Discussion Forum

Area  of the region {(x,y)}ε $R^{2}:y\geq \sqrt{\mid x+3\mid},5y\leq (x+9)\leq15$ } is equal to

Answer:

Explanation:

Here, {(x,y)}ε $R^{2}:y\geq \sqrt{\mid x+3\mid},5y\leq (x+9)\leq15$ }

$\therefore  y\geq \sqrt{x+3}$

$\Rightarrow$    $y\geq\begin{cases}\sqrt{x+3},& when & x \geq -3\\\sqrt{-x-3}& when & x \leq-3\end{cases}$

or           $y^{2}\geq\begin{cases}x+3 & when& x \geq -3\\-3-x & when & x \leq -3\end{cases}$

shown as

Also,  $5y\leq (x+9)\leq 15$

$\Rightarrow$ $(x+9)\geq 5y$ and $x\leq 6$

shown as 

$\therefore$      { $(x,y)\epsilon R^{2}:y\geq\sqrt{\mid x+3 \mid}, 5y\leq(x+9)\leq15$}

$\therefore$  Required area

                               = Area of trapezium ABCD - Area of ABE under parabola- Area of CDE under parabola

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

                  $=\frac{15}{2}-[\frac{(-3-x)^{\frac{3}{2}}}{-\frac{3}{2}}]_{-4}^{-3}-[\frac{(x+3)^{\frac{3}{2}}}{\frac{3}{2}}]^{1}_{-3}$

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

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