The area enclosed by y^2=8x and y=√2,x that lies outside the triangle formed by y=√2,x, x=…

Question

The area enclosed by $y^2=8x$ and $y=\sqrt2\,x$ that lies outside the triangle formed by $y=\sqrt2\,x$, $x=1$, $y=2\sqrt2$, is equal to:

Accepted Answer

$$A\left( {2,2\sqrt 2 } \right),B\left( {1,2\sqrt 2 } \right),C\left( {1,\sqrt 2 } \right)$$

Area $$ = \int\limits_0^{4\sqrt 2 } {\left( {{y \over {\sqrt 2 }} - {{{y^2}} \over 8}} \right)dy - } $$ area ($$\Delta$$BAC)

$$ = \left[ {{{{y^2}} \over {2\sqrt 2 }} - {{{y^3}} \over {24}}} \right]_0^{4\sqrt 2 } - {1 \over 2} \times AB \times BC$$

$$ = 8\sqrt 2  - {{32 \times 4\sqrt 2 } \over {24}} - {1 \over 2} \times 1 \times \sqrt 2 $$

$$ = 8\sqrt 2  - {{16\sqrt 2 } \over 3} - {{\sqrt 2 } \over 2}$$

$$ = {{\sqrt 2 } \over 6}(48 - 32 - 3) = {{13\sqrt 2 } \over 6}$$

Solutions · 1