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For the vector field $\mathbf{F}(x,y,z) = \langle 5y+\sin x, z^2+\cos y, x^3\rangle$, I need to find the integral using Stokes' Theorem $\int_C\mathbf{F}\cdot\mathrm{d}\mathbf{r}$ where $C$ is the curve $\mathbf{r}(t)=\langle \sin t, \cos t, 2\sin t\cos t\rangle, t\in[0,2\pi]$.

I have gotten the curl as $\langle -2z,-3x^2,-5\rangle$. But I was troubled in parametrising the surface.

When I suggest that the curve is contained in the surface parametrised by $\mathbf{r}(u,v)=\langle u\sin v, u\cos v, 2u\sin v\cos v\rangle$, I have $\mathbf{r}_u\times\mathbf{r}_v=\langle 2u\cos^3 v, 2u\sin^3 v,-u\rangle$ which changes direction in the domain of $v\in[0,2\pi]$. How should I correctly parametrise the surface?

Thank you very much.

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  • $\begingroup$ A normal vector need not be constant and in general isn't, but it should always be on one side of the surface which yours is. $\endgroup$ – Triatticus Jan 20 '18 at 8:40
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More simply, we can take as surface $2xy-z=0$ with downward unit normal vector $$\mathbf{n}=\frac{\langle 2y,2x,-1\rangle}{\sqrt{1+4x^2+4y^2}}$$ (note that the closed path is clockwise oriented). Then, by Stokes' theorem, the given line integral is equal to $$\iint_{x^2+y^2\leq 1}\underbrace{\langle -2z,-3x^2,-5\rangle}_{\text{curl}(\mathbf{F})} \cdot \underbrace{\frac{\langle 2y,2x,-1\rangle}{\sqrt{1+4x^2+4y^2}}}_{\mathbf{n}}\cdot \underbrace{\sqrt{1+4x^2+4y^2}dxdy}_{dS}$$ where $z=2xy$. Finally, it is easy to verify that it boils down to $$\iint_{x^2+y^2\leq 1}(-8xy^2-6x^3+5)dx dy=5\iint_{x^2+y^2\leq 1}dx dy=5\pi.$$

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