Question

\(\text{The area of the region enclosed between the curves } 4x^2 = y \text{ and } y = 4 \text{ is:}\)

• A. \( 16 \) sq. units
• B. \( \frac{32}{3} \) sq. units
• C. \( \frac{8}{3} \) sq. units
• D. \( \frac{16}{3} \) sq. units

Hint:

When calculating the area between curves, always identify the points of intersection and set up the integral based on the difference between the functions. If symmetry exists (as in this case), you can often simplify the calculation by considering only one side and doubling the result. Make sure to properly evaluate the integral and check the units for consistency.

Answer 1

The Correct Option is D

To find the area enclosed between the curves \(4x^2 = y\) and \(y = 4\), we proceed as follows:

Rewrite \(4x^2 = y\) as \(x^2 = \frac{y}{4}\), giving:

\[x = \pm \sqrt{\frac{y}{4}}\]

The curves intersect at \(y = 4\). Therefore, we need to find the area bounded by these curves from \(y = 0\) to \(y = 4\).

The area is given by:

\[\text{Area} = 2 \int_{0}^{4} \sqrt{\frac{y}{4}} \, dy\]

Simplifying the integrand:

\[\text{Area} = 2 \int_{0}^{4} \frac{\sqrt{y}}{2} \, dy = \int_{0}^{4} \sqrt{y} \, dy\]

Evaluate the integral:

\[\int_{0}^{4} y^{1/2} \, dy = \left[ \frac{2}{3} y^{3/2} \right]_{0}^{4} = \frac{2}{3} \times (4)^{3/2} = \frac{2}{3} \times 8 = \frac{16}{3}\]

Thus, the area enclosed between the curves is \(\frac{16}{3}\) sq. units.

Answer 2

To find the area enclosed between the curves \( 4x^2 = y \) and \( y = 4 \), we proceed as follows:

Rewrite \( 4x^2 = y \) as \( x^2 = \frac{y}{4} \), giving:

\[ x = \pm \sqrt{\frac{y}{4}} \]

The curves intersect at \( y = 4 \). Therefore, we need to find the area bounded by these curves from \( y = 0 \) to \( y = 4 \).

The area is given by:

\[ \text{Area} = 2 \int_{0}^{4} \sqrt{\frac{y}{4}} \, dy \]

Simplifying the integrand:

\[ \text{Area} = 2 \int_{0}^{4} \frac{\sqrt{y}}{2} \, dy = \int_{0}^{4} \sqrt{y} \, dy \]

Evaluate the integral:

\[ \int_{0}^{4} y^{1/2} \, dy = \left[ \frac{2}{3} y^{3/2} \right]_{0}^{4} = \frac{2}{3} \times (4)^{3/2} = \frac{2}{3} \times 8 = \frac{16}{3} \]

Conclusion:

Thus, the area enclosed between the curves is \( \frac{16}{3} \) square units.