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 $0\le v\le\dfrac\pi4$. The limits on $u$ should be obvious. We find the upper limit for $v$ by solving for $v$ where the sphere and cone intersect:

$\begin{cases}x^2+y^2+z^2=1\\z=\sqrt{x^2+y^2}\end{cases}\implies x^2+y^2=\dfrac12$

$\implies(\cos u\sin v)^2+(\sin u\sin v)^2=\dfrac12$

$\implies\sin^2v=\dfrac12$

$\implies\sin v=\dfrac1{\sqrt2}\implies v=\dfrac\pi4$

Take the normal vector to $S$ to be

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

(orientation does not matter here)

Then the area of $S$ is

$\displaystyle\iint_S\mathrm dA=\iint_S\|\vec r_u\times\vec r_v\|\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{\pi/4}\int_0^{2\pi}\sin v\,\mathrm du\,\mathrm dv$

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