[Solved] The Area Enclosed Between The Curves Y2 = 4x And X2 = 4y Is

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Explanation:

GATE ME 2009 Images-Q25

y2 = 4x, x2 = 4y

or \({y^2} = 4x = 4\sqrt {4y} = 8\sqrt y\;\)

⇒ y4 – 64y = 0

then y = 0, 4

Similarly put y = 0, 4

In curve x2 = 4y

we get x = 4, 0

So, point of intersection are (0,0) and (4,4)

\(\begin{array}{l} Area = \mathop \smallint \limits_0^4\ \mathop \smallint \limits_{2\sqrt x }^{\frac{{{x^2}}}{4}} dydx = \mathop \smallint \limits_0^4 \left[ y \right]_{2\sqrt x }^{\frac{{{x^2}}}{4}}dx\\ = \mathop \smallint \limits_0^4 \left[ {\frac{{{x^2}}}{4} - 2\sqrt x } \right]dx = \left[ {\frac{{{x^3}}}{{12}} - \frac{{4{x^{\frac{3}{2}}}}}{3}} \right]_0^4\\ = \left| {\frac{{{4^3}}}{{12}} - \frac{{4{{\left( 4 \right)}^{\frac{3}{2}}}}}{3}} \right| = \frac{{16}}{3} \end{array}\;\)

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