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

$\frac{2}{3}$

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$\frac{4}{3}$

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$\frac{5}{3}$

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Solution

The correct option is C
$\frac{4}{3}$

Area =2 $\int_{0}^{1} [(1 - 3 y^{2}) - (- 2 y^{2})] d y = \frac{4}{3}$

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Similar questions

x = - 2 y^{2}

x = 1 - 3 y^{2}

x = - 2 y^{2}, x = 1 - 3 y^{2}

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

x = y^{2}

x = 3 - 2 y^{2}

x + 2 y^{2} = 0

x + 3 y^{2} = 1

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