<s>[INST] 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 $V.$ 
 [/INST] 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 $$\iint_\mathcal C 2(x^2 + y^2) \, dS = \int_0^{2 \pi} \int_0^{\sqrt{1 - \sqrt 3 /2}} 2r^2 ||F_r(r, \theta) \times F_\theta(r, \theta)|| \cdot r \, dr \, d \theta \text{ and}$$ $$\iint_\mathcal N 2(x^2 + y^2) \, dS = \int_0^{2 \pi} \int_0^{(\sqrt 3 - 1)/2} 2r^2 ||G_r(r, \theta) \times G_\theta(r, \theta)|| \cdot r \, dr \, d \theta. \phantom{\text{ and }}$$
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