The area of the region - R - x - y - 5 x 2- y - 2 x 2-9 isA- 9 -3 square unitsB- 11 -3 square unitsC- 6 -3 square unitsD- 12 -3 square units

Required area =2 ∫ ^√(3)_0(2x^2+9 5x^2)dx =2 ∫ ^√(3)_0(9 3x^2)dx =2[9x x^3]^√(3)_0=12√(3) sq. units ...

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Area between Two Curves

The area of t...

Question

The area of the region $: R = {(x, y) : 5 x^{2} \leq y \leq 2 x^{2} + 9}$ is

9 \sqrt{3}

square units

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12 \sqrt{3}

square units

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11 \sqrt{3}

square units

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6 \sqrt{3}

square units

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Solution

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Similar questions

5 x^{2} - y = 0

2 x^{2} - y + 9 = 0

The area inside the parabola $5 x^{2} - y = 0$ but outside the parabola $2 x^{2} - y + 9 = 0$ , is

y = cos x

y = sin x

0 \leq x \leq \frac{π}{2}

x = - 2 y^{2}

x = 1 - 3 y^{2}

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