The area of the region bounded by the curve x = y^2 and the line x = 4 is equal to : Option 1: 16/3 square units Option 2: 32/3 square units Option 3: 20/3 square units Option 4: 4/3 square units

Answer: 32/3 square units. Solution: Curve x = y^2 meets x = 4 at y = ± 2. By symmetry, area = 2 _0^2 (4 - y^2) dy = 2 [ 4y - y^3/3 ]_0^2 = 2 ( 8 - 8/3 ) = 2 * 16/3 = 32/3 sq units.

Calculus

Application of Integrals

Easy

The area of the region bounded by the curve $x = y^2$ and the line $x = 4$ is equal to :

$\frac{16}{3}$ square units

$\frac{32}{3}$ square units

$\frac{20}{3}$ square units

$\frac{4}{3}$ square units

✅ Correct Option: 2

Curve $x = y^2$ meets $x = 4$ at $y = \pm 2$ .

By symmetry, area $= 2 \int_{0}^{2} (4 - y^2) \, dy$

$= 2 \left[ 4y - \frac{y^3}{3} \right]_{0}^{2}$

$= 2 \left( 8 - \frac{8}{3} \right) = 2 \cdot \frac{16}{3} = \frac{32}{3}$ sq units.

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Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

The area of the region bounded by the curve $x = y^2$ and the line $x = 4$ is equal to :