A = \frac{6a^2}{6} - \frac{8a^2}{6} + \frac{3a^2}{6} = \frac{6a^2 - 8a^2 + 3a^2}{6} = \frac{a^2}{6}\]

\[\boxed{A = \frac{a^2}{6} \text{ square units}}\]

Verification

Let's verify for $a = 6$ (to get a nice integer area):

\[A = \frac{36}{6} = 6 \text{ square units}\]

The curve passes through $(6, 0)$ and $(0, 6)$.

Alternative method: Symmetry

Due to symmetry about the line $y = x$, we could also compute:

\[A = 2\int_0^a y \, dx - a^2\]

But our direct method is simpler.

Summary

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

\[\boxed{A = \frac{a^2}{6} \text{ square units}}\]

Key steps:

1. Identify intercepts: $(a, 0)$ and $(0, a)$
2. Express $y$ in terms of $x$: $y = (\sqrt{a} - \sqrt{x})^2 = a - 2\sqrt{ax} + x$
3. Set up integral: $A = \int_0^a (a - 2\sqrt{ax} + x) \, dx$
4. Evaluate term by term: $a^2 - \frac{4a^2}{3} + \frac{a^2}{2}$
5. Simplify: $\frac{6a^2 - 8a^2 + 3a^2}{6} = \frac{a^2}{6}$

Key insight:

The curve $\sqrt{x} + \sqrt{y} = \sqrt{a}$ forms a concave arc connecting $(a, 0)$ to $(0, a)$, and the area under it is exactly $\frac{1}{6}$ of the square $[0, a] \times [0, a]$ which has area $a^2$.