Surface integral Problem statement
$$
\mbox{Calculate the surface integral}\quad
\int_{Y}\ y\,\sqrt{z\,}\,\sqrt{4x^{2} + 4y^{2} + 1\,}\,\,{\rm d}S
$$
where $Y$ is the surface
$\left\{\left(x,y,z\right)\ \ni\ z=x^2+y^2\,,\ y \geq 0\,,\  y \geq -x\,,\
1 \leq x^{2} + y^{2} \leq 4\right\}$.
Progress
I am trying to visualize the surface step by step. So $z=x^2+y^2$ is an infinite paraboloid, $y\geq 0$ gives us only half of this paraboloid(one side of the y-axis), $y\geq -x$ does not restrict anything(?) and $1\leq x^2+y^2\leq 4$ is either: The area between the two circles of radius 1 and 2 OR(since $z=x^2+y^2$) equivalent to $1\leq z\leq 4$(How large are the circles then...?)
Some guidance would be helpful.. 
 A: A hint:
Your surface can be parametrized by introducing polar coordinates in the $(x,y)$-plane as follows:
$$(r,\phi)\ \mapsto\  (r\cos\phi,\  r\sin\phi,\  r^2)\ .$$
Now you have to do three things: First find the exact parameter domain in the $(r,\phi)$-plane,  then express the integrand and $dS$ in terms of $r$ and $\phi$, and finally integrate.
A: $\newcommand{\+}{^{\dagger}}%
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$\ds{\int_{Y}\ y\root{z}\root{4x^{2} + 4y^{2} + 1}\,\dd{\rm S}}$

$\ds{Y \equiv\braces{\pars{x,y,z}\ \ni\ z=x^2+y^2\,,\ y \geq 0\,,\  y \geq -x\,,\
1 \leq x^{2} + y^{2} \leq 4}}$

The surface $\ds{Y}$ is parametrized by $\ds{x}$ and $\ds{y}$ as
  $\ds{\vec{r} = x\,\hat{x} + y\,\hat{y} + \pars{x^{2} + y^{2}}\,\hat{z}}$ such that
  \begin{align}
\dd{\rm S} &=\verts{\partiald{\vec{r}}{x}\times\partiald{\vec{r}}{y}}\,\dd x\,\dd y
=\verts{\pars{\hat{x} + 2x\,\hat{z}}\times\pars{\hat{y} + 2y\,\hat{z}}}\,\dd x\,\dd y
=\verts{\hat{z} - 2y\,\hat{y} + 2x\,\hat{x}}\,\dd x\,\dd y
\\[3mm]&=\root{4x^{2} + 4y^{2} + 1}\,\dd x\,\dd y
\end{align}

\begin{align}
&\color{#00f}{\large\int_{Y}\ y\root{z}\root{4x^{2} + 4y^{2} + 1}\,\dd{\rm S}}
=\int_{Y}\ y\root{x^{2} + y^{2}}\pars{4x^{2} + 4y^{2} + 1}\,\dd x\,\dd y
\\[3mm]&=\left.\int_{0}^{2\pi}\dd\theta\int_{0}^{\infty}\dd\rho\,\rho
\bracks{\rho\sin\pars{\theta}\rho\pars{4\rho^{2} + 1}}
\vphantom{\Huge A^A}\right\vert
_{\sin\pars{\theta}\ \geq\ 0 \atop {\vphantom{\Huge A}\sin\pars{\theta}\ \geq\ -\cos\pars{\theta} \atop {\vphantom{\Huge A}1\ \leq\ \rho\ \leq 2}}}
\\[3mm]&=\int_{0}^{3\pi/4}\dd\theta\int_{1}^{2}\dd\rho\,\rho^{3}
\sin\pars{\theta}\pars{4\rho^{2} + 1}
=\pars{1 + {\root{2} \over 2}}\int_{1}^{2}\pars{4\rho^{5} + \rho^{3}}\,\dd\rho
\\[3mm]&=\pars{1 + {\root{2} \over 2}}\,{183 \over 4}
=\color{#00f}{\large\pars{2 + \root{2}}{183 \over 8}} \approx 78.1001
\end{align}
A: Here's your surface:

I generated it with the following Mathematica code.  If you can make sense of the code, particularly the ParametricPlot3D command, then you can probably see how the $y=-x$  affects the surface and what your bounds of integration will be after you set up the surface integral as an iterated integral.
bounds = ContourPlot3D[{y == 0, y == -x, z == 1, z == 4},
 {x, -2, 2}, {y, -2, 2}, {z, 0, 5},
 BoxRatios -> Automatic, ContourStyle -> Opacity[0.3],
 Mesh -> None];
surface = ParametricPlot3D[{r*Cos[t], r*Sin[t], r^2}, {t, 0, 3 Pi/4}, 
  {r, 1, 2}, BoundaryStyle -> Thick];
Show[{bounds, surface}]

After a bit of algebra, you should find yourself getting to
$$\int _1^4\int _0^{3\pi/4}r^3 \left(1+4
   r^2\right) \sin (t)\,dt\,dr = \frac{11175}{8} \left(2+\sqrt{2}\right).$$
