JEE MAIN - Mathematics (2022 - 29th June Morning Shift - No. 5)
The area enclosed by y2 = 8x and y = $$\sqrt2$$ x that lies outside the triangle formed by y = $$\sqrt2$$ x, x = 1, y = 2$$\sqrt2$$, is equal to:
$${{16\sqrt 2 } \over 6}$$
$${{11\sqrt 2 } \over 6}$$
$${{13\sqrt 2 } \over 6}$$
$${{5\sqrt 2 } \over 6}$$
Explanation
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$$A\left( {2,2\sqrt 2 } \right),B\left( {1,2\sqrt 2 } \right),C\left( {1,\sqrt 2 } \right)$$
Area $$ = \int\limits_0^{4\sqrt 2 } {\left( {{y \over {\sqrt 2 }} - {{{y^2}} \over 8}} \right)dy - } $$ area ($$\Delta$$BAC)
$$ = \left[ {{{{y^2}} \over {2\sqrt 2 }} - {{{y^3}} \over {24}}} \right]_0^{4\sqrt 2 } - {1 \over 2} \times AB \times BC$$
$$ = 8\sqrt 2 - {{32 \times 4\sqrt 2 } \over {24}} - {1 \over 2} \times 1 \times \sqrt 2 $$
$$ = 8\sqrt 2 - {{16\sqrt 2 } \over 3} - {{\sqrt 2 } \over 2}$$
$$ = {{\sqrt 2 } \over 6}(48 - 32 - 3) = {{13\sqrt 2 } \over 6}$$
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