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

Answer: (D)

$\frac{16}{3}$ 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:

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

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