Question

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.

Answer 1

Parameterize this surface (call it
`S`) by

`\vec r(u,v)=\cos u\sin v\,\vec\imath+\sin u\sin v\,\vec\jmath+\cos v\,\vec k`

with
`0\le u\le2\pi` and
`0\le v\le\frac\pi4`. The limits on
`u` should be obvious. We find the upper limit for
`v` by solving for
`v` where the sphere and cone intersect:

`\begin{cases}x^2+y^2+z^2=1\\z=√(x^2+y^2)\end{cases}\implies x^2+y^2=\frac12`

`\implies(\cos u\sin v)^2+(\sin u\sin v)^2=\frac12`

`\implies\sin^2v=\frac12`

`\implies\sin v=\frac1{\sqrt2}\implies v=\frac\pi4`

Take the normal vector to
`S` to be

`\vec r_u*\vec r_v=-\cos u\sin^2v\,\vec\imath-\sin u\sin^2v\,\vec\jmath-\cos v\sin v\,\vec k`

(orientation does not matter here)

Then the area of
`S` is

`\displaystyle\iint_S\mathrm dA=\iint_S\|\vec r_u*\vec r_v\|\,\mathrm du\,\mathrm dv`

`=\displaystyle\int_0^(\pi/4)\int_0^(2\pi)\sin v\,\mathrm du\,\mathrm dv`

`=\displaystyle2\pi\int_0^(\pi/4)\sin v\,\mathrm dv=\boxed{(2-\sqrt2)\pi}`