Find the area bounded by the parabola `x =4 -y ^(2) and ` y -axis.

Answer

To find the area bounded by the parabola \( x = 4 - y^2 \) and the y-axis, we can follow these steps:

### Step 1: Understand the parabola
The equation \( x = 4 - y^2 \) represents a parabola that opens to the left. To find the points where it intersects the y-axis, we set \( x = 0 \).

### Step 2: Find the intersection points
Setting \( x = 0 \):
\[ 0 = 4 - y^2 \implies y^2 = 4 \implies y = \pm 2 \]
So, the parabola intersects the y-axis at the points \( (0, 2) \) and \( (0, -2) \).

### Step 3: Determine the area to be calculated
The area we want to find is bounded by the parabola and the y-axis, from \( y = -2 \) to \( y = 2 \).

### Step 4: Set up the integral for the area
The area \( A \) can be calculated using the integral:
\[ A = \int_{-2}^{2} (4 - y^2) \, dy \]
Here, \( 4 - y^2 \) is the function that gives the x-coordinate of the parabola for a given y.

### Step 5: Calculate the integral
Now we will compute the integral:
\[ A = \int_{-2}^{2} (4 - y^2) \, dy \]
This can be split into two integrals:
\[ A = \int_{-2}^{2} 4 \, dy - \int_{-2}^{2} y^2 \, dy \]

Calculating the first integral:
\[ \int_{-2}^{2} 4 \, dy = 4[y]_{-2}^{2} = 4(2 - (-2)) = 4 \times 4 = 16 \]

Calculating the second integral:
\[ \int_{-2}^{2} y^2 \, dy = \left[ \frac{y^3}{3} \right]_{-2}^{2} = \frac{(2)^3}{3} - \frac{(-2)^3}{3} = \frac{8}{3} - \left(-\frac{8}{3}\right) = \frac{8}{3} + \frac{8}{3} = \frac{16}{3} \]

### Step 6: Combine the results
Now substituting back into the area formula:
\[ A = 16 - \frac{16}{3} = \frac{48}{3} - \frac{16}{3} = \frac{32}{3} \]

### Final Answer
Thus, the area bounded by the parabola \( x = 4 - y^2 \) and the y-axis is:
\[ \boxed{\frac{32}{3}} \]