If the area of the region bounded by the curves y 2 - 2 y = - x and x + y = 0 is A , then 8 A =

Given,

Equations of curve $y^2-2y=-x$ and $x+y=0$

Now solving above equations we get, $x=0,-3$

Now plotting the area we get, 

Now from above diagram required area will be

$A={\int }_0^3\left(\left(2y-y^2\right)+y\right)dy$

$\Rightarrow A={\int }_0^3\left(3y-y^2\right)dy$

$\Rightarrow A={\left[\frac{3y^2}{2}-\frac{y^3}{3}\right]}_0^3$

$\Rightarrow A=\left[\frac{27}{2}-\frac{27}{3}\right]$

$\Rightarrow A=\frac{27}{6}=\frac{9}{2}$ sq. units

So, $A=\frac{9}{2}\Rightarrow 8A=9\times 4=36$.