The area of the region bounded by the curve `sqrtx+sqrty=sqrta(x,ygt0)` and the coondinate axes is

Answer

To find the area of the region bounded by the curve \( \sqrt{x} + \sqrt{y} = \sqrt{a} \) (for \( x, y > 0 \)) and the coordinate axes, we can follow these steps:

### Step 1: Rearranging the equation
We start with the equation of the curve:
\[ \sqrt{x} + \sqrt{y} = \sqrt{a} \]
We can isolate \( \sqrt{y} \):
\[ \sqrt{y} = \sqrt{a} - \sqrt{x} \]
Now, squaring both sides gives:
\[ y = (\sqrt{a} - \sqrt{x})^2 \]

### Step 2: Expanding the equation
Expanding the right-hand side:
\[ y = a - 2\sqrt{a}\sqrt{x} + x \]
This can be rewritten as:
\[ y = x + a - 2\sqrt{a}\sqrt{x} \]

### Step 3: Setting up the integral for area
To find the area under the curve from \( x = 0 \) to \( x = a \), we will integrate the function \( y \):
\[ \text{Area} = \int_0^a (a + x - 2\sqrt{a}\sqrt{x}) \, dx \]

### Step 4: Splitting the integral
We can split the integral into three parts:
\[ \text{Area} = \int_0^a a \, dx + \int_0^a x \, dx - 2\sqrt{a} \int_0^a \sqrt{x} \, dx \]

### Step 5: Calculating each integral
1. The first integral:
\[ \int_0^a a \, dx = a \cdot x \bigg|_0^a = a^2 \]
2. The second integral:
\[ \int_0^a x \, dx = \frac{x^2}{2} \bigg|_0^a = \frac{a^2}{2} \]
3. The third integral:
\[ \int_0^a \sqrt{x} \, dx = \frac{2}{3} x^{3/2} \bigg|_0^a = \frac{2}{3} a^{3/2} \]

### Step 6: Combining the results
Now substituting back into the area formula:
\[ \text{Area} = a^2 + \frac{a^2}{2} - 2\sqrt{a} \cdot \frac{2}{3} a^{3/2} \]
This simplifies to:
\[ \text{Area} = a^2 + \frac{a^2}{2} - \frac{4}{3} a^2 \]
Combining the terms:
\[ \text{Area} = a^2 \left(1 + \frac{1}{2} - \frac{4}{3}\right) \]
Finding a common denominator (6):
\[ \text{Area} = a^2 \left(\frac{6}{6} + \frac{3}{6} - \frac{8}{6}\right) = a^2 \left(\frac{1}{6}\right) \]

### Final Result
Thus, the area of the region bounded by the curve and the coordinate axes is:
\[ \text{Area} = \frac{a^2}{6} \]