The area of the region bounded by y2 = 8x and y2 = 16(3 - x) is equal to:

Question

Accepted Answer

${c_1}:{y^2} = 8x$

${c_2}:{y^2} = 16(3 - x)$

Solving c1 and c2

$48 - 16x = 8x$

$x = 2$

$$\therefore$$ $$y =  \pm \,4$$

$$\therefore$$ Area of shaded region

$$ = 2\int\limits_0^4 {\left\{ {\left( {{{48 - {y^2}} \over {16}}} \right) - \left( {{{{y^2}} \over 8}} \right)} \right\}dy} $$

$$ = {1 \over 8}\left[ {48y - {y^3}} \right]_0^4 = 16$$

Solutions · 1