Question

Use Stokes' Theorem to evaluate S curl F · dS. F(x, y, z) = 5y cos(z) i + ex sin(z) j + xey k, S is the hemisphere x2 + y2 + z2 = 4, z ≥ 0, oriented upward. Step 1 Stokes' Theorem tells us that if C is the boundary curve of a surface S, then curl F · dS S = C F · dr Since S is the hemisphere x2 + y2 + z2 = 4, z ≥ 0 oriented upward, then the boundary curve C is the circle in the xy-plane, x2 + y2 = 4 Correct: Your answer is correct. seenKey 4 , z = 0, oriented in the counterclockwise direction when viewed from above. A vector equation of C is r(t) = 2 Correct: Your answer is correct. seenKey 2 cos(t) i + 2 Correct: Your answer is correct. seenKey 2 sin(t) j + 0k with 0 ≤ t ≤ 2π.

Answer 1

By Stokes' theorem, the integral of the curl of $\vec F$ across $S$ is equal to the integral of $\vec F$ along the boundary of $S$, call it $C$. Parameterize $C$ by

$\vec r(t)=2\cos t\,\vec\imath+2\sin t\,\vec\jmath$

with $0\le t\le2\pi$. So we have

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

$=\displaystyle\int_0^{2\pi}(10\sin t\cos 0\,\vec\imath+e^{2\cos t}\sin0\,\vec\jmath+2\cos t\,e^{2\sin t}\,\vec k)\cdot(-2\sin t\,\vec\imath+2\cos t\,\vec\jmath)\,\mathrm dt$

$=\displaystyle\int_0^{2\pi}-20\sin^2t\,\mathrm dt$

$=\displaystyle-10\int_0^{2\pi}(1-\cos2t)\,\mathrm dt=\boxed{-20\pi}$