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The following is an old exam problem (Calc III). It looks simple and technical, but I end up with a difficult integral and I guess I have a mistake somewhere.

We are given the vector field $F(x,y,z)=(4z+2xy,x^2+z^2,2yz+x)$. We are asked to calculate the line integral $\int_{C} \vec{F} \cdot d\vec{r}$, where $C$ is the intersection of the conic $z=\sqrt{x^2+y^2}$ and the cylinder $x^2+(y-1)^2=1$.

Stokes' Theorem allows us to replace the required integral with $\int_{S} \text{Curl}\vec{F} \cdot \hat{n} dS$, where $S$ is a surface bounded by $C$, and $\hat{n}$ is a normal to that surface.

The curl is $\text{Curl}\vec{F}=(0,3,0)$, so the integral simplifies to $3 \int_{S} (0,1,0) \cdot \hat{n} dS$.

I choose the surface to be $(x,y,\sqrt{x^2+y^2})$ with $x^2+(y-1)^2 \le 1$. I choose the parametrization $x=r\cos\theta, y=1+r\sin\theta$, and ended up with the integral $\int_{0}^{1} \int_{0}^{2\pi} \frac{r^2 \sin\theta +r}{\sqrt{r^2+1+2r\sin\theta}} dr d\theta$. I know how to solve similar integrals but this specific one seems non-elementary.

What am I doing wrong?

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    $\begingroup$ I think it's easier to stick with the line integral form, if you realize that Stokes' theorem implies that subtracting a curl-free field $\vec{G}$ from $\vec{F}$ will give the same integral over a closed loop. So you can drop certain parts of $\vec{F}$ from the integral immediately. $\endgroup$ – Semiclassical Sep 2 '14 at 17:48
  • $\begingroup$ @Semiclassical - Thanks, that's a useful and clever remark. We can replace $\vec{F}$ with $(4z,0,x)$ and use the parametrization suggested by Jyrki - $(x,y,z)=(\sin(2\theta), 2\sin^2\theta, 2\sin \theta), \theta \in [0,\pi]$. $\endgroup$ – Ofir Sep 2 '14 at 18:48
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    $\begingroup$ You can actually go a step further: What's the curl of $(z,0,x)$? (Another way to look at it is, what's the simplest function which has the same curl as $\vec{F}$ itself?) $\endgroup$ – Semiclassical Sep 2 '14 at 18:50
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A possibility is to express the circle $$ x^2+(y-1)^2=1 $$ using polar coordinates in the form $$ r=2\sin\phi,\quad 0\le\phi\le\pi. $$ This is an often recurring exercise (I have used it many times). This gives the surface a parametrization $$ \vec{r}=(x,y,z)=(2t\cos\phi\sin\phi,2t\sin^2\phi,2t\sin\phi) $$ with $\phi\in[0,\pi], t\in[0,1]$.

Unless I made a mistake we get $$ \begin{aligned} \vec{n}\,dS&=\frac{d\vec{r}}{dt}\times\frac{d\vec{r}}{d\phi}\,dt\,d\phi\\ &=4t\sin^2\phi(-\cos\phi,-\sin\phi,1)\,dt\,d\phi. \end{aligned} $$ Looks like this leads to a friendly integral.

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  • $\begingroup$ No mistake - this clever parameterization works (leads to $-8$). Thank you. $\endgroup$ – Ofir Sep 2 '14 at 18:52
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I'll expand on my remarks in comments a little. As observed in the OP, we're interested in a line integral $\int_C \mathbf{F}\cdot d\mathbf{r}$ for the prescribed closed contour $C$ and vector field $\mathbf{F}$. By Stokes' theorem, we may equate this to the net flux of $\nabla\times \mathbf{F}=3\hat{y}$ over the region $S$ of which $C$ is the boundary, i.e. $3\int_S (d\mathbf{S})_y$. Jyrki's answer gives a sensible parametrization for this integral, and so I will not consider it explicitly.

Instead, let's note an implication of Stokes' theorem: Any vector field with curl of $3\hat{y}$ will produce the same integral; in particular, the vector field $\mathbf{G}=3z\hat{x}$ satisfies this. To see why this works at the level of the line integral, observe that $$\mathbf{F}-\mathbf{G}=(4z+2xy,x^2+z^2,2yz+x)$$ may be recognized as the gradient $\nabla h$ of the scalar field $h(\mathbf{x})=x^2 y+x z+yz^2$. Then the fundamental theorem of line integrals implies that the integral over $\nabla h$ vanishes since $C$ is a closed loop, so the line integrals of $\mathbf{F}$ and $\mathbf{G}$ over $C$ must coincide.

With that in mind, all we need to compute is $\int_C \mathbf{G}\cdot d\mathbf{r}=\int_C 3z\,dx$. To take advantage of the symmetry of the curve under $x\to -x$, we parametrize the curve as $$x(\phi)=\sin \phi,\quad y(\phi)=1-\cos\phi,\quad z(\phi)=\sqrt{x^2+y^2}=2\sin \frac{\phi}{2},\quad \phi\in[0,2\pi).$$

The resulting trigonometric integral is elementary and reproduces the correct answer of $-8$.

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