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:
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:
(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...
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$\begingroup$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 $$ \frac{\pi r}{6h^2}\left(\sqrt{r^2+4h^2}^3-r^3\right)\tag{2} $$
$\endgroup$ 1 $\begingroup$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.
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