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This question is loosely related to a question I asked earlier today about surface parametrisation.

I have the vector field $\boldsymbol{v}=(2z,x,y)$ and want to find the surface integral

$$ \int_S (\nabla \times \boldsymbol{v}) \cdot d\boldsymbol{A} $$

where $S$ is the surface of the paraboloid $z=x^2+y^2$ underneath the plane $z=y$, bounded by the curve $C$ formed by the intersection of the paraboloid $z=x^2+y^2$ and the plane $z=y$.

I realise that there are simpler surfaces to pick (the disc) or I could use Stokes' Theorem and find the line integral over $C$, but my question specifically concerns the surface described above.

I can use the fact that we have a level set

$$ f=x^2+y^2-z=0 \implies \nabla f = (2x, 2y, -1)$$ which describes our surface and the normals to it. Then by a well known result,

$$ \begin{split} \int_S (\nabla \times \boldsymbol{v}) \cdot d\boldsymbol{A} &= \int_A \frac{ (\nabla \times \boldsymbol{v}) \cdot \nabla f}{\boldsymbol{e}_3 \cdot \nabla f} \, dx \, dy \\ &=\int_A(1-2x-4y) \, dx \, dy \end{split} $$

but I'm not really sure where to go from here (i.e. how to pick the limits of integration). I was thinking about parameterising the projection of $C$ in the $xy$ plane using polars but this seems completely wrong as I'm considering the surface of the paraboloid, not the disc. But I have no idea how you'd find the limits to use here/how you would even parameterise the paraboloid's surface to do the integral.

There are a lot of questions similar to this one on this site, but having read a lot of them, I'm still unsure as to whether a problem like this always gets reduced to a region in the $xy$ plane or if sometimes, you have to do some kind of a 3d parameterisation. The answer for me seems to be no, having looked at the proof of the result I've used above, and also simply due to the fact that the integral only has $dx \, dy$ at the end - no $dz$.

I'd be very grateful if anyone could shed some light on this - thanks.

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    $\begingroup$ Note that your surface is not closed, so you cannot use the notation $\oint$. $\endgroup$ – Kuifje Feb 5 '17 at 3:49
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Option 1:

The projection of the intersection of $z=x^2+y^2$ and $z=y$ in the $xy$ plane is $x^2+y^2=y$, i.e., the circle centered at $(0,1/2)$ with radius $1/2$, or in polar coordinates, $r=\sin \theta$. So you can parametrize the surface $S$ as follows: $$ \begin{cases} x=x \\ y=y \quad \quad \quad \quad \mbox{with }(x,y)\in D=\{(r,\theta)\;|\; 0 \le \theta \le \pi, \; 0 \le r \le \sin \theta \}\\ z=x^2+y^2 \end{cases} $$ It follows that $$ \Phi = \iint_S \nabla \times \vec{v} \; dS = \iint_D \nabla \times \vec{v}(x,y)\cdot \pmatrix{1\\0\\2x}\times \pmatrix{0\\1\\2y}\; dA = \iint_D \pmatrix{1\\2\\1} \cdot \pmatrix{-2x\\-2y\\1}\; dA \\= \iint_D 1-2x-4y \;dA $$ This matches your answer, except for the $4$ coefficient for $y$. If I am not mistaken, $ \nabla \times \vec{v}=(1,2,1)$. Now what you need to do is integrate over $D$ (or $A$ with your notation), as defined above.

Switching to polar coordinates yields $$ \Phi = \int_0^{\pi}\int_0^{\sin \theta}(1-2r \cos \theta-4r\sin\theta) r dr d\theta = -\frac{\pi}{4} $$

Option 2:

Alternatively, if you are familiar with the divergence theorem: $$ \Phi = \iint_S \nabla \times \vec{v} \; dS = \underbrace{\iiint_V \nabla\cdot \nabla \times \vec{v} \; dV}_{=0} - \iint_{S_2} \nabla \times \vec{v} \; dS $$ where $S_2$ is the surface that closes $S$, i.e., the part of the plane $z=y$ on top of $D$, which we can parametrize with $$ \begin{cases} x=x \\ y=y \quad \quad \quad \quad \mbox{with }(x,y)\in D=\{(r,\theta)\;|\; 0 \le \theta \le \pi, \; 0 \le r \le \sin \theta \}\\ z=y \end{cases} $$ It follows that $$ \Phi = - \iint_{S_2} \nabla \times \vec{v} \; dS = - \iint_D \nabla \times \vec{v}(x,y)\cdot \pmatrix{1\\0\\0}\times \pmatrix{0\\1\\1}\; dA = - \iint_D \pmatrix{1\\2\\1} \cdot \pmatrix{0\\-1\\1}\; dA = \iint_D dA = \frac{\pi}{4} $$

When we use this theorem, we conventionally parametrize the surface "outwards", i.e., in the opposition direction as in Option $1$, hence the sign difference.

Option 3:

Last but not least, you can use Stoke's theorem: $$ \Phi = \iint_S \nabla \times \vec{v} \; dS = \oint_C \vec{v} \cdot d\vec{r} $$ where $C$ is the curve at the intersection between the paraboloid and the plane. Using the parametrization proposed by @levap in the question that you asked earlier : $$ \begin{cases} x=\frac{\cos t}{2} \\ y=\frac{1+\sin t}{2} \quad \quad \quad \quad \mbox{with } t\in [0,2\pi] \\ z=\frac{1+\sin t}{2} \end{cases} $$ It follows that $$ \Phi = \oint_C \vec{v} \cdot d\vec{r} = \int_0^{2\pi} \frac{1}{4} \pmatrix{2(1+\sin t)\\ 1+ \cos t \\ 1 + \sin t}\cdot \pmatrix{-\sin t\\ \cos t \\ \cos t}\; dt = -\frac{\pi}{4} $$


Isn't Calculus fun ? :)

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  • $\begingroup$ Thank you - very helpful answer (especially your version 1). Indeed, at some point I did parametrise $D$ with polars like you did, and got to the integrand in terms of $x$ and $y$, but for some reason it did not come to me that I could simply use the standard $x \rightarrow r \cos \theta$ and $ y \rightarrow r \sin \theta$ - perhaps it's the original off-centre circle polar parameterisation which confused me. $\endgroup$ – mathphys Feb 5 '17 at 15:22

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