My question is in connection with the study of the triple integrals in spherical coordinates. Supposed we have a surface $S$ defined by the equation $z=\sqrt{16-x^2-y^2}$. What would be the area of the surface determined by the points $A,B,C,D$? In case $$\mbox{length of arc from A to D }=\mbox{length of arc from B to C }$$ and $$\mbox{length of arc from D to C }=\mbox{length of arc from A to B }$$ it seems that the idea that I've on the books is the area should be $$\mbox{Area=(length of arc from A to D)$\times$ (length of arc from D to C)}.$$ I've been stuck as to how that area happened (assuming it is correct).
A little tip would mean a lot for me and thank you in advance. Please see the figure below
Since you didn't give us the explicit coordinates of $A, B, C,$ and $D$, it's tough to know what you're claiming. It appears to be this:
Given a quadrilateral on the surface of a sphere, where the angle at vertex $D$ is 90 degrees (I'm guessing from visual appearance here), the area of the quadrilateral is the product of the lengths of the two sides that meet at vertex $D$.
That's just false, for if the quadrilateral is very small, then the surface of the sphere is approximately a euclidean plane, and in the Euclidean plane, we can consider the quad with vertices $ D = (0,0), A=(1, 0), C = (0, 1)$, and $B = (x, y)$, where $x, y > 0$. The product of the two edges at $D$ is $1$, but if $(x, y) = (1, 2)$, for instance, then the actual area is $1.5$. [If this square seems too large to you, you can scale everything by a factor $s$; the area then becomes $1.5s^2$, which is still different from $1.0s^2$.]
You might say "No...I meant a geodesic quad with four equal angles, just the way it's drawn, and all of them being 90 degrees!" But the Gauss-Bonnet theorem tells you that the area of such a quad must be zero.
"Well...maybe with just all four angles equal then!" Fine. Let's look at four points nicely distributed about the north pole, with latitude $\phi$ and longitudes $0, \pi/2, \pi,$ and $3\pi/2$, where $\phi = 0$ would put the points all at the north pole. Does THAT quad have an area that's the product of two edge-lengths? (I'm going to consider a unit sphere rather than a radius $4$ sphere to make describing things easier.) Well, let's look at the situation where $\phi = \dfrac{\pi}{2}$. All four points are on the equator, hence the distances between adjacent ones are $1/4$ of the length of the equator: $\pi/2$. So the product of edge-lengths is $$ A_1 = \frac{\pi^2}{4} \approx 2.46 . $$ But the area of the "quad", which is the whole upper hemisphere, hence half the area of the sphere, $$ A_2 = 2\pi \approx 6.28. $$ These numbers are evidently different.
So I think that the claim about products of edge-lengths is likely to be false in any general situation that matches the one you've drawn.