Question

Let $y=f\left(x\right)=\{\begin{array}{cc}\sqrt{x+3},& -3\le x<-2\\ 1+\sqrt{x+2},& -2\le x<-1\\ 2+\sqrt{x+1},& -1\le x\le 0\end{array}$.
If the area enclosed between $|y|=f(-|x|)$ and $x^2+y^2=5$ is $p+\pi q$ sq. units, then the value of $(p+q)$ is

Answer 1

Area bounded by the $y=f\left(x\right)$ and $x-$axis from $x=-3$ to $x=0$
$=\underset{-3}{\overset{-2}{\int }}\sqrt{x+3}dx+\underset{-2}{\overset{-1}{\int }}(1+\sqrt{x+2})dx+\underset{-1}{\overset{0}{\int }}(2+\sqrt{x+1})dx={\left[\frac{2}{3}\right(x+3{)}^{3/2}]}_{-3}^{-2}+{[x+\frac{2}{3}(x+2{)}^{3/2}]}_{-2}^{-1}+{[2x+\frac{2}{3}(x+1{)}^{3/2}]}_{-1}^0=\frac{2}{3}[1-0]+[-1+2+\frac{2}{3}(1-0\left)\right]+[0+2+\frac{2}{3}(1-0\left)\right]=\frac{2}{3}+\frac{5}{3}+\frac{8}{3}=5$

Required area
$=4(5-\frac{1}{4}\times \pi r^2)=4(5-\frac{\pi }{4}\times 5)=20-5\pi \Rightarrow p=20,q=-5\therefore p+q=15$