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The area in s...

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The area (in sq. units) of the region bounded by the curves $x^{2} + 2 y - 1 = 0, y^{2} + 4 x - 4 = 0 and y^{2} - 4 x - 4 = 0,$ in the upper half plane is

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Solution

Given: $c_{1} : y = \frac{- x^{2} + 1}{2}, c_{2} : y^{2} = - 4 (x - 1), c_{3} : y^{2} = 4 (x + 1)$

Required area $= 2 \int_{0}^{1} (2 \sqrt{1 - x} - \frac{1 - x^{2}}{2}) d x$
$= - \frac{4 (1 - x)^{\frac{3}{2}}}{\frac{3}{2}} - x + {\begin{matrix} \frac{x^{3}}{3} \end{matrix} ∣ ∣ ∣}_{0}^{1} = (\frac{8}{3}) - 1 + (\frac{1}{3}) = 2$

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