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https://sites.science.oregonstate.edu/math/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/flux/flux.html

"Find the flux of the vector field <y,x,z> in the negative z direction through the part of the surface z=g(x,y)=16-x^2-y^2 that lies above the xy plane" (quoted verbatim)

I'm tryng to do the question at the bottom but I'm messing up on the integral part of $\iint_{S} F \cdot dS = \iint_{D} F \cdot (r_s \times r_\theta) dA$ (here $s$ is the radiuS)

For my normal, since I thought $r = \langle r \cos \theta , r \sin \theta, 16-r^2 \rangle$, I thought it would be something like this: $$ n = \begin{vmatrix} i & j & k \\ \cos \theta & \sin \theta & -2r \\ -r \sin \theta & r \cos \theta & 0 \end{vmatrix} $$

And $F = \langle r\sin\theta, r\cos\theta , 16-r^2 \rangle$?

I'm able to get down to how the normal vector is $n = \langle 2r^2 \cos \theta , -2r^2 \sin \theta, r \rangle$ and then $F\cdot n = 16r-r^3$ but that doesn't seems to work in the integral because I get something that isn't $128\pi$ (I'll be flipping the sign at the end so that I get the downward direction)

From my equation of $r(s, \theta)$ I have $s \in [0,4]$ and $\theta \in [0,2\pi]$ and I know that $$ \int_{0}^{2\pi} \int_{0}^{4} 16r - r^3 dr d \theta = 128\pi$$ but I don't think that's how the $dA$ should work in cylindrical coordinates? Shouldn't it be $r \ dr \ d\theta$? But then $$ \int_{0}^{2\pi} \int_{0}^{4} (16r - r^3) r dr d \theta \neq 128\pi$$

So my question is, if my $F\cdot n$ is correct, then how do I set up the integral in cylindrical coordinates to get the right answer.

Ally
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  • The question at which bottom? Type it instead of asking us to open links and read that stuff. – Kurt G. Sep 17 '24 at 09:14
  • Added the question verbatim. – Ally Sep 17 '24 at 09:39
  • Thanks. But the author of that page does more by peforming all the calculation steps. Comparing this with the integral you got your's doesn't match. You should be able to trace down where earlier you messed up. – Kurt G. Sep 17 '24 at 09:43
  • Well, somehow I'm not seeing it. Is my $F \cdot n$ not the same as his? (I'm ysing cylindrical from the start, rather than with the $x$'s and $y$'s) – Ally Sep 17 '24 at 09:49
  • You keep asking me to go through the steps in that link. Why should I? The hint I gave you should be enough to look for an earlier error. – Kurt G. Sep 17 '24 at 10:12
  • Not only does the normal vector you got look incorrect, it also won't make your life easier when you do all the calculations in cylindrical coordinates from the beginning. Converting vectors between systems can be tricky. I suggest to follow the author's route and take appropriate coordinates only when you have to perform the integration. – Kurt G. Sep 17 '24 at 11:00
  • Although I could follow the website, I want to try it this way since both methods should ultimately be equivalent. In my added calculations, I'm off by a factor of $r$ somewhere. – Ally Sep 17 '24 at 17:49
  • I am telling you that your normal vector $\mathbf{n}=\langle 2r^2 \cos \theta , -2r^2 \sin \theta, r \rangle$ is wrong because it cannot depend on $\theta$ as the surface has that rotational symmetry. You absolutely want to go the hard route that causes a lot of more work. Hint: the dot product $\mathbf{F}\cdot\mathbf{n}$ is a scalar and much easier to convert to cylindrical coordinates than the individual vectors. See the link in my previous comment if you still want to do the latter. – Kurt G. Sep 17 '24 at 17:56
  • I think your point about the dot product being a scalar and being easier to convert between different coordinate systems than individual vectors is probably why I'm having this issue. – Ally Sep 17 '24 at 18:08
  • Not probably. Certainly. – Kurt G. Sep 17 '24 at 20:03
  • In general, and assuming one isn't already experienced in such things, how would you know that you can't (easily) convert vectors between coordinate systems? It seems that one could (easily) convert coordinates when everything is in terms of x, y, z's to cylindrical or spherical, but that's not the case for vectors. I am thankful for your help so far but it seems a lot of my issues could have been fixed had I somehow known that changing coordinates for vectors wouldn't have (easily) worked in the first place. – Ally Sep 17 '24 at 20:50

2 Answers2

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Too long for a comment:

  • The author's correct expression for $\mathbf{F\cdot n}$ is $$\tag1 \mathbf{F\cdot n}=-4xy-z=-4xy-16+x^2+y^2\,. $$

  • When we convert this to cylindrical coordinates we get $$\tag2 \mathbf{F\cdot n}=-4r^2\sin\theta\cos\theta-16+r^2\,. $$

  • Now I give you hints to go your hard route: to get the correct expression for the normal vector in cylindrical coordinates you express the surface as a level set of the function $$\tag3 f(x,y,z)=16-x^2-y^2-z=16-r^2-z\,. $$

  • We know that $\nabla f$ is a normal vector. Hence $$\tag4 \mathbf{n}=\pmatrix{-2x\\-2y\\-z}\quad\text{ (author) or, going your route, }\quad\mathbf{n}=\pmatrix{-2r\\0\\-z}\,. $$

  • To get $\mathbf{F}=(y,x,z)$ in cylindrical coordinates you have to use a conversion formula in the link I already gave:

\begin{align}\tag5 F^x&=F^r\cos\theta-rF^\theta\sin\theta\,,\\[2mm]\tag6 F^y&=F^r\sin\theta+rF^\theta\cos\theta\,,\\[2mm]\tag7 F^z&=F^z\,, \end{align} or equivalently, \begin{align}\tag8 F^r&=F^x\cos\theta+F^y\sin\theta\,,\\[2mm]\tag9 F^\theta&=-\frac1rF^x\sin\theta+\frac 1rF^y\cos\theta\,, \\[2mm]\tag{10} F^z&=F^z\,. \end{align}

  • You will get an $\mathbf F=(F^r,F^\theta,F^z)$ in cylindrical coordinates which will lead to the correct dot product from above:

$$\tag{11}\boxed{\quad\mathbf{F}=\pmatrix{2r\sin\theta\cos\theta\\-\sin^2\theta+\cos^2\theta\\z}\,.\quad}$$

  • This is another useful post where it is explained in detail how vectors differ from scalars or points on the manifold when converted to different coordinates systems.
Kurt G.
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Your expression for $F \cdot n$ is not correct. Remember that when you are working in different coordinate systems, you should be able to recover the same expressions for all vectors and expressions. If you convert $16 r - r^3$ back to cartesian coordinates, you see that it does not match up to the formulas in the link.

Recheck your (converted) expressions for $F$ and $n$. It could help to transform the calculated values in the link, and see if you get the same expressions when starting in cylindrical coordinates.

BrakkoFTW
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  • I added my calculations for $F$ and $n$. Somewhere, I'm off by a factor of $r$ but I'm not sure where it is since wouldn't $x=r \cos \theta$ and $y= r \sin \theta$ – Ally Sep 17 '24 at 17:41