Surface integral of piecewise volume boundary? How should I go about solving the surface integral
$$\iint_{\partial V} 2(x^2+y^2) \, dS,$$ where $V$ is the region bounded by the paraboloid $$z=\frac12-x^2-y^2$$ and the cone $$z^2=x^2+y^2?$$
I have done surface integrals before, but I am unsure how to proceed with finding the boundary of the solid $V.$ 

The outward surface normals for the two surfaces are given by the gradients of the expressions:
$$z+x^2+y^2-\frac{1}{2}=0 \implies \vec{n} = \frac{1}{\sqrt{3-4z}}\langle 2x, 2y, 1\rangle$$
$$x^2+y^2-z^2 = 0 \implies \vec{n} = \frac{1}{2\sqrt{2}|z|}\langle 2x, 2y, -2z \rangle$$
and consider the two vector fields
$$F_1 = \sqrt{3-4z} \langle x, y , 0 \rangle $$
$$F_2 = 2\sqrt{2} |z| \langle x, y, 0 \rangle$$
The integral of the first one on the upper surface and the second one of the second surface gives us the integrand in the problem. If we consider the disk created by the two surfaces' plane of intersection, the integral on that disk for both of those vector fields is $0$. Thus we can use the divergence theorem to calculate the integral:
$$\iint_{\partial V} 2(x^2+y^2)\:dS = \iiint_{V_1} 2\sqrt{3-4z}\:dV + \iiint_{V_2} 4\sqrt{2} z \:dV$$
The easiest way to do these integrals is in cylindrical coordinates
$$\int_0^{2\pi} \int_0^{\frac{-1+\sqrt{3}}{2}} \int_{\frac{-1+\sqrt{3}}{2}}^{\frac{1}{2}-r^2} 2r\sqrt{3-4z}\:dz\:dr\:d\theta = \frac{2\pi}{3} \int_0^{\frac{-1+\sqrt{3}}{2}} r\left(5-2\sqrt{3}\right)^{\frac{3}{2}} - r\left(4r^2+1\right)^{\frac{3}{2}} \:dr = \frac{\pi\left(5-2\sqrt{3}\right)^2\left(\sqrt{3}-1\right)^2}{12} - \frac{\pi\left(5-2\sqrt{3}\right)^{\frac{5}{2}}}{30}$$
$$\int_0^{2\pi} \int_0^{\frac{-1+\sqrt{3}}{2}} \int_0^z 2\sqrt{2} zr\:dr\:dz\:d\theta = 2\sqrt{2}\pi \int_0^{\frac{-1+\sqrt{3}}{2}} z^3\:dz = \frac{\pi\left(\sqrt{3}-1\right)^4}{16\sqrt{2}}$$
and the final answer is the sum of these two.
