The area of the region bounded by `x^(2)=y-2, y=4, y=6` and the Y-axis in the first quadrant is

Options

A. `2/3`
B. `2/3(8-sqrt(2))`
C. `2/3(8-2sqrt(2))`
D. `3/2(8-sqrt(2))`

Answer

To find the area of the region bounded by the curves \(x^2 = y - 2\), \(y = 4\), \(y = 6\), and the Y-axis in the first quadrant, we will follow these steps:

### Step 1: Understand the curves and the area to be calculated
The equation \(x^2 = y - 2\) can be rewritten as \(y = x^2 + 2\). This is a parabola that opens upwards. The lines \(y = 4\) and \(y = 6\) are horizontal lines. We need to find the area between these curves in the first quadrant, bounded by the Y-axis.

### Step 2: Determine the limits of integration
The area of interest is between \(y = 4\) and \(y = 6\). We will integrate with respect to \(y\) from \(y = 4\) to \(y = 6\).

### Step 3: Express \(x\) in terms of \(y\)
From the equation \(x^2 = y - 2\), we can express \(x\) as:
\[ x = \sqrt{y - 2} \]
This is valid for \(y \geq 2\), which is satisfied in our range from \(y = 4\) to \(y = 6\).

### Step 4: Set up the integral for the area
The area \(A\) can be calculated using the integral:
\[ A = \int_{y=4}^{y=6} x \, dy = \int_{4}^{6} \sqrt{y - 2} \, dy \]

### Step 5: Perform the substitution
Let \(t = y - 2\). Then, \(dy = dt\) and the limits change as follows:
- When \(y = 4\), \(t = 4 - 2 = 2\)
- When \(y = 6\), \(t = 6 - 2 = 4\)

Thus, the integral becomes:
\[ A = \int_{2}^{4} \sqrt{t} \, dt \]

### Step 6: Integrate
The integral of \(\sqrt{t}\) is:
\[ \int \sqrt{t} \, dt = \frac{2}{3} t^{3/2} \]
Now we evaluate the definite integral:
\[ A = \left[ \frac{2}{3} t^{3/2} \right]_{2}^{4} \]
Calculating this gives:
\[ A = \frac{2}{3} \left( 4^{3/2} - 2^{3/2} \right) \]

### Step 7: Calculate the values
Calculating \(4^{3/2}\) and \(2^{3/2}\):
- \(4^{3/2} = (2^2)^{3/2} = 2^3 = 8\)
- \(2^{3/2} = 2 \sqrt{2}\)

Substituting back into the area expression:
\[ A = \frac{2}{3} (8 - 2\sqrt{2}) \]

### Final Result
Thus, the area of the region bounded by the curves is:
\[ A = \frac{2}{3} (8 - 2\sqrt{2}) = \frac{16 - 4\sqrt{2}}{3} \]