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

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

We have

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

$\implies\nabla\times\vec F(x,y,z)=4\,\vec k$

Parameterize the ellipse $S$ by

$\vec s(u,v)=\dfrac{u\cos v}{\sqrt5}\,\vec\imath+\dfrac{u\sqrt5\sin v}4\,\vec\jmath$

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

Take the normal vector to $S$ to be

$\dfrac{\partial\vec s}{\partial\vec u}\times\dfrac{\partial\vec s}{\partial\vec v}=\dfrac u4\,\vec k$

Then the flux of the curl is

$\displaystyle\iint_S4\,\vec k\cdot\dfrac u4\,\vec k\,\mathrm dA=\int_0^{2\pi}\int_0^1u\,\mathrm du\,\mathrm dv=\boxed{\pi}$