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 $V.$ 
 A: I assume the surface in question lies above the plane $z = 0;$ however, if not, you can easily adapt this solution to the case that the surface contains both the upper and lower nappe of the cone.
Graphing the two surfaces shows that the surface for which we would like to compute the surface area resembles an ice cream cone: it consists of the cap $\mathcal C$ of the paraboloid $z = \frac 1 2 - x^2 - y^2$ and part of the upper nappe $\mathcal N$ of the cone $z^2 = x^2 + y^2.$ Ultimately, we will seek to compute $$\iint_{\partial V} 2(x^2 + y^2) \, dS = \iint_\mathcal C 2(x^2 + y^2) \, dS + \iint_\mathcal N 2(x^2 + y^2) \, dS.$$
Observe that $\mathcal C$ and $\mathcal N$ intersect if and only if $z^2 = x^2 + y^2$ and $z = \frac 1 2 - x^2 - y^2$ if and only if $z = \frac 1 2 - z^2$ if and only if $z^2 + z - \frac 1 2 = 0$ if and only if $z = \frac{\sqrt 3 - 1}{2}.$ Consequently, we have that $$\mathcal C = \biggl \{(x, y, z) \,|\, z = \frac 1 2 - x^2 - y^2 \text{ and } \frac{\sqrt 3 - 1}{2} \leq z \leq \frac 1 2 \biggr \} \text{ and}$$ $$\mathcal N = \biggl \{(x, y, z) \,|\, z^2 = x^2 + y^2 \text{ and } 0 \leq z \leq \frac{\sqrt 3 - 1}{2} \biggr \}. \phantom{\text{ and butt }}$$ Geometrically, the cap is a "deformation" of disk in the $xy$-plane, so we may parametrize $\mathcal C$ by polar coordinates $F(r, \theta) = \bigl \langle r \cos \theta, r \sin \theta, \frac 1 2 - r^2 \bigr \rangle$ for $0 \leq r \leq \sqrt{1 - \frac{\sqrt 3}{2}}$ and $0 \leq \theta \leq 2 \pi.$
Likewise, the upper nappe $\mathcal N$ of the cone can be parametrized most easily by polar coordinates $G(r, \theta) = \langle r \cos \theta, r \sin \theta, r \rangle$ for $0 \leq r \leq \frac{\sqrt 3 - 1}{2}$ and $0 \leq \theta \leq 2 \pi.$
Can you finish the solution from here? Use the definition of the surface integrals $$\iint_\mathcal C 2(x^2 + y^2) \, dS = \int_0^{2 \pi} \int_0^{\sqrt{1 - \sqrt 3 /2}} 2r^2 ||F_r(r, \theta) \times F_\theta(r, \theta)|| \cdot r \, dr \, d \theta \text{ and}$$ $$\iint_\mathcal N 2(x^2 + y^2) \, dS = \int_0^{2 \pi} \int_0^{(\sqrt 3 - 1)/2} 2r^2 ||G_r(r, \theta) \times G_\theta(r, \theta)|| \cdot r \, dr \, d \theta. \phantom{\text{ and }}$$
A: 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.
