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As a preamble, if we have a surface given by $z=z(x,y)$ in $\mathbb{R}^3$, i.e. $z =z(x,y) = \sqrt{a^2-x^2-y^2}$ then $z- z(x,y)=0$ gives us a constant surface of a scalar field $f$, the $grad$ of which will tell us the normal vector to our surface: $\nabla f = \mathbf{k} -\frac{∂z}{∂x}\mathbf{i}-\frac{∂z}{∂y}\mathbf{j}$, the norm of which is $\sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2}$



A hemisphere of radius $a$ (with the $z$ axis as the axis of symmetry) is described by the equation $x^2+y^2+z^2 (=r^2) = a^2$ and has a surface area given by

$$\iint dS = \iint \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2}dA$$

If you project onto the $xy$ plane.

This formula is obtained by considering an element of surface area $\mathbf{dS}$ and its projected area $\mathbf{dA}$, and relating the two by $dA = \cos(\alpha) dS = \mathbf{\hat{n}}\cdot\mathbf{k}\, dS \rightarrow dS = dA/\mathbf{\hat{n}}\cdot\mathbf{k} = \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2}dA $ using $\mathbf{n}$, the normal to the surface, from the preamble and $\mathbf{k}$ the unit vector in the $z$ direction. enter image description here

For the above example we can use $z =\sqrt{a^2-x^2-y^2}$ to find our partial derivatives, and then evaluate the integral using plane polar co-ordinates ($0 \leq \theta \leq 2\pi, 0 \leq r \leq a $) to reach an answer of $2\pi a^2$ as expected.

However, if we use a normal vector to this surface expressed in 3D polar co-ordinates: $$\mathbf{\hat{n}} = \mathbf{r}/\|\mathbf{r}\| = (x\mathbf{i} + y\mathbf{j} + z\mathbf{r} )/ \|\mathbf{r}\| = \sin\phi \cos\theta \mathbf{i}+\sin\phi \sin\theta \mathbf{j}+\cos\phi \mathbf{k}$$

our logic above the included diagram should still hold: $dS = dA/\mathbf{\hat{n}}\cdot\mathbf{k}$ (here projecting onto the $xy$ i.e. $r,\theta$ plane). This yields $dS = dA/\cos\phi$ for our 3d polar case, which has got me stuck on the next step of evaluating

$$\iint dS = \iint \sec\phi dA = \iint \sec(\phi) rdrd\theta$$

It is intuitively true that as the angle $\phi$ increases, our surface area element on the hemisphere becomes more and more vertical w.r.t. the $xy$ plane and so this $1/\cos\phi$ makes our corresponding area element larger, but how should I actually evaluate the integral?

$\iint \sec\phi dA = \iint \sec(\phi) rdrd\theta$ is certainly ringing alarm bells, which makes me think that projecting on the $r,\theta$ plane in the first place was a mistake, which is a problem not encountered when working in 3D cartesians.

Is there such a thing as projecting onto the "$\theta,\phi$ plane" and then integrating over those two variables? - if this were to be a valid method I believe it could solve my problem here. Please do ask for further clarification if it is needed.

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  • $\begingroup$ If you are doing this in spherical coordinates, please evaluate $|r'_{\theta} \times r'_{\phi}|$ which should come to $a^2 \sin\phi$. Then integrate $\int_0^{2\pi} \int_0^{\pi/2} a^2 \sin\phi \ d\phi \ d\theta$ $\endgroup$
    – Math Lover
    Apr 12 at 17:31
  • $\begingroup$ @MathLover this is indeed the simplest way to arrive at an answer - unfortunately I am tasked with obtaining an answer specifically by projecting onto the $xy$ plane, hence my otherwise obtuse reasoning. $\endgroup$
    – Poo2uhaha
    Apr 12 at 17:35
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    $\begingroup$ I think $\cos\phi$ must be $\sqrt{1^2-r^2}/1$, and $\sec\phi$ is $1/\cos\phi$. This, AFAICT, gives you an integral which is improper but still can be evaluated. $\endgroup$ Apr 12 at 17:39
  • $\begingroup$ @bob.sacamento this is a good idea, but could you clarify where your expression for $\cos \phi$ comes from? I.e. we can use $adj/hyp$ but the value of $r$, the radial distance from the origin, is a constant on this hemispherical surface, so I do not understand the $\sqrt{1^2-r^2}$ term (which would also therefore be constant). $\endgroup$
    – Poo2uhaha
    Apr 12 at 17:44
  • $\begingroup$ As I understand it, the radius of the sphere is 1. If it is, instead, some constant value $R$, then you have $\sqrt{R^2-r^2}/R$ instead. $r$ is the distance from the origin in the $xy$ plane. After that, it's just trigonometry. $\endgroup$ Apr 12 at 18:09
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First of all, they are usually called spherical coordinates, not polar, when you are in $3D$. The radial variable is usually called $\rho$ instead of $r$. The element of area on the surface of a sphere is the product of two infinitesimal lengths. $\rho d\phi$ is the "changing latitude" length, and we usually say $r = \rho \sin \phi$ for the horizontal radius at that "latitude", so that $\rho \sin \phi d\theta$ is the "east-west" length, giving a surface area element $dS = \rho^2\sin\phi d\phi d\theta$. On this surface, $\rho = a$. So yes, you can absolutely integrate that way.

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  • $\begingroup$ Thank you for refining my notation; if we are asked specifically for an answer obtained as a result of projecting onto the $xy$ plane, is there any suitable approach using spherical co-ordinates, or do we have to use cartesians to avoid my problem detailed in the post? $\endgroup$
    – Poo2uhaha
    Apr 12 at 17:31
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    $\begingroup$ You can use cylindrical coordinates, which are polar coordinates with a $z$ axis added. $\endgroup$ Apr 12 at 17:32
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Trigger warning: in my explanation I'll be using the physics convention when referring to spherical coordinates.

I see your confusion. I think that yeah, as you point out $\iint \sec\phi dA = \iint \sec(\phi) rdrd\theta$ is a mistake - not in the sense that there's anything logically wrong but it doesn't actually provide us any neat way in calculating the surface area since $\theta \neq \phi$ of course so you have 3 degreees of freedom when you can really just do with 2. For this, it would probably be best reverting to cartesian coordinates - or even better parameterising the surface and finding the normal from there. Hope this helps!

enter image description here

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    $\begingroup$ physics notation bruh go back to your moments of inertia $\endgroup$
    – Poo2uhaha
    Apr 12 at 18:12
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    $\begingroup$ bruh moment, haha get it!! bruh moment of inertia $\endgroup$
    – jambajuice
    Apr 12 at 18:13
  • $\begingroup$ how would u recommend I parameterise the surface if spherical polars lead me into a rut? $\endgroup$
    – Poo2uhaha
    Apr 12 at 18:15
  • $\begingroup$ Essentially, we can deduce that this surface can be described by $x=ρsin(\theta)cos(\phi)$, $y=ρsin(\theta)sin(\phi)$, $z=ρcos(\theta)$, where all of this can just be encapsulated in $r(\theta,\phi)$ (describing the position vector in terms of two free degrees of freedom). From this - we can calculate $r'_{\theta} \times r'_{\phi}$ where $r'_{\theta}$ and $r'_{\phi}$ denote the cross product of partial derivatives of the position vector wrt to the independent variables. $\endgroup$
    – jambajuice
    Apr 12 at 18:22
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    $\begingroup$ " I'll be using the physics convention" YEA! $\endgroup$ Apr 12 at 21:18

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