Use a surface integral to find the surface area of the portion of the sphere xUse a surface integral to find the surface area of the portion of the sphere x^2 +y^2 +z^2 = 1 inside the cone z^2 = x^2 + y^2 above the xy-plane.2 +y2 +z2 = 1 inside the cone z2 = x2 + y2 above the xy-plane.

Parameterize this surface (call it [tex]S[/tex]) by[tex]\vec r(u,v)=\cos u\sin v\,\vec\imath+\sin u\sin v\,\vec\jmath+\cos v\,\vec k[/tex]with [tex]0\le u\le2\pi[/tex] and [tex]0\le v\le\dfrac\pi4[/tex]. The limits on [tex]u[/tex] should be obvious. We find the upper limit for [tex]v[/tex] by solving for [tex]v[/tex] where the sphere and cone intersect:[tex]\begin{cases}x^2+y^2+z^2=1\\z=\sqrt{x^2+y^2}\end{cases}\implies x^2+y^2=\dfrac12[/tex][tex]\implies(\cos u\sin v)^2+(\sin u\sin v)^2=\dfrac12[/tex][tex]\implies\sin^2v=\dfrac12[/tex][tex]\implies\sin v=\dfrac1{\sqrt2}\implies v=\dfrac\pi4[/tex]Take the normal vector to [tex]S[/tex] to be[tex]\vec r_u\times\vec r_v=-\cos u\sin^2v\,\vec\imath-\sin u\sin^2v\,\vec\jmath-\cos v\sin v\,\vec k[/tex](orientation does not matter here)Then the area of [tex]S[/tex] is[tex]\displaystyle\iint_S\mathrm dA=\iint_S\|\vec r_u\times\vec r_v\|\,\mathrm du\,\mathrm dv[/tex][tex]=\displaystyle\int_0^{\pi/4}\int_0^{2\pi}\sin v\,\mathrm du\,\mathrm dv[/tex][tex]=\displaystyle2\pi\int_0^{\pi/4}\sin v\,\mathrm dv=\boxed{(2-\sqrt2)\pi}[/tex]