Question

The area bounded by the curves \(x = 4 - y^2\) and the Y-axis is:

• A. \(16 \, \text{sq. unit}\)
• B. \(\frac{32}{3} \, \text{sq. unit}\)
• C. \(\frac{16}{3} \, \text{sq. unit}\)
• D. \(32 \, \text{sq. unit}\)

Hint:

To compute areas bounded by curves, set the integral limits based on where the curve intersects the axes, then evaluate the integral.

Answer 1

The Correct Option is B

Step 1: The given equation is \( x = 4 - y^2 \). The area is bounded by the curve and the Y-axis. We first find the limits of integration by setting \( x = 0 \):

\[ 0 = 4 - y^2 \quad \implies \quad y^2 = 4 \quad \implies \quad y = \pm 2 \]

Step 2: The area is the integral of \( x = 4 - y^2 \) from \( y = -2 \) to \( y = 2 \):

\[ \text{Area} = \int_{-2}^{2} \left( 4 - y^2 \right) dy \]

Step 3: Compute the integral:

\[ \text{Area} = \left[ 4y - \frac{y^3}{3} \right]_{-2}^{2} \]

Step 4: Substituting the limits:

\[ \text{Area} = \left( 4(2) - \frac{(2)^3}{3} \right) - \left( 4(-2) - \frac{(-2)^3}{3} \right) = 16 - \frac{16}{3} \] \[ \text{Area} = \frac{48}{3} - \frac{16}{3} = \frac{32}{3} \]