# Confusion in using double integrals & projection to calculate hemispherical surface area

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. 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.

• 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 $\int_0^{2\pi} \int_0^{\pi/2} a^2 \sin\phi \ d\phi \ d\theta$ Apr 12 at 17:31
• @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. Apr 12 at 17:35
• 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. Apr 12 at 17:39
• @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). Apr 12 at 17:44
• 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. Apr 12 at 18:09

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.

• 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? Apr 12 at 17:31
• You can use cylindrical coordinates, which are polar coordinates with a $z$ axis added. Apr 12 at 17:32

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! • physics notation bruh go back to your moments of inertia Apr 12 at 18:12
• bruh moment, haha get it!! bruh moment of inertia Apr 12 at 18:13
• how would u recommend I parameterise the surface if spherical polars lead me into a rut? Apr 12 at 18:15
• 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. Apr 12 at 18:22
• " I'll be using the physics convention" YEA! Apr 12 at 21:18