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I've been reading "An Introduction to Dynamic Meteorology" by Holton, and come a across a mathematical problem I can't solve. Consider a surface $A$ in $\mathbb{R}^3$ and the surface integral \begin{equation}\iint_A \cos\theta\, dA,\end{equation} where $\theta$ is angle between the origin and a point on the surface. Let $A_e$ describe the surface you get by projecting $A$ into the $\theta=\frac{\pi}{2}$ plane (the "equatorial plane", just the $x,y$ plane in cartesian coordinates). The problem is then to show that the above surface integral gives the area of $A_e$.

I was able to show this in the case that $A$ lies on the surface of a sphere with radius $R$ by parametrising $A$ in spherical coordinates as \begin{equation}\mathbf{p}(\theta,\phi)=R\cdot\hat{\mathbf{r}}(\theta,\phi),\end{equation} for $\theta,\phi\in A_\theta,A_\phi$ respectively, and noting that $A_e$ can then be parametrised as \begin{equation}\mathbf{p}_e=R\sin\theta \cdot \hat{\mathbf{r}}\left(\frac{\pi}{2},\phi\right),\end{equation} with $\theta,\phi\in A_\theta,A_\phi$ as before. Working through the integrals, and recalling $\frac{\partial \hat{\mathbf{r}}}{\partial \phi}=\sin\theta\cdot \hat{\boldsymbol{\phi}}$ you can show that \begin{align} \iint_A \cos\theta\,dA &= \int_{A_\phi}\int_{A_\theta} R^2\sin\theta\cos\theta\,d\theta\,d\phi \\ &=\int_{A_\phi}\int_{A_\theta} \left| \frac{\partial \mathbf{p}_e}{\partial \theta}\times \frac{\partial \mathbf{p}_e}{\partial \phi} \right| \,d\theta\,d\phi \\ &= \iint_{A_e} \,dA_e. \end{align}

I tried to extend this argument to a general surface $A$ not necessarily lying on the surface of the sphere by parametrising it as \begin{equation}\mathbf{p}(s,t)=r(s,t)\cdot\hat{\mathbf{r}}(\theta(s,t),\phi(s,t)),\end{equation} for $s,t\in A_s,A_t$. $A_e$ can then be parametrised as \begin{equation}\mathbf{p}_e(s,t)=r(s,t)\sin\theta(s,t)\cdot\hat{\mathbf{r}}\left(\frac{\pi}{2},\phi(s,t)\right)\end{equation} again with $s,t\in A_s,A_t$. I couldn't get this to work however, I just ended up with a mess of algebra. Does anyone have a general solution to this problem?

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  • $\begingroup$ I don't think $\cos\theta dA$ gives the projection onto the plane $xy$. Consider points with $x,y$ zero or almost zero. Your integral for the sphere is not correct, $\cos\theta$ doesn't incorporate to the expression of the jacobian; the integral properly managed is zero (as expected). Maybe $\theta$ is the colatitude. $\endgroup$ – Rafa Budría Jun 3 '17 at 7:15
  • $\begingroup$ "Is NOT the colatitude" I meant. $\endgroup$ – Rafa Budría Jun 3 '17 at 7:28
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$\theta$ cannot depend on the position of the surface, since that will vary if we shift the surface, but the projection onto a given plane should not vary.

Instead, $\theta$ should be the angle between the surface normal and the normal to the horizontal plane. We can see this in the following vertical slice through:

This shows an infinitesimal chunk of the surface, which is approximately planar. We see that the angle between the normal to the horizontal and the normal to the surface is the same as the angle between the surface and the horizontal. Hence the projected area onto the plane is $dA \cos{\theta}$.

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  • $\begingroup$ Yes I re-read the text book again and $\theta$ is in fact the angle between the surface normal and the vertical like you say! Only in the special case where $A$ lies on the surface of a sphere centred at the origin does $\theta$ coincide with how I originally described it, the book had a diagram labelling $\theta$ this way which is why I got confused! I also did some thinking along the lines of your diagram, but doesn't your argument only work if your projecting a small (infinitesimal) rectangle onto the plane when one side of the rectangle is already parallel to the plane? $\endgroup$ – Ewan Jun 4 '17 at 1:44
  • $\begingroup$ Besides, its actually small (infinitesimal) parallelograms we need to think about right, not rectangles? $\endgroup$ – Ewan Jun 4 '17 at 1:53
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    $\begingroup$ It might seem that way initially, but the result holds for the projection of any shape in one plane inclined to another: you can slice it into rectangles with their sides parallel to the horizontal plane (up to very small contributions at the edges that disappear as you take the limit of smaller rectangles). You can turn parallelograms into rectangles by slicing & moving one side. So basically, prove it for a rectangle of finite size in an inclined plane first, extend to any finite shape in that plane, and then you can apply it to infinitesimal chunks of surface that are approximately planar. $\endgroup$ – Chappers Jun 4 '17 at 1:56

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