If the area of the region bounded by the curves y ^ 2 - 2y = - x...

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

$A = {∫}_{0}^{3}\left(\left(2y - {y}^{2}\right) + y\right)dy$

$⇒ A = {∫}_{0}^{3}\left(3y - {y}^{2}\right)dy$

$⇒ A = {\left[\dfrac{3{y}^{2}}{2} - \dfrac{{y}^{3}}{3}\right]}_{0}^{3}$

$⇒ A = \left[\dfrac{27}{2} - \dfrac{27}{3}\right]$

$⇒ A = \dfrac{27}{6} = \dfrac{9}{2}$ sq. units

So, $A = \dfrac{9}{2} ⇒ 8A = 9 \times 4 = 36$.