25

The following problem comes from a vector calculus exam.

Let $$ S = \left\{ (x,y,z) \in \mathbb{R}: z = e^{1 - (x^2 + y^2)^2}, z > 1 \right\} $$ be an embedded surface with the orientation corresponding to the positive $\bf{OZ}$ direction, and let $$\mathbf F:\mathbb{R}^3\to \mathbb{R}^3; (x,y,z)\mapsto (x e^{y^2}, 2ye^{x^2}, 5-3z) $$ be a vector field. Calculate $\bf F$'s flux through the surface $S$, $$\iint_S \langle \mathbf{F}\cdot d\mathbf{S}\rangle$$

I am yet to find an effective way to solve it. So far, I have encountered solutions in terms of the likes of error functions or $$\int e^{x^2}dx$$ which is far from what one would expect at that level. Most of our attempts have been by using cylindrical coordinates and both Stokes's and Gauss's divergence theorems, but these awkward terms keep coming back.

For those who might have a go at it, let me give you the vector $n$ normal to the surface $$n= (-r^4 \cos\theta e^{1-r^4}, r^4\sin \theta e^{1-r^4}, r)$$ with the surface's cylindrical coordinates $(x,y,z) = (r\cos\theta, r\sin\theta, e^{1-r^4})$.

Source

Here is the original question in Spanish.

exam question

Rodrigo de Azevedo
  • 23,223
Sam Skywalker
  • 1,034
  • 1
    No disrespect but are you sure the question asks to find the flux itself and not the surface integral of the CURL of the vector field? The question seems to have been tailored so that the surface integral of the curl is easily calculated through applying stokes' theorem and it comes down to evaluating the integral of $\cos\left(\theta\right)\sin\left(\theta\right)e^{\sin^2\left(\theta\right)}$ along the unit circle ; which is exactly the type of question you'd expect in this exam. – Km356 Jun 12 '19 at 08:10
  • Yes, I am. I have added a picture of the question as it was asked in the exam, in Spanish. If I must be honest, part of me believes there must be something wrong with the question. Several PhD candidates have had a go at it, to no avail... and that's why I set a bounty! My main problem with using Stokes and carrying the thing to the border is that I know no viable integral to compute there. – Sam Skywalker Jun 12 '19 at 09:36
  • Also, thanks for editing it. – Sam Skywalker Jun 12 '19 at 10:41
  • I just would like to note two things based on experience: number one is that plugging values into $erf(x)$ is pretty easy, as long as your not plugging any values between -2 and 2 the answer is going to something that is very close to either positive one or negative one based on whether the input is positive or negative(check its graph), and number two is that trying a different/new change of variables might help find a solution in terms of more basic functions. – Malzahar Dec 22 '19 at 18:34
  • @Essam, thanks for your comment. However, I am far from familiar with error functions. In my attempt to solve the problem they appeared, but I couldn't go any further from that point and I supposed I was doing something wrong ($er\ f(x)$ did not appear in that course, targeted at second-years at uni). What change of variables would you suggest? – Sam Skywalker Dec 23 '19 at 21:35
  • 1
    With standard polar coordinates,$$\iint_S\langle\mathbf F,d,\mathbf S\rangle=\int_0^1\int_0^{2\pi}\left[4u^5e^{1+u^2\sin^2v-u^4}\cos^2v+8u^5e^{1+u^2\cos^2v-u^4}\sin^2v+\left(5-3e^{1-u^4}\right)u\right],dv,du$$...any takers? Mathematica can actually evaluate the $v$-integral in terms of a Bessell function but struggles with the rest. – user170231 Jul 18 '23 at 18:39
  • I’m still interested. I’m up for a discussion about ‘the rest’ @RodrigodeAzevedo – Sam Skywalker Jul 19 '23 at 09:08
  • Have you thought about using divergence theorem? You can close the surface by introducing the "base" $$z=1,x^2+y\leq 1$$ –  Jul 19 '23 at 15:37
  • @MatthewH. If you can do it and write an answer, the bounty is yours – Rodrigo de Azevedo Jul 19 '23 at 15:39
  • With the divergence theorem, the flux we want is$$2\pi+\iiint_R\left(2e^{x^2}+e^{y^2}-3\right),dz,dx,dy$$ where $R = \left{(x,y,z) \mid x^2+y^2\le1\land1\le z\le e^{1-\left(x^2+y^2\right)^2}\right}$. – user170231 Jul 19 '23 at 16:47
  • 3
    This problem was definitely a mistake somewhere and I can think of a few fixes. If the requested integral was intended to be curl $F$, then Stokes' theorem could be used to shift the integral onto the disk (a little known application of Stokes' theorem that bypasses Divergence theorem), and the answer would be $0$. The alternative is the surface could be $z=e^{1-(x^2+y^2)}$, then we could rewrite the vector field on the surface as $$F' = \frac{1}{z}(xe^{1-x^2},2ye^{1-y^2},5z-3z^2)$$ which is $1/z$ times a conservative field. This form works well with multivariable integration by parts. – Ninad Munshi Jul 19 '23 at 17:15
  • 3
    @NinadMunshi most likely. The professor who set the problem at the time did confirm after the exam that it was correctly stated. I thought of asking him about this problem, but he’s dead and deeply missed. – Sam Skywalker Jul 20 '23 at 11:56
  • 3
    That is a shame and I'm sorry for the loss. May he rest in peace. He has left quite a legacy for us on this forum! I was even thinking of posing the problem to some of my coworkers for a cash prize. – Ninad Munshi Jul 20 '23 at 16:30
  • Thank you for your kind words. If you get a satisfactory answer, I’d love to hear it. – Sam Skywalker Jul 21 '23 at 17:25
  • $2\pi+3\pi\int_0^1(e^{t/2}I_0(t/2)-1)(e^{1-t^2}-1)dt\approx 8.35919$ where $I_0(z)$ is the modified Bessel function of the first kind. – Bob Dobbs Jul 25 '23 at 14:08

1 Answers1

2

According to @NinadMunshi's useful feedback, the other answer of mine (which is fortunately deleted now by the MSE moderators on my request) was incorrect. In there, I tried to split the field $\bf F$ into two fields $\bf F_1$ (antisymmetric in $x,y$) and $\bf F_2$ (symmetric in $x,y$). Then I incorrectly deduced that the flux of $\bf F_1$ should be zero due to its antisymmetry, which is not correct since $d\bf S$ shares the same antisymmetry with $\bf F_1$, so that the flux is non-zero.

The surface integration vector $dS$ can be written as $$ d{\bf S}=rdrd\phi \left(4r^3z\cos\phi\hat x+8r^3z\sin\phi \hat y+\hat z\right). $$ where $z=e^{1-r^4}$. Therefore, $$ {\bf F}\cdot{d\bf S}=drd\phi \left(4r^5\cos^2\phi e^{1+r^2\sin^2\phi-r^4}+8r^5\sin^2\phi e^{1+r^2\cos^2\phi-r^4}+5r-3re^{1-r^4} \right) $$ and we have $$ \iint_S{\bf F}\cdot{d\bf S} {= \int_0^1\int_0^{2\pi} drd\phi \left(4r^5\cos^2\phi e^{1+r^2\sin^2\phi-r^4}+8r^5\sin^2\phi e^{1+r^2\cos^2\phi-r^4}+5r-3re^{1-r^4} \right) \\= 5\pi-\frac{3e\pi\sqrt{\pi}}{2}\text{erf}(1)+\int_0^1\int_0^{2\pi} d\phi dr \left(6r^2\cos^2\phi e^{1+r\sin^2\phi-r^2}\right) \\= 5\pi-\frac{3e\pi\sqrt{\pi}}{2}\text{erf}(1)+\int_0^1\int_0^{2\pi} d\phi dr \left(3r^2[1+\cos2\phi] e^{1+\frac{r}{2}-r^2}e^{-\frac{r}{2}\cos2\phi}\right) \\= 5\pi-\frac{3e\pi\sqrt{\pi}}{2}\text{erf}(1)+6\pi e\int_0^1 r^2 e^{\frac{r}{2}-r^2}\left[I_0\left(-\frac{r}{2}\right)-I_1\left(-\frac{r}{2}\right)\right]dr, } $$ where $I_n(x)$ is the modified Bessel function of the first kind.

Mostafa Ayaz
  • 33,056