Question

The area (in sq. units) of the region bounded by the curves \( y = x^2 \) and \( y = 8 - x^2 \) is

• A. \(\frac{32}{3}\)
• B. \(\frac{16}{3}\)
• C. \(\frac{64}{3}\)
• D. \(\frac{128}{3}\)

Hint:

To find the area between curves, determine intersection points and integrate the difference of the upper and lower functions. For even integrands, use symmetry to simplify.

Answer 1

The Correct Option is C

Find the intersection points: \[ x^2 = 8 - x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x = \pm 2 \] The curve \( y = 8 - x^2 \) lies above \( y = x^2 \) for \( x \in [-2, 2] \), since \( 8 - x^2 \geq x^2 \). The area is: \[ A = \int_{-2}^2 \left( (8 - x^2) - x^2 \right) \, dx = \int_{-2}^2 (8 - 2x^2) \, dx \] Since the integrand is even, compute: \[ A = 2 \int_0^2 (8 - 2x^2) \, dx = 2 \left[ 8x - \frac{2}{3}x^3 \right]_0^2 = 2 \left( 8 \cdot 2 - \frac{2}{3} \cdot 8 \right) = 2 \left( 16 - \frac{16}{3} \right) = 2 \cdot \frac{32}{3} = \frac{64}{3} \] Option (3) is correct. Options (1), (2), and (4) do not match.