Question

The area of the region bounded by the curve \[ y = x^2 \quad \text{and the line} \quad y = 16 \quad \text{is:} \]

• A. \( \frac{256}{3} \) sq. units
• B. 64 sq. units
• C. \( \frac{128}{3} \) sq. units
• D. \( \frac{32}{3} \) sq. units

Hint:

For symmetric curves, compute the area for half the region and then double it to find the full area.

Answer 1

The Correct Option is A

To calculate the area, we need to find the points where the curve intersects the line \( y = 16 \). From the equation \( y = x^2 \), we set \( x^2 = 16 \), so \( x = \pm 4 \). Thus, the area is given by the integral from \( -4 \) to \( 4 \) of \( y = x^2 \): \[ A = \int_{-4}^{4} x^2 \, dx \] Since the curve is symmetric, we can compute the area from 0 to 4 and double the result: \[ A = 2 \int_0^4 x^2 \, dx \] Now, evaluate the integral: \[ A = 2 \left[ \frac{x^3}{3} \right]_0^4 = 2 \times \left( \frac{4^3}{3} - 0 \right) = 2 \times \frac{64}{3} = \frac{128}{3} \] Thus, the area of the region is \( \frac{256}{3} \) sq. units.