1. Equation of an ellipse centred at (0,b)

The usual expression of the equation of an ellipse centred at the origin (0,0) of the coordinates on x-axis and y-axis is:
x²/a² + (y-b)²/b² = 1.
Where b is the major diameter, and a the minor diameter.

Translated from (0,0) to (0,b), and rotated at 90 degrees ( to be convenient to integrate), becomes:

x²/a² + (y-b)²/b² = 1.

2. The integral of this ellipse

I = ∫ x dy = a ∫ sqrt[1 - ((y-b)²/b²)] dy
y runs from 0 to m
Let's chamge the variable y as: (y - b)/b = t. Therefore:
I = a ∫ sqrt[1 - (t²)] b dt = a b ∫ sqrt[1 - (t²)] dt
t runs from -1 to (m-b)/b
t = sin z, then dt = cos z
I = a b ∫ cos z cos z dz = a b ∫ cos² z dz =
(a b/2) ∫ (1 + cos 2z)/2 dz = (a b/2) [ z + (1/2)sin 2z]
z runs from - π/2 to arcsin[(m-b)/b]
I = (a b/2) { arcsin[(m-b)/b] + π/2 + (1/2)sin 2[arcsin[(m-b)/b] - 0}

I = (ab/4){2arcsin[(m-b)/b]+π+sin (2[arcsin[(m-b)/b])}

Remark:
if m = 2b (above y-axis half) then:
I = (a b/4) { 2 (π/2) + π + sin 2[π/2] } = I = (a b/4) { 2 (π/2) + π + 0 }
I = a b π/2
With the below y-axis half, we have the total area of the ellipse π ab.

Example of an elliptical storage tank: To measure the volume of liquid in the tank, we use a depth gauge (measuring stick) to determine the amount of liquid in the tank.
2b = h
2a = w
The area of the ellipse side is:
A = (hw/8) { 2 arcsin[(m-b)/b] + π + sin 2[arcsin[(m-b)/b] }
With a length l, we have the expression of the volume:

V = (hwl/8) { 2 arcsin[(m-b)/b] + π + sin 2[arcsin[(m-b)/b] },
in unit of h, w, and l.

This is the volume of a liquid in a tank with length "l", width "w" and depth "h". The area related to the width and depth is delimited by an ellipse of major diameter 2b = h and minor diameter 2a = w