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:
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: $$\int_{t=-\frac{3}{4}\pi}^{0} \int_{r=1}^{3} r\sin(t)r^2(1+4r^2) \mathrm{d}r\mathrm{d}t$$ $$=\left(\int_{t=-\frac{3}{4}\pi}^{0} \sin(t) \mathrm{d}t\right)\left(\int_{r=1}^{3} r^3(1+4r^2) \mathrm{d}r\right)=-\frac{758}{3}\left(2+\sqrt{2}\right)$$