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

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.