Question

Use the surface integral in​ Stokes' Theorem to calculate the circulation of the field Bold Upper F equals x squared Bold i plus 4 x Bold j plus z squared Bold kF=x2i+4xj+z2k around the curve​ C: the ellipse 25 x squared plus 16 y squared equals 525x2+16y2=5 in the​ xy-plane, counterclockwise when viewed from above.

Answer 1

Stokes' theorem says the integral of the curl of
`\vec F` over a surface
`S` with boundary
`C` is equal to the integral of
`\vec F` along the boundary. In other words, the flux of the curl of the vector field is equal to the circulation of the field, such that

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

We have

`\vec F(x,y,z)=x^2\,\vec\imath+4x\,\vec\jmath+z^2\,\vec k`

`\implies\\abla*\vec F(x,y,z)=4\,\vec k`

Parameterize the ellipse
`S` by

`\vec s(u,v)=(u\cos v)/(\sqrt5)\,\vec\imath+\frac{u\sqrt5\sin v}4\,\vec\jmath`

with
`0\le u\le1` and
`0\le v\le2\pi`.

Take the normal vector to
`S` to be

`(\partial\vec s)/(\partial\vec u)*(\partial\vec s)/(\partial\vec v)=\frac u4\,\vec k`

Then the flux of the curl is

`\displaystyle\iint_S4\,\vec k\cdot\frac u4\,\vec k\,\mathrm dA=\int_0^(2\pi)\int_0^1u\,\mathrm du\,\mathrm dv=\boxed{\pi}`