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The area (in square unit) of the portion enclosed by the curve $\sqrt{2x}+ \sqrt{2y} = 2 \sqrt{3}$ and the axes of reference is

  1. $2$
  2. $4$
  3. $6$
  4. $8$
in Geometry by Boss (17.5k points)
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1 Answer

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The above equation can be re-written as follows:

$\sqrt{x} + \sqrt{y} = \sqrt{6}$

This depicts the graphical notation of the above curve:

The curve can be re-written as follows:

$\sqrt{y} = \sqrt{6} - \sqrt{x}$

$y = 6 + x + \left ( 2\ast \sqrt{6}\ast \sqrt{x}\right )$

The value of x varies from x = 0  to x = 6

Area under the curve  = $\int_{0}^{6}\left [6 + x + \left ( 2\ast \sqrt{6}\ast \sqrt{x}\right )\right ]dx$

                                       =  $\left [ 6\ast x + {\frac{x^{2}}{2}} + 3\ast\sqrt{6}\ast x^{\frac{3}{2}}\right ]$

On computing this, you get Area = 6.

Thus, answer is option (C).

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