The area (in square units) of the region bounded by the curve $x = y^{2} and x = 3 - 2 y^{2}$ is

$\frac{3}{2}$

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$2$

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$3$

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$4$

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Solution

The correct option is D
$4$

$Solving the equations X = 1, y = \pm 1 ∴ Area = \int_{- 1}^{1} (3 - 2 y^{2} - y^{2}) d y = 2 [3 y - y^{3}]_{0}^{1} = 4$

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Similar questions

The area (in square units) of the region bounded by the curve $x = y^{2} and x = 3 - 2 y^{2}$ is

x = - 2 y^{2}

x = 1 - 3 y^{2}

y = | x - 1 |

y = 3 - | x |

The area (in square units) bounded by the curves $x = - 2 y^{2} a n d x = 1 - 3 y^{2}$ is

x^{2} = 2 y - x

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