Surface area under a curve on a sphere I have an incomplete spiral pattern (as shown) on a sphere of radius R.
I want to find the area under the curve made by the highest points of the incomplete sprial.
I have tried solving this in these ways:

*

*Thinking about how one finds an area under a curve on a circle. One would make slices of some width w and find the height value of Y in each bin and just multiply Y*w to get that bin contribution. Add all the contributions and get the area under the curve.
I am not able to translate this idea for a sphere, how to bin this?

*Another way is to consider a disk(in Y-Z plane) of width w, some r distance from the center of the sphere. Now depending on this r, the number of points lying on that disk will vary. One can find the center of this disk and find at which angles are the "top" two points are, find the arc length, and multiply with width w. Do all the disks and add them and that's your area under the curve.
This didn't work because there are multiple(which are not the part of the topmost curve) contributing to the area, which is wrong.

Any help is greatly appreciated.
These are the coordinates in $(x,y,z)$ format:
https://pastebin.com/Att1Ep1r

P.S.: This is my first time asking questions on stack exchange. Please let me know if I am doing something wrong.
 A: Comment
In 3D we look at 2D/3D Fibonacci spherical spiral packing and observe..
Fibonacci packing
The spiral closely resembles a log spiral in 2D and corresponding loxodrome in 3D.
From the time of Greeks it was known from the sunflower dividing a circle into two arcs whose ratio is the
$$ Golden Number\text{ or }Ratio = \phi $$
then the shorter arc of rhombus subtends an angle $\phi$ of about $137.5^{\circ}$
I took half this angle and interpreted $ \tan^{-1}\phi^2 $ to mean a 2D log spiral appearance making a constant angle between curve and radius:
$$ \frac{r}{r_i}=   e^\left({\dfrac {\theta}{\phi^2} }\right) $$

which shows several $137.5^{\circ}$ rhombuses.
After further comments, I shall upload another 3D Loxodrome image.
I am lead to this conclusion because:

*

*that a spiral and a visible tendency exist unmistakably.


*The helices are involutes at $ r\to \infty$ the involutes have constant widths to comfortably accommodate future addition in a Fibonacci order.


*If we ignore packing differences for clock wise and anticlockwise which exist, Nature could not have changed from the Greek to the present time.
A: I'm not sure exactly what you're asking, but if you have a space curve, say lying on the the graph of $z=f(x,y)$,
$$
r(t)=(x(t), y(t), z(t)), \ z(t)=f(x(t),y(t)), \ a\leq t\leq b,
$$
and you want the (signed) area of the vertical surface between the plane $z=0$ and the curve, you can integrate $f(x,y)$ over $(x(t), y(t))$ with respect to arclength along $(x(t), y(t))$
$$
\int_a^bf(x(t),y(t))\sqrt{(x'(t))^2+(y'(t))^2}dt.
$$
See figure 6.13 here (or any vector calculus book) for a picture of what I'm describing.
