The area bounded by the curve `x = 4 - y^(2)` and the Y-axis is

Options

A. 16 sq units
B. 32 sq units
C. `(32)/(3)` sq units
D. `(16)/(3)` sq units

Explanation

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

### Step 1: Understand the curve
The equation \( x = 4 - y^2 \) represents a sideways parabola that opens to the left. The vertex of this parabola is at the point (4, 0). The curve intersects the Y-axis when \( x = 0 \).

### Step 2: Find the intersection points
To find the intersection points with the Y-axis, set \( x = 0 \):
\[ 0 = 4 - y^2 \]
This simplifies to:
\[ y^2 = 4 \implies y = \pm 2 \]
Thus, the curve intersects the Y-axis at the points (0, 2) and (0, -2).

### Step 3: Set up the integral for the area
The area bounded by the curve and the Y-axis can be found by integrating the function from \( y = -2 \) to \( y = 2 \). We express \( x \) in terms of \( y \):
\[ x = 4 - y^2 \]
The area \( A \) can be calculated as:
\[ A = \int_{-2}^{2} (4 - y^2) \, dy \]

### Step 4: Calculate the integral
Now we compute 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:
\[ \int_{-2}^{2} 4 \, dy = 4[y]_{-2}^{2} = 4(2 - (-2)) = 4(4) = 16 \]

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 5: Combine the results
Now, substituting back into the area equation:
\[ A = 16 - \frac{16}{3} \]
To combine these, convert 16 into a fraction:
\[ 16 = \frac{48}{3} \]
So,
\[ A = \frac{48}{3} - \frac{16}{3} = \frac{32}{3} \]

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