Question

Evaluate the line integral ∮F•dr by evaluating the surface integral in​ Stokes' Theorem with an appropriate choice of S. Assume that C has a counterclockwise orientation when viewed from above. F = x^2 - y^2, z^2 - x^2, y^2 - z^2.C is the boundary of the square |x| ≤ 16​, |y| ≤ 16 in the plane z = 0.

Answer 1

The vector field

`\vec F(x,y,z)=\langle x^2-y^2,z^2-x^2,y^2-z^2\rangle`

has curl

`\\abla*\vec F(x,y,z)=\langle2y-2z,0,2y-2x\rangle`

Stokes' theorem says the line integral of
`\vec F` along
`C` is equal to the integral of the curl of
`\vec F` over a surface
`S` with
`C` as its boundary. Parameterize
`S` by

`\vec s(x,y)=\langle x,y,0\rangle`

with
`-16\le x\le16` and
`-16\le y\le16`. Take the normal vector to
`S` to be

`(\partial\vec s)/(\partial x)*(\partial\vec s)/(\partial y)=\langle0,0,1\rangle`

Then the line integral reduces to

`\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\\abla*\vec F)\cdot\mathrm d\vec S`

`=\displaystyle\int_(-16)^(16)\int_(-16)^(16)\langle2y-2z,0,2y-2x\rangle\cdot\langle0,0,1\rangle\,\mathrm dx\,\mathrm dy`

`=\displaystyle2\int_(-16)^(16)\int_(-16)^(16)(y-x)\rangle\,\mathrm dx\,\mathrm dy=\boxed0`