I feel like this should be really easy, but I'm not sure if I'm doing it correctly so I'm going to give it a go here, and if I'm not very good at maths (I'm not) then you can hopefully correct me!

Suppose I want to paint the surface of a parabolic dish, how would I calculate that area?

To make it slightly more clear, this type of thing:

Picture of radio telescope

Where a cross section of it would be some function:

$y = a x^2\quad$ (1)

from $y = 0$ (what I will call the vertex) to some point $y=h$ at the opening of the dish.

I thought then that the radius at a certain height would be:

$r = \sqrt{\frac{y}{a}}\quad$ (2)

and therefore the radius at the top, $d$, would be given by the equation:

$d = \sqrt{\frac{h}{a}}$

or in other words:

$a = \frac{h}{d^2}$

So I rewrote (1) and (2) using that:

$y = \frac{h}{d^2} x^2 \quad$ (3)

$r = \sqrt{\frac{d^2y}{h}}\quad$ (4)

Now I can find the areas of small strips around the dish, and integrate. And this is where I get a bit stuck. I initially did the area of rectangles, but of course the strips (if you unfold them) are actually like this:

picture of cone and an unfolded bit of the curved surface

(of course they're not exactly like that because that's a cone and we have a dish, but they're similar in how they're both not rectangles for almost the same reason, if that makes sense)

And I have no idea how to express that area mathematically...

  • $\begingroup$ The practical answer is to measure the diameter of the opening of the dish $d$, compute the area of a circle with that diameter $\frac \pi 4 d^2$, and buy a little more paint than that. The true area won't be too much more. $\endgroup$ – Ross Millikan Jun 27 '15 at 15:29

The formula for the surface area of a curve rotated about the $y$-axis (see the Theorem of Pappus) is $$ \begin{align} \int_0^r\overbrace{\vphantom{\sqrt{y'^2}}2\pi x}^{\text{length of strip}}\overbrace{\sqrt{1+y'^2}\,\mathrm{d}x}^{\text{width of strip}} &=\int_0^r2\pi x\sqrt{1+4a^2x^2}\,\mathrm{d}x\\ &=\frac{\pi}{4a^2}\int_0^{4a^2r^2}\sqrt{1+t}\,\mathrm{d}t\\[6pt] &=\frac{\pi}{6a^2}\left(\sqrt{1+4a^2r^2}^3-1\right)\tag{1} \end{align} $$ where $t=4a^2x^2$. You can use $h=ar^2$ to get $$ \frac{\pi r}{6h^2}\left(\sqrt{r^2+4h^2}^3-r^3\right)\tag{2} $$

  • $\begingroup$ As $a\to0$, the expression in $(1)$ tends to $\pi r^2$. $\endgroup$ – robjohn Jun 28 '15 at 1:46

As you describe it, the origin is at the center of the dish, $x$ is radial, and $y$ is up the tower. If you know the radius of the dish $r$ and the height of the edge $h$, you have $h=ar^2, a=\frac h{r^2}$ Now consider a small piece of the curve from $x$ to $x+dx$. The area swept out as you rotate this part of the curve is the arc length between $x$ and $x+dx$ times $2 \pi x$. The element of arc length is $ds=\sqrt{dx^2+dy^2}=dx\sqrt{1+(\frac {dy}{dx})^2}$ So you want $\int_0^r dx2 \pi x\sqrt{1+(\frac {dy}{dx})^2}=\int_0^r dx2 \pi x\sqrt{1+(2ax)^2}$

Added: Alpha integrates this to $\frac {\pi (4a^2r^2+1)^{3/2}}{6a^2}$, which should reduce to $\pi r^2$ in the limit $a \to 0$, but I can't get it to do so. The integral above does reduce that way nicely.

  • $\begingroup$ I specifically didn't want it in terms of $a$ which was why I eliminated $a$ in my equations. $h$ and $r$ are measurable you see, so I wanted the area in terms of those. But never mind, we could just substitute back in. $\endgroup$ – Jamie Twells Jun 27 '15 at 15:50
  • $\begingroup$ Actually, could you expand on why $2\pi x \mathrm{d}s$ is the area? Why is it not more complicated than that? I mean, imagine a rectangle with dimensions $2 \pi x$ and $\mathrm{d}s$. If you fold that rectangle around to meet, then lie it on the dish, it leaves a V shape where the bottom corners meet but the top corners have a little gap, surely? $\endgroup$ – Jamie Twells Jun 27 '15 at 16:00
  • $\begingroup$ So does this eventually give $A = \frac{\pi}{hd}(2hx^2+d^2)^\frac{3}{2}$ where $d$ is radius at the opening (widest radius) and $h$ is the height $\endgroup$ – Jamie Twells Jun 27 '15 at 16:57
  • $\begingroup$ For your second comment, the overlap area is second order in differentials-it is proportional to $ds^2$ so doesn't contribute. For the integral, Alpha gets $\frac {2 \pi (1+ar^2)^{3/2}}{3a}$. I haven't checked if that matches your solution. You shouldn't have an $x$ in your result. For comparison, the area of a spherical cap is $2 \pi h R$, where $R$ is the radius of the sphere. Your shape is not very different. To get $R$ you have $R^2=r^2+(R-h)^2$ $\endgroup$ – Ross Millikan Jun 27 '15 at 19:45
  • $\begingroup$ yes sorry should have been $d^2$ not $x^2$ $\endgroup$ – Jamie Twells Jun 27 '15 at 19:48

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