Question

Problem set 11, question 6B-4. Really stuck with this, I keep getting the wrong integrand when using the method shown below.

@phi
@baru
@anonymous
Can you please help as you've helped me before/appear to have been working on this section of the course recently. Thanks!

Answer 1

I used the formula for ndS for a general surface on which I know the normal vector N, \[ndS=\dfrac{N}{N*k}dxdy\] to test my knowledge of the formula but I get the wrong integrand for the surface integral and it is undefined for z=0 which doesn't make any sense. Is there a reason why the formula isn't working in this case?

Answer 2

Sorry, should have said that the question was from the section on Surface Integrals and Flux and the question is;
\[Find \int\limits\int\limits_{S}F.dS\] where F=yj and S=the half of the sphere \[x^{2}+y^{2}+z^{2}=a^{2}\] for which \[y \ge 0\] oriented so that n points away from the origin.

Thanks for any help/guidance you can give.

Answer 3

using \[ n\ dS=\dfrac{N}{N \cdot k}\ dx\ dy \]
will not work for a plane perpendicular to the the x-y plane
(we would have to modify the approach to integrate for e.g. dx dz)

but for the sphere, we will have problems. maybe we could divide the surface into regions where dx dy works, and then switch to dx dz ?

Answer 4

I've not done that before, so could you please give me some hints on how I would do it that way? Also why can the formula not be used for a sphere, because they seem to suggest that it can in a worked example (with a full sphere)?

Answer 5

ok, using N= <x,y,z> in
\[ n\ dS=\dfrac{N}{N \cdot k}\ dx\ dy \]
I get
\[ \frac{<x,y,z>}{z} \ dx \ dy \]
and
\[ \int \int_S \frac{<0,y,0> \cdot <x,y,z>}{z} \ dx \ dy \\
\int \int_S \frac{y^2 }{z} \ dx \ dy
\]
the limits are defined by the "shadow" of the half-sphere onto the x-y plane
if we look at just the curve above the x-y plane
x ranges from \( -a \cos \theta \text{ to }+a \cos \theta \)
or double the result of 0 to \(+a \cos \theta \)
and y ranges from 0 to a
double this to account for the lower surface below the x-y plane
\[ 4 \int_0^a \int_0^{\sqrt{a^2-y^2}} \frac{y^2}{\sqrt{a^2-x^2-y^2}} \ dx \ dy \]

Answer 6

Thanks for that. That's a really clever way of thinking it through and it works, I get the correct answer. It still just leaves me with two questions though. Firstly I do not understand why your expressions for ndS and F.ndS are undefined for z=0.

Secondly, why does using your method give me a different integrand to the integrand produced by working F.ndS out directly via the formulas for n and dS on a sphere,\[n=\dfrac{<x,y,z>}{a}\] and \[dS=a^2\sin (\phi)d \phi d \theta\]

Any thoughts?

Answer 7

good questions.
on the 2nd, I'm thinking the difference arises because in one case we are integrating over the "actual surface" , while in the 2nd, we are integrating over the "shadow" of the surface projected onto the x-y plane. But it's not clear (to me) how to translate from one form to the other and show how/why the integrands are different.

Answer 8

When I convert from spherical to rectangular coordinates in the formulas for n and dS on a sphere I get that \[F.ndS=\dfrac{y^2}{a}dxdy\] and computing the integral with the same bounds as the integral you found (and multiplying it by 4 as you did) gives an answer of \[\dfrac{\pi a^3}{4}\] which doesn't match the correct answer of \[\dfrac{2 \pi a^3}{3}\] The two integrands (F.ndS) are different but I still can't work out why. If you have any more ideas then that would be great.

Answer 9

Sorry, another (possibly silly question but I'm a bit rusty on this). How can/do you split the bounds of your double integral?

Answer 10

Here is a write-up I did to clarify my thinking.

Answer 11

Thanks so much for that, it's great. I have a few questions (sorry!).

1) Firstly, I think equation 15 should equal\[a^2\sin(\phi)\cos(\phi)\]which would impact on the rest of the calculation and this is one of the reasons why I think that the two integrands are different when I use your method rather than my original method.

2) Why doesn't equation 15 equal\[a^2\sin(\phi)\]as dxdy does on a spherical shell (without the d phi d theta)?

3) I'm still struggling with the symmetry explanation between equations 8 and 9. I understand it in terms of Flux and fluid passing through the surface but I don't understand the 'y^2 is the same sign' and 'similarly, there is symmetry about the x-axis' comments.

4) I still don't understand why your expressions for ndS and F.ndS are undefined for z=0.

Thanks again for your time and help on this one, I really appreciate it!

Answer 12

1) Firstly, I think equation 15 should equal
a^2sin(ϕ)cos(ϕ)

Can you post your work ?

I don't understand your 2nd question.

3) ***I don't understand the 'y^2 is the same sign' ***

I am wrong with that comment. y^2 being positive is irrelevant (we are told y>=0)
If you like, you can change the bounds on the inner integral (over x) from 0 to sqr(a^2-y^2) to -sqr(a^2 -y^2) to +sqr(a^2 - y^2)
if you don't want to use symmetry.

The symmetry between the "quarter sphere" above the x-y plane and the quarter sphere below the x-y plane is harder to explain.

Answer 13

Hi Phi, thanks for that.

1) \[\dfrac{\partial(x,y)}{\partial(\phi,\theta)}=\det\Bigg(\left[\begin{matrix}x_\phi & x_\theta \\ y_\phi & y_\theta \end{matrix}\right]\Bigg)=\det\Bigg(\left[\begin{matrix}a \cos (\phi)\cos (\theta) & -a \sin (\phi)\sin (\theta) \\ a \cos(\phi)\sin(\theta) & a\sin (\phi)\cos (\theta)\end{matrix}\right]\Bigg)\]\[=a^2\sin(\phi)\cos(\phi) \cos^2(\theta)+a^2\sin(\phi)\cos(\phi)\sin^2(\theta)\]\[=a^2\sin(\phi)\cos(\phi)(\cos^2(\theta)+\sin^2(\theta))\]\[=a^2\sin(\phi)\cos(\phi)\] If this is right and you have also made a mistake in equation 16 when you are expressing z in spherical coordinates (I think that z should be \[a \cos(\phi)\] as it is when you define it above equation 15), then your final integral in equation 17 is correct but you've got to it by an incorrect method. Do you agree?

2) I thought it should be \[a^2 \sin (\phi)\]because you used this in equation 5.

3) I do want to stick with the symmetry explanation as I need to understand it this way for exams etc so If you have any thoughts on how to explain the symmetry further then that would be great.

Thanks again!

Answer 14

1) Oh, yes you are correct. I mis-typed up my notes.
However, I also messed up where I replaced z with \( a \cos \theta\). That should be \( a \cos \phi\)

2) I think the confusion arises because we must distinguish between integrating over dS (the actual surface, usually described in polar coords) and integrating over the "shadow" of the surface projected onto the x-y plane.

btw, try doing the problem by projecting the surface onto the x-z plane. That might be more intuitive? i.e. use N dot F / N dot j

3) When integrating over the "shadow" , all the problems I've seen have the surface above the x-y plane. When the surface is below the x-y plane, we have to adjust the sign of N dot F / N dot k. The z-coord also changes sign, and we get back the same expression as for the surface above the x-y plane.
More intuitively, the flux is <0,y,0> independent of z, so it seems that the "top" and "bottom" half of the volumes should give the same answer.

Answer 15

for what it's worth, updated and new material.

Answer 16

Hi Phi,

Sorry for the delay, I've been away. Thanks so much for documenting and explaining all of this, I've got a much better understanding of the topic now.

Really apprieciate all of your help.

Thanks!