Surface Area of a Sphere in Spherical Coordinates

[laser1]

Homework Statement

Description

Relevant Equations

N/A

My problem is when doing the surface integral of the ice cream bit. In the solution manual, it simply states that $d\mathbf{a}=r\sin \theta d\phi dr \hat {\boldsymbol \theta}$. The way I solved this problem was to take $\mathbf{\vec{r}}=(r\sin \theta \cos \phi, r\sin \theta \sin \phi, r\cos \theta)$, and then, as $\theta$ is a constant, take the partials with respect to $r$ and $\phi$, and then do the cross product. Following this, from the image, I guessed that the normal was in the $\boldsymbol{\hat{\theta}}$ direction, as it pointed in the direction of increasing $\theta$.

My result of the cross product was $(-r\sin \theta \cos \theta \cos \phi, -r\sin \theta \cos \theta \sin \phi, r\sin^2 \theta)$. I factored out $r\sin \theta$ and then computed $\boldsymbol {\hat{\theta}}$ by taking the partial derivative of $\mathbf{\vec{r}}$ with respect to $\theta$. Thus, $d\mathbf{a}=\vec{\mathbf{r}_\phi} \times \vec{\mathbf{r}_R} d\phi dr = r\sin \theta d\phi dr \boldsymbol{\hat{\theta}}$.

My problem is that this took a lot of time and space. Is there any other way to know that $d\mathbf{a}=r\sin \theta d\phi dr \boldsymbol{\hat{\theta}}$, aside from memorising? Was memorising it the intended method? I mean, I don't see how anyone remembers the above, it doesn't seem standard to me.

 

[PeroK]

The differential in spherical coordinates is:
$$dr \ \mathbf{\hat r} + rd\theta \ \boldsymbol{\hat \theta} + r\sin \theta d\phi \ \boldsymbol{\hat \phi}$$And, yes, Latex doesn't seem to like making the angles boldface. There must be a trick to make it work. In any case, on a surface of constant $\theta$, this reduces to the surface differential:
$$dr \ \mathbf{\hat r} + r\sin \theta d\phi \ \boldsymbol{\hat \phi}$$So, the directional surface element on this surface is:
$$d\mathbf{A} = \pm (dr \ \mathbf{\hat r} \times r\sin \theta d\phi \ \boldsymbol{\hat \phi}) = \pm r\sin \theta dr d\phi \ \boldsymbol{\hat \theta}$$In this case, you want the flux out through the surface, so $\boldsymbol{\hat n} = +\boldsymbol{\hat \theta}$.

 

[pasmith]

yes, Latex doesn't seem to like making the angles boldface. There must be a trick to make it work.

\boldsymbol{\hat\theta} $\boldsymbol{\hat\theta}$

 

[Orodruin]

Spherical coordinates are orthogonal. In orthogonal coordinates the surface element corresponding to the $y_1$ coordinate surface is given by*:
$$
d\vec S = h_2 h_3 \hat y_1 dy_2 dy_3
$$
where $h_a$ is the scale factor for $y_a$.

Therefore, for the surface element of the $\theta$ coordinate surface in spherical coordinates is:
$$
d\vec S = h_r h_\varphi \hat\theta \, dr\, d\varphi
$$
Now use that the scale factors are $h_r = 1$ and $h_\varphi = r\sin\theta$.

* This of course requires knowledge of the appropriate form of the surface element in orthogonal coordinates. See, eg, section 1.6.2 of my book.

 

[Orodruin]

I would also like to make some commercial for the use of differential forms here, even if it is probably out of the scope of the OP's curriculum.

The volume form in 3D Euclidean space being $\eta = dx\wedge dy \wedge dz$ is easily converted to spherical coordinates $\eta = r^2 \sin(\theta) dr \wedge d\theta \wedge d\varphi$. On a surface $S$ with $\theta$ constant the surface flux integral of $\vec v$ is given by
$$
\int_S \eta\left(\vec v, \vec E_\varphi, \vec E_r\right) = \int_S \eta_{a\varphi r} v^a d\varphi \, dr
= \int_S \eta_{\theta\varphi r} v^\theta d\varphi \, dr
= \int_S r^2 \sin(\theta) v^\theta d\varphi \, dr
$$
where $\vec E_a = \partial \vec r/\partial y^a$ and $v^\theta$ is the contravariant component of $\vec v$ in the $\theta$ direction, i.e, $\vec v \cdot \nabla \theta = \vec v \cdot \frac{\hat \theta}{r}$. This entirely avoids the use of scale factors whatsoever and readily generalises to non-orthogonal coordinates (and more than 3 dimensions ... and to curved spaces ...).

 

[laser1]

Now use that the scale factors are hr=1 and hφ=rsin⁡θ.

* This of course requires knowledge of the appropriate form of the surface element in orthogonal coordinates. See, eg, section 1.6.2 of my book.

Yes I am not sure why scale factors are those. What is your book?

 

[PeroK]

Yes I am not sure why scale factors are those. What is your book?

My advice is to keep going with Griffiths. You've nearly got to electrostatics now. All the mathematics that he uses are in that first section on vector calculus.

 

[Orodruin]

Yes I am not sure why scale factors are those. What is your book?

The scale factors are by definition the magnitudes of the vectors $\partial\vec r/\partial y^a$.

My book is just an example. This should be covered by any introductory text on vector analysis.

 

[PeroK]

The scale factors are by definition the magnitudes of the vectors $\partial\vec r/\partial y^a$.

My book is just an example. This should be covered by any introductory text on vector analysis.

The question in the OP is from Griffiths EM. There's a 60-page introduction to vector calculus at the start of the book. That only covers spherical and cylindrical coordinates and not more general curvilinear coordinates. The question here is one of the final problems in the first section.

My advice remains that the OP should do perhaps a couple more problems and then get stuck into electrostatics!

 

[Orodruin]

I am not a big fan of Griffiths vector introduction. To me it has always read a bit more like summary than introduction. The lack of generalized curvilinear theory being one of the shortcomings.

In the end, it depends on the goals of the OP: Do they want to understand the vector analysis on a deeper level (which goes into the direction of what they asked about) or do they want to go on with learning EM theory (perhaps with a somewhat weaker grasp on the vector analysis part).

Perhaps I am biased by having taken more focused vector analysis before EM theory (using Griffiths - we just skipped the introductort vector part as it had been covered by another course).

 

[laser1]

My advice is to keep going with Griffiths. You've nearly got to electrostatics now. All the mathematics that he uses are in that first section on vector calculus.

At the moment I am actually on chapter 3. My approach is to do all the problems within the text (not sure what to call them, but the ones not at the end) while learning, then as I move onto the next chapter, I concurrently do the end of problem chapters from the previous chapters. I think (?) this is supposed to increase retention.

 

[vela]

The diagram below is useful for a less mathy way of coming up with the expression for $dA$:

Credit: https://tikz.net/spherical_volume/

 

[PeroK]

The diagram below is useful for a less mathy way of coming up with the expression for $dA$:

View attachment 355418
Credit: https://tikz.net/spherical_volume/

I'm not sure what's not "mathsy" about that!

 

[vela]

I am not a big fan of Griffiths vector introduction. To me it has always read a bit more like summary than introduction. The lack of generalized curvilinear theory being one of the shortcomings.

That's not surprising. In the US, students should have seen this stuff a year before in Calc III. Griffiths is likely just including the chapter as a summary/refresher.

 

[Orodruin]

That's not surprising. In the US, students should have seen this stuff a year before in Calc III. Griffiths is likely just including the chapter as a summary/refresher.

Sure, here (Sweden) too. In the physics program at my university we have dedicated courses on linear algebra, single variable calculus, multivariable calculus, vector analysis, mathematical methods in physics, differential equations and transforms (among others) before students take electrodynamics. This is why I say the Griffiths chapter seems insufficient as an introduction to the subject and will likely not cover everything in the detailed required to reach a deeper understanding. For that a dedicated text on vector analysis would be preferable. That was kind of my point.