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

$\nabla\times\vec F(x,y,z)=\langle2y-2z,0,2y-2x\rangle$

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

$\dfrac{\partial\vec s}{\partial x}\times\dfrac{\partial\vec s}{\partial y}=\langle0,0,1\rangle$

Then the line integral reduces to

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_{-16}^{16}\int_{-16}^{16}\langle2y-2z,0,2y-2x\rangle\cdot\langle0,0,1\rangle\,\mathrm dx\,\mathrm dy$

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