Here, $$\{ (x,y) \in {R^2}:y \ge \sqrt {\left| {x + 3} \right|} ,\,5y \le (x + 9) \le 15\} $$

$$\therefore$$ $$y \ge \sqrt {x + 3} $$

$$ \Rightarrow y \ge \left\{ \matrix{ \sqrt {x + 3} ,\,when\,x \ge - 3 \hfill \cr \sqrt { - x - 3} ,\,when\,x \le - 3 \hfill \cr} \right.$$

or, $${y^2} \ge \left\{ \matrix{ x + 3,\,when\,x \ge - 3 \hfill \cr - 3 - x,\,when\,x \le - 3 \hfill \cr} \right.$$

Shown as

Also, $$5y \le (x + 9) \le 15$$

$$ \Rightarrow (x + 9) \ge 5y$$ and $$x \le 6$$

Shown as

$$\therefore$$ $$\{ (x,y) \in {R^2}:y \ge \sqrt {\left| {x + 3} \right|} ,\,5y \le (x + 9) \le 15\} $$

$$\therefore$$ Required area = Area of trapezium ABCD $$-$$ Area of ABE under parabola $$-$$ Area of CDE under parabola

$$ = {1 \over 2}(1 + 2)(5) - \int_{ - 4}^{ - 3} {\sqrt { - (x + 3)} dx - \int_{ - 3}^1 {\sqrt {(x + 3)} dx} } $$

$$ = {{15} \over 2} - \left[ {{{{{( - 3 - x)}^{3/2}}} \over { - {3 \over 2}}}} \right]_{ - 4}^{ - 3} - \left[ {{{{{(x + 3)}^{3/2}}} \over {{3 \over 2}}}} \right]_{ - 3}^1$$

$$ = {{15} \over 2} + {2 \over 3}[0 - 1] - {2 \over 3}[8 - 0]$$

$$ = {{15} \over 2} - {2 \over 3} - {{16} \over 3} = {{15} \over 2} - {{18} \over 3} = {3 \over 2}$$