Question

Evaluate the surface integral of vector field f (x,y,z) = x i + y j+(x+y) k over the portion s of the paraboloid z = x^2 + y^2 lying above the disk x^2 + y^2 ? 1. use outward pointing normals.

Answer 1

Parameterize the surface
`S` by


`\mathbf r(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+u^2\,\mathbf k`

where we use the polar coordinate transformation
`x(u,v)=u\cos v` and
`y(u,v)=u\sin v`, from which it follows that
`x(u,v)^2+y(u,v)^2=r^2=z(u,v)`. For this region, we have
`0\le u\le1` and
`0\le v\le2\pi`.

The surface integral is then


`\displaystyle\iint_S\mathbf F\cdot\mathrm dS=\int_(u=0)^(u=1)\int_(v=0)^(v=2\pi)\mathbf F(\mathbf r(u,v))\cdot(\mathbf r_u*\mathbf r_v)\,\mathrm dv\,\mathrm du`

where
`\mathbf r_u` and
`\mathbf r_v` are the respective partial derivatives of
`\mathbf r(u,v)`. We have


`\mathbf F(\mathbf r(u,v))=u\cos v\,\mathbf i+u\sin v\,\mathbf j+u(\cos v+\sin v)\,\mathbf k`


`\mathbf r_u=\cos v\,\mathbf i+\sin v\,\mathbf j+2u\,\mathbf k`

`\mathbf r_v=-u\sin v\,\mathbf i+u\cos v\,\mathbf j`

`\implies\mathbf r_u*\mathbf r_v=-2u^2\cos v\,\mathbf i-2u^2\sin v\,\mathbf j+u\,\mathbf k`

Now, presumably you know what a paraboloid looks like. An outward-pointing normal vector to the surface would probably refer to any normal that points downward, so check what happens to the cross product above at, let's say, the point
`(u,v)=(0.1,0)` (something close to the origin; the origin itself won't work because
`\mathbf r(0,0)=\mathbf 0`):


`\mathbf r_u*\mathbf r_v\bigg|_((u,v)=(0.1,0))=-0.02\,\mathbf i+0.01\,\mathbf k`

Now in Cartesian space, the point
`(-0.02,0,0.01)` would lie above the paraboloid, since


`(-0.02)^2+0^2=0.0004>0`

which means this normal vector has an inward orientation. To rectify this, swap the sign and replace
`\mathbf r_u*\mathbf r_v` with
`\mathbf r_v*\mathbf r_u` (which differs by a sign because the cross product is anticommutative).


`\implies\mathbf F(\mathbf r(u,v))\cdot(\mathbf r_v*\mathbf r_u)=u^2(\cos v+\sin v)-2u^3`

So the value of the surface integral is


`\displaystyle\iint_S\mathbf F\cdot\mathrm dS=\int_(u=0)^(u=1)\int_(v=0)^(v=2\pi)(u^2(\cos v+\sin v)-2u^3)\,\mathrm dv\,\mathrm du=\pi`