What is the area bounded by the curve `sqrt(x)+sqrt(y)=sqrt(a)(x,yge0)` and the coordinate axes?

Options

A. `(5a^(2))/6`
B. `(a^(2))/3`
C. `(a^(2))/2`
D. `(a^(2))/6`

Answer

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

### Step 1: Identify the Points of Intersection
To find the points where the curve intersects the axes, we can set \(x = 0\) and \(y = 0\) separately.

1. **When \(x = 0\)**:
\[ \sqrt{0} + \sqrt{y} = \sqrt{a} \implies \sqrt{y} = \sqrt{a} \implies y = a \]
This gives the point \((0, a)\).

2. **When \(y = 0\)**:
\[ \sqrt{x} + \sqrt{0} = \sqrt{a} \implies \sqrt{x} = \sqrt{a} \implies x = a \]
This gives the point \((a, 0)\).

### Step 2: Rewrite the Equation for \(y\)
From the equation \(\sqrt{x} + \sqrt{y} = \sqrt{a}\), we can express \(y\) in terms of \(x\):
\[ \sqrt{y} = \sqrt{a} - \sqrt{x} \]
Squaring both sides gives:
\[ y = (\sqrt{a} - \sqrt{x})^2 = a - 2\sqrt{a}\sqrt{x} + x \]

### Step 3: Set Up the Integral for Area
The area \(A\) bounded by the curve and the axes can be calculated using the integral:
\[ A = \int_{0}^{a} y \, dx \]
Substituting the expression for \(y\):
\[ A = \int_{0}^{a} (a - 2\sqrt{a}\sqrt{x} + x) \, dx \]

### Step 4: Evaluate the Integral
Now we evaluate the integral:
\[ A = \int_{0}^{a} a \, dx - 2\sqrt{a} \int_{0}^{a} \sqrt{x} \, dx + \int_{0}^{a} x \, dx \]

1. **First Integral**:
\[ \int_{0}^{a} a \, dx = a \cdot x \bigg|_{0}^{a} = a^2 \]

2. **Second Integral**:
\[ \int_{0}^{a} \sqrt{x} \, dx = \frac{2}{3} x^{3/2} \bigg|_{0}^{a} = \frac{2}{3} a^{3/2} \]
Thus,
\[ -2\sqrt{a} \cdot \frac{2}{3} a^{3/2} = -\frac{4}{3} a^2 \]

3. **Third Integral**:
\[ \int_{0}^{a} x \, dx = \frac{x^2}{2} \bigg|_{0}^{a} = \frac{a^2}{2} \]

### Step 5: Combine the Results
Now we combine all these results:
\[ A = a^2 - \frac{4}{3} a^2 + \frac{1}{2} a^2 \]
Finding a common denominator (which is 6):
\[ A = \frac{6a^2}{6} - \frac{8a^2}{6} + \frac{3a^2}{6} = \frac{6 - 8 + 3}{6} a^2 = \frac{1}{6} a^2 \]

### Final Answer
Thus, the area bounded by the curve and the coordinate axes is:
\[ \boxed{\frac{a^2}{6}} \]