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:

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

Answer 2

Area Bounded by $x = 4 - y^2$ and the Y-axis

We need to find the area bounded by the curve $x = 4 - y^2$ and the y-axis.

Step 1: Find the intersection points with the y-axis

To find where the curve intersects the y-axis, we set $x = 0$:

$0 = 4 - y^2$

$y^2 = 4$

$y = \pm 2$

The curve intersects the y-axis at the points $(0, 2)$ and $(0, -2)$.

Step 2: Set up the integral for the area

The area bounded by the curve and the y-axis can be found by integrating with respect to $y$ from $-2$ to $2$. The width of a horizontal strip is $x$, and the height is $dy$. The area $dA = x dy$.

The total area $A$ is given by:

$$ A = \int_{-2}^{2} x dy $$

Step 3: Substitute the expression for $x$

Substitute $x = 4 - y^2$ into the integral:

$$ A = \int_{-2}^{2} (4 - y^2) dy $$

Step 4: Evaluate the integral

Since the function $4 - y^2$ is even, we can write:

$$ A = 2 \int_{0}^{2} (4 - y^2) dy $$

Integrate with respect to $y$:

$$ A = 2 \left[ 4y - \frac{y^3}{3} \right]_{0}^{2} $$

Evaluate at the limits:

$$ A = 2 \left[ \left( 4(2) - \frac{(2)^3}{3} \right) - \left( 4(0) - \frac{(0)^3}{3} \right) \right] $$

$$ A = 2 \left[ \left( 8 - \frac{8}{3} \right) - (0) \right] $$

$$ A = 2 \left[ \frac{24 - 8}{3} \right] $$

$$ A = 2 \left[ \frac{16}{3} \right] $$

$$ A = \frac{32}{3} $$

Step 5: State the final answer

The area bounded by the curve $x = 4 - y^2$ and the y-axis is $\frac{32}{3}$ square units.

Final Answer: (B) $\frac{32}{3}$ sq. unit