Volume Integral of this set I'm not sure about this exercise. Be:
$$E=\left\{(x,y,z) \in \mathbb{R^3} : x\geq 0, y\geq 0, 0\leq z\leq \frac{1}{\sqrt{x^2+y^2}}-1\right\}$$
Find:
$$\int_{E} z\, \max\{x,y\}\: dx \, dy \, dz$$ 
My idea is that since $x,y\geq 0$, one of the two variables will be greater than the other in a section of the quadrant divided by the bisector and vice versa. For example, let's suppose that $x\geq y$, the  section of the volume would be something like:
$$\left\{(x,y,z)\in \mathbb{R^3}  : 0\leq x\leq \infty, 0\leq y\leq x,0\leq z\leq \frac{1}{\sqrt{x^2+y^2}}-1\right\}$$
The volume of this would be:
$$\int_0^\infty\int_0^x\int_0^{\frac{1}{\sqrt{x^2+y^2}}-1} zx\: dz\, dy\, dx$$ 
Then the volume of $E$ would be this volume plus the volume of the case where $y\geq x$, does this look correct to you?

Partial answer:
The upper limit of integration for the $x$ variable doesn't actually go to infinity. Note that the bounds for $z$ imply an additional constraint on the range of $x$ and $y$:
$$\begin{align}
0\le z\le\frac{1}{\sqrt{x^2+y^2}}-1
&\implies0\le\frac{1}{\sqrt{x^2+y^2}}-1\\
&\implies1\le\frac{1}{\sqrt{x^2+y^2}}\\
&\implies\sqrt{x^2+y^2}\le1\\
&\implies x^2+y^2\le1\\
&\implies0\le x\le1\land0\le y\le\sqrt{1-x^2}.\\
\end{align}$$
With the correct integration bounds for your triple integral, and strategically splitting up the integration intervals to eliminate the max functions, we find:
$$\begin{align}
I
&=\iiint_{E}\mathrm{d}^{3}v\,\left[z\max{\left(x,y\right)}\right]\\
&=\int_{0}^{1}\mathrm{d}x\int_{0}^{\sqrt{1-x^{2}}}\mathrm{d}y\int_{0}^{\frac{1}{\sqrt{x^2+y^2}}-1}\mathrm{d}z\,z\max{\left(x,y\right)}\\
&=\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\int_{0}^{\sqrt{1-x^{2}}}\mathrm{d}y\int_{0}^{\frac{1}{\sqrt{x^2+y^2}}-1}\mathrm{d}z\,z\max{\left(x,y\right)}\\
&~~~~~+\int_{\frac{1}{\sqrt{2}}}^{1}\mathrm{d}x\int_{0}^{\sqrt{1-x^{2}}}\mathrm{d}y\int_{0}^{\frac{1}{\sqrt{x^2+y^2}}-1}\mathrm{d}z\,z\max{\left(x,y\right)}\\
&=\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\int_{0}^{\frac{1}{\sqrt{x^2+y^2}}-1}\mathrm{d}z\,z\max{\left(x,y\right)}\\
&~~~~~+\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\int_{x}^{\sqrt{1-x^{2}}}\mathrm{d}y\int_{0}^{\frac{1}{\sqrt{x^2+y^2}}-1}\mathrm{d}z\,z\max{\left(x,y\right)}\\
&~~~~~+\int_{\frac{1}{\sqrt{2}}}^{1}\mathrm{d}x\int_{0}^{\sqrt{1-x^{2}}}\mathrm{d}y\int_{0}^{\frac{1}{\sqrt{x^2+y^2}}-1}\mathrm{d}z\,z\max{\left(x,y\right)}\\
&=\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\int_{0}^{\frac{1}{\sqrt{x^2+y^2}}-1}\mathrm{d}z\,xz\\
&~~~~~+\int_{0}^{\frac{1}{\sqrt{2}}}\mathrm{d}x\int_{x}^{\sqrt{1-x^{2}}}\mathrm{d}y\int_{0}^{\frac{1}{\sqrt{x^2+y^2}}-1}\mathrm{d}z\,yz\\
&~~~~~+\int_{\frac{1}{\sqrt{2}}}^{1}\mathrm{d}x\int_{0}^{\sqrt{1-x^{2}}}\mathrm{d}y\int_{0}^{\frac{1}{\sqrt{x^2+y^2}}-1}\mathrm{d}z\,xz.\\
\end{align}$$
Presumably you can take the helm from there?
