The area bounded by the curve $x=y^2+4y$ with $y-$ axis

The given curve is $x=y^2+4y$

This curve will pass through the origin
When $x=0$, $y^2+4y=0\Rightarrow y=0,y=-4$

The rough sketch of the given curve is as below

The required area is the area of the shaded region

The area of the shaded region is the area bounded by the curve $x=y^2+4y$ and lines $y=0,y=-4,y-axis$

Hence, the area of the shaded region is $A=\left|{\int }_{-4}^0\left(y^2+4y\right)dy\right|$

It implies 

$\begin{array}{rcl}A& =& \left|{\int }_{-4}^0\left(y^2+4y\right)dy\right|\\ & =& \left|{\left[\frac{y^3}{3}+2y^2\right]}_{-4}^0\right|\\ & =& \left|0+\frac{64}{3}-32\right|\\ & =& \frac{32}{3}\end{array}$

Therefore, the area of the required region is $\boxed{\frac{32}{3}}$