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Applying Stokes' theorem to a surface, I obtained the following equations, $$R_y - Q_z = xe^{y}-e^{x}\cos(z)$$ $$P_z - R_x = -2y\sin(z)-e^{y}$$ $$ Q_x-P_y = e^{x} \sin(z) - 2\cos(z)$$ where the field $$\vec F = 2y\cos(z)\hat i +e^{x} \sin(z) \hat j + xe^{y}\hat k$$ How do I solve this system for $P, Q$ and $R$?

gofvonx
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Artemisia
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If you just need the value of surface integral (rather than verification of Stoke's theorem), it can be computed as follows: $C: x^2+y^2=9,\, z=0$ is the curve of common boundary of the surfaces $x^2+y^2+z^2\leq 9$ and $z\geq 0$ (surface bounded by $C$) with the parametric equations $x=3\cos t,\ \ y=3\sin t,\ \ ( z=0)\, t\in[0, 2\pi]$. On the curve, $P=6\sin t$, $Q=0$ and $R=3\cos t e^{3\sin t}$ with $dx=-3\sin t~dt$ and $dy=3\cos t~dt$. Thus this leads to (if I computed correctly) $\int_C Pdx+Qdy+Rdz=\int_0^{2\pi}(-18\sin^2t)~dt=-18\pi$.

daulomb
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  • Ah great! Thanks. I just needed help with the solution for the partial derivative equations. – Artemisia Nov 10 '13 at 15:45
  • How do I do this sum if I need to verify Stokes' theorem? – Artemisia Nov 10 '13 at 15:53
  • In order to evaluate the RHS we can make use of the divergence theorem by obtaining a closed surface. This can be done by closing the upper semi-sphere (say $S_1$) by the plane $S_2:z=0$. In this case applying the divergence theorem along with the fact that $\nabla\cdot curl\vec F=0$. Eventually you will have $\int\int_{S_1}curl\vec F\cdot d\vec S=-\int\int_{S_2}curl\vec F\cdot d\vec S$ with $\vec n= (0, 0, -1)$ is the outer unit normal to $S_2$. – daulomb Nov 10 '13 at 16:26
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    The last integral will be equal to $\displaystyle\int\int_{x^2+y^2\leq 9}(-2)dydx=-18\pi$. This verifies Stoke's theorem. – daulomb Nov 10 '13 at 16:46