Surface of revolution between two circles I want to compute the surface of revolution between thoose two circles $(x_1,y_1)$ and $(x_2,y_2)$. I assume that the radius of the circles follow a well known law $x(y)$.
What I don't understand is that in my book, they say that an infinitesimal surface is : $2 \pi r(y) ds$ with $ds=\sqrt{dx^2+dy^2}$.
But I would choose $2 \pi r(y) dy$. Indeed if I want to compute the volume I would write : $ \pi r^2(y) dy$, so why should I take $ds$ for the surface and $dy$ for the volume, I don't understand.
(I know that the fact I take $dy$ for the volume is not a justification to take $dy$ for the surface but what I want to know is...how can I know which differential element is the good one ???).
How to be sure of the good differential element to take ?

 A: It really comes down to the same reason why you don't just use $dx$ or $dy$
to measure the length of a curve.
Instead, the element of length measurement is $ds = \sqrt{dx^2 + dy^2}$.
Consider the two curves $y = f(x) = 2$ and
$y = g(x) = 2+\sin\left(200\pi x\right)$
from $(x,y) = (0,2)$ to $(x,y) = (1,2).$
For the area between either curve and the $x$ axis, the integrals
$\int_0^1 f(x)\,dx$ and $\int_0^1 g(x)\,dx$ are perfectly correct and
both give the result $2.$
But the length of the curve $(x,g(x))$ is clearly much longer than
the length of $(x,f(x))$.
You have to follow the path of the curve in the direction the curve
actually follows, not just in either the $x$ direction or the $y$ direction
alone, to measure the length of the curve.
The reason it really comes down to the same reason is that
the area of the shaded surface in your figure is 
$2\pi x$ times the length of the $(x,y)$ curve between the two surfaces,
not $2\pi x$ times the distance between the surfaces.
After all, that surface is generated by taking that particular bit of the curve and sweeping it around the $y$ axis, 
not by taking some vertical line segment and sweeping it around the $y$ axis.
A: $dA =2 \pi r(y) ds$ makes no sense of consistency of used notation.
With $(x,y)$ notation of axes $ ds =\sqrt{dx^2+dy^2}$  and surface area $ dA = 2 \pi x ds  = 2 \pi x(y) ds, $ and
with $(r,z)$ notation of axes  $ ds =\sqrt{dr^2+dz^2}$ and  surface area $dA =  2 \pi r ds =  2 \pi r(z) ds$ 
