Answer
$ \frac{32}{3}$
Work Step by Step
The area between two curves $y=f(x)$ and $y=g(x)$ between $x=a$ and $x=b$ is given by the formula:
$$A=\int_a^b |f(x)-g(x)|dx.$$
Given
$$x+y=0, \quad x=y^2+3 y$$
find the intersection points
\begin{aligned}
y^2+3 y&= -y\\
y^2+4y&=0\\
y(y+4)&=0
\end{aligned}
Then $y=0,\ \ y=-4$,
Since $ -y\geq y^2+3y $ for $-4\leq y\leq 0$, then
\begin{aligned}
\text{Area} & =\int_{-4}^0 (-y-y^2-3 y)dy\\
&= \left(\frac{-1}{3}y^3 -2y^2\right)\bigg|_{-4}^0\\
&= \frac{32}{3}
\end{aligned}
