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=\frac13√(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(\frac13√(x^2+y^2)\right)^2=4\implies x^2+y^2=\frac{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{√(10)}\le v\le\pi`.

Take the normal vector to
`S` to be

`(\partial\vec s)/(\partial v)*(\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{√(10)}}^\pi\left\|(\partial\vec s)/(\partial v)*(\partial\vec s)/(\partial u)\right\|\,\mathrm dv\,\mathrm du`

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