Question

G find the area of the surface over the given region. use a computer algebra system to verify your results. the sphere r(u,v) = a sin u cos v i + a sin u sin v j + a cos u k, where 0 ≤ u ≤ π and 0 ≤ v ≤ 2π

Answer 1

Presumably you should be doing this using calculus methods, namely computing the surface integral along
`\mathbf r(u,v)`.

But since
`\mathbf r(u,v)` describes a sphere, we can simply recall that the surface area of a sphere of radius
`a` is
`4\pi a^2`.

In calculus terms, we would first find an expression for the surface element, which is given by


`\displaystyle\iint_S\mathrm dS=\iint_S\left\|(\partial\mathbf r)/(\partial u)*(\partial\mathbf r)/(\partial v)\right\|\,\mathrm du\,\mathrm dv`


`(\partial\mathbf r)/(\partial u)=a\cos u\cos v\,\mathbf i+a\cos u\sin v\,\mathbf j-a\sin u\,\mathbf k`

`(\partial\mathbf r)/(\partial v)=-a\sin u\sin v\,\mathbf i+a\sin u\cos v\,\mathbf j`

`\implies(\partial\mathbf r)/(\partial u)*(\partial\mathbf r)/(\partial v)=a^2\sin^2u\cos v\,\mathbf i+a^2\sin^2u\sin v\,\mathbf j+a^2\sin u\cos u\,\mathbf k`

`\implies\left\|(\partial\mathbf r)/(\partial u)*(\partial\mathbf r)/(\partial v)\right\|=a^2\sin u`

So the area of the surface is


`\displaystyle\iint_S\mathrm dS=\int_(u=0)^(u=\pi)\int_(v=0)^(v=2\pi)a^2\sin u\,\mathrm dv\,\mathrm du=2\pi a^2\int_(u=0)^(u=\pi)\sin u`

`=-2\pi a^2(\cos\pi-\cos 0)`

`=-2\pi a^2(-1-1)`

`=4\pi a^2`

as expected.