Question

Use a parameterization of the lower portion cut from the sphere x squared plus y squared plus z squared equals 4x2+y2+z2=4 by the conez equals StartFraction 1 Over StartRoot 3 EndRoot EndFraction StartRoot x squared plus y squared EndRootz=13x2+y2 to express the area of the surface as a double integral. Then evaluate the integral.

Answer 1

Substituting $z$ from the cone's equation,

$z=\dfrac13\sqrt{x^2+y^2}$

into the equation of the sphere,

$x^2+y^2+z^2=4$

gives the intersection of the two surfaces,

$x^2+y^2+\left(\dfrac13\sqrt{x^2+y^2}\right)^2=4\implies x^2+y^2=\dfrac{18}5$

which is a circle of radius $\sqrt{\frac{18}5}$ centered at $\left(0,0,\frac13\sqrt{\frac{18}5}\right)$.

We parameterize this part of the sphere outside the cone (call it $S$) by

$\vec s(u,v)=\langle2\cos u\sin v,2\sin u\sin v,2\cos v\rangle$

with $0\le u\le2\pi$ and $\cos^{-1}\frac1{\sqrt{10}}\le v\le\pi$.

Take the normal vector to $S$ to be

$\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=\langle4\cos u\sin^2v,4\sin u\sin^2v,4\cos v\sin v\rangle$

Then the area of $S$ is

$\displaystyle\iint_S\mathrm dA=\int_0^{2\pi}\int_{\cos^{-1}\frac1{\sqrt{10}}}^\pi\left\|\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}\right\|\,\mathrm dv\,\mathrm du$

$=\displaystyle2\pi\int_{\cos^{-1}\frac1{\sqrt{10}}}^\pi4\sin v\,\mathrm dv=\boxed{\frac{40+4\sqrt{10}}5\pi}$