Compute $\int_Yy\sqrt{z}\sqrt{1+4x^2+4y^2} \ d\mathbf{S}$. 
Compute the surfaceintegral $$\int_Yy\sqrt{z}\sqrt{1+4x^2+4y^2} \
 d\mathbf{S},$$ where the surface $Y$ is given by $z=x^2+y^2, \ y \leq x, \
 y\leq 0$ and $1\leq x^2+y^2\leq9.$

Looking in my book on the page on surface integrals, they only cover flux integrals using double integrals. And suddenly this question pops up. I hate it when the questions you face are not representet by the book, no example on there is of this kind.
What is this? What am I computing here and how what is an effective system to use in order to solve this kinds of integrals?
EDIT: The area on the xy plane is as follows:

 A: I think the
$$(x,\,y)=(r\cos(t),\,r\sin(t))$$
parametrization will be good (I can't imagine it, so we will see if it works).
I'm not so good in the integral names, but I think the $dS$ in your case is not the vector $\vec{n}\mathrm{d}S=\mathrm{d}\vec{S}$:
$$dS=\left|\frac{\partial \vec{r}}{\partial r} \times \frac{\partial \vec{r}}{\partial t}\right| \mathrm{d}r \mathrm{d}t$$
So the position vector is:
$$\vec{r}=(x,\,y,\,z)=(r\cos(t),\,r\sin(t),\,r^2)$$
The two partial derivate is:
$$\frac{\partial \vec{r}}{\partial r}=(\cos(t),\,\sin(t) ,\,2r)$$
$$\frac{\partial \vec{r}}{\partial t}=(-r\sin(t),\,r\cos(t),\,0)$$
Now we need to calculate their cross's absolute value:
$$\left|\frac{\partial \vec{r}}{\partial r} \times \frac{\partial \vec{r}}{\partial t}\right|=\left|(-2r^2\cos(t),\,2r^2\sin(t),\,r)\right|=\sqrt{4r^4\cos^2(t)+4r^4\sin^2(t)+r^2}$$
$$=\sqrt{4r^4+r^2}=\sqrt{r^2}\sqrt{1+4r^2}$$
The integrand can be rewritten as:
$$y\sqrt{z}\sqrt{1+4(x^2+y^2)}=r\sin(t)\sqrt{r^2}\sqrt{1+4r^2}$$
Multiplying the $2$ together the integral will become:
$$\iint r\sin(t)r^2(1+4r^2)\mathrm{d}r\mathrm{d}t$$
I think the variables must be in the following intervals: $r \in [1,3]$ and $t \in [-\frac{3}{4}\pi,0]$
So:
$$\int_{t=-\frac{3}{4}\pi}^{0} \int_{r=1}^{3}  r\sin(t)r^2(1+4r^2) \mathrm{d}r\mathrm{d}t$$
$$=\left(\int_{t=-\frac{3}{4}\pi}^{0} \sin(t) \mathrm{d}t\right)\left(\int_{r=1}^{3} r^3(1+4r^2) \mathrm{d}r\right)=-\frac{758}{3}\left(2+\sqrt{2}\right)$$
