write $\iiint_E \hspace{1mm}dV$ in 6 forms. where $E = \left\{ (x, y, z)|0\leq z\leq x+y, x^2\leq y\leq \sqrt{x},0\leq x\leq 1\right\}$ write $\iiint_E \hspace{1mm}dV$ in 6 forms. where $E = \left\{ (x, y, z)\hspace{1mm}|0\leq z\leq x+y, x^2\leq  y\leq \sqrt{x},0\leq  x\leq 1\right\}$
As you can see two forms are easy.
$$\iiint_E \hspace{1mm}dV = \int_0^{1}\int_{x^2}^{\sqrt{x}}\int_0^{x+y} \hspace{1mm}dz\hspace{1mm}dy\hspace{1mm}dx =  \int_0^{1}\int_{y^2}^{\sqrt{y}}\int_0^{x+y} \hspace{1mm}dz\hspace{1mm}dx\hspace{1mm}dy$$
After that I am having difficulty visualizing the 3D graph 
 A: The $6$ different orders of integration can be worked out straight-forwardly:
x-y-z
This is what you did first. The bounds are
$$\int_0^1 \int_{x^2}^{\sqrt x} \int_0^{x+y}$$
x-z-y
Again $x$ is independent, so $x\in[0,1]$. Now we bind $z\in [0, x+y]$. For this, $y$ is at most $\sqrt x$, so $z\in [0,x+\sqrt x]$.
This results in $y \ge z-x$ in addition to $x^2\le y\le \sqrt x$. Hence $y \in [\max(z-x, x^2), \sqrt x]$ and we obtain
$$\int_0^1 \int_0^{x+\sqrt x} \int_{\max(z-x, x^2)}^{\sqrt x}$$
y-x-z $y\in[0,1]$ is clear, then $y^2 \le x \le \sqrt y$ and finally $z$ stays in $[0,x+y]$
$$\int_0^1 \int_{y^2}^{\sqrt y} \int_0^{x+y}$$
y-z-x $y\in [0,1]$. Now $z\in [0,x+y]$ i.e. $z\in[0,y + \sqrt y]$ as above. Then $x\in [\max(z-y, y^2), \sqrt y]$ as above
$$\int_0^1 \int_0^{y+\sqrt y} \int_{\max(z-y, y^2)}^{\sqrt y}$$
z-x-y $z\in [0, 2]$. Then $x\in [0,1]$ but such that $y=z-x\le \sqrt x$, i.e. $x+\sqrt x \ge z$ that is $x\ge z + \frac12 - \sqrt{z+\frac14} \ge 0$. On the other side $x^2 + x \le z$ i.e. $x\le \sqrt{z+\frac14}-\frac12 \le 1$. We thus get the most ugly form:
$$\int_0^2 \int_{z+1/2-\sqrt{z+1/4}}^{\sqrt{z+1/4}-1/2}\int_{\max(z-x, x^2)}^{\sqrt x}$$
z-y-x This is analogous to above (we can interchange $x$ and $y$):
$$\int_0^2 \int_{z+1/2-\sqrt{z+1/4}}^{\sqrt{z+1/4}-1/2}\int_{\max(z-y,y^2)}^{\sqrt y}$$
