The area of the smaller reion enclosed by the curves y² = 8x + 4 and x² + y² + $4 \sqrt[]{3} x - 4 = 0$ is equal to

Option 1 -

$\frac{1}{3} (2 - 12 \sqrt[]{3} + 8 π)$

Option 2 -

$\frac{1}{3} (2 - 12 \sqrt[]{3} + 6 π)$

Option 3 -

$\frac{1}{3} (4 - 12 \sqrt[]{3} + 8 π)$

Option 4 -

$\frac{1}{3} (4 - 12 \sqrt[]{3} + 6 π)$

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 3

Detailed Solution:

${(x + 2 \sqrt[]{3})}^{2} + y^{2} = 4^{2}$

$y^{2} = 8 (x + \frac{1}{2})$

Point of intersection are (0, 2) and (0, -2)

$2 \int_{0}^{2} [(\sqrt[]{16 - y^{2}} - 2 \sqrt[]{3}) - \frac{(y^{2} - 4)}{8}] d y$

$= \frac{1}{3} (8 π + 4 - 12 \sqrt[]{3})$

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The area of the smaller reion enclosed by the curves y² = 8x + 4 and x² + y² + $4 \sqrt[]{3} x - 4 = 0$ is equal to

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The area of the smaller reion enclosed by the curves y2 = 8x + 4 and x2 + y2 + 4 3 x − 4 = 0 is equal to

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The area of the smaller reion enclosed by the curves y² = 8x + 4 and x² + y² + $4 \sqrt[]{3} x - 4 = 0$ is equal to