Find the area of the region enclosed by the given curves .
`x=4-y^(2), x=0`

Answer

To find the area of the region enclosed by the curves \( x = 4 - y^2 \) and \( x = 0 \), we will follow these steps:

### Step 1: Identify the curves and their intersection points
The curves given are:
1. \( x = 4 - y^2 \) (a parabola opening to the left)
2. \( x = 0 \) (the y-axis)

To find the points of intersection, we set \( x = 0 \):
\[ 0 = 4 - y^2 \]
This simplifies to:
\[ y^2 = 4 \]
Taking the square root gives us:
\[ y = 2 \quad \text{and} \quad y = -2 \]
Thus, the points of intersection are \( (0, 2) \) and \( (0, -2) \).

### Step 2: Set up the integral for the area
The area \( A \) between the curves can be found by integrating the function \( x = 4 - y^2 \) with respect to \( y \) from \( y = -2 \) to \( y = 2 \):
\[ A = \int_{-2}^{2} (4 - y^2) \, dy \]

### Step 3: Compute the integral
Now we will calculate the integral:
\[ A = \int_{-2}^{2} (4 - y^2) \, dy \]
This can be split into two separate integrals:
\[ A = \int_{-2}^{2} 4 \, dy - \int_{-2}^{2} y^2 \, dy \]

Calculating the first integral:
\[ \int_{-2}^{2} 4 \, dy = 4y \bigg|_{-2}^{2} = 4(2) - 4(-2) = 8 + 8 = 16 \]

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

### Step 4: Combine the results
Now we can find the area:
\[ A = 16 - \frac{16}{3} \]
To subtract these, we convert 16 into a fraction:
\[ 16 = \frac{48}{3} \]
Thus:
\[ A = \frac{48}{3} - \frac{16}{3} = \frac{32}{3} \]

### Final Answer
The area of the region enclosed by the curves \( x = 4 - y^2 \) and \( x = 0 \) is:
\[ \boxed{\frac{32}{3}} \]