A quarter sphere with a radius of $10 \text{ units}$.
Please help, also remember the sides.
I used the normal formula of the total surface area of a sphere and divided it by $4$, then added half the area of a circle but it wasn't equal to the correct answer.
[Math] Surface area of quarter of a Sphere
areageometrysolid-geometry
Related Solutions
You might want to look for a treatise by Archimedes on "the method", in which he does this computation using Cavalieri's principle (more or less) and a rather clever mechanical argument. There's a good case to be made that in doing so, he was inventing some of the key ideas of calculus, and the proofs he offers don't really meet the modern standard of rigor, but nonetheless, it's a remarkable achievement.
In this answer, it is shown geometrically that the area of a strip of sphere between two parallels of latitude is the same as the area of the orthogonal projection of that strip onto the cylinder with the same radius as the sphere and whose axis is parallel to the north-south axis of the sphere.
Finding the area of this cylinder is just multiplying its height, $2r$, by its circumference, $2\pi r$. Thus, the area of the sphere is $4\pi r^2$.
Your son's procedure seems to be summing the circumferences of the green strips on the sphere, but there must be some accounting for the widths of these strips. This seems to be the problem he is encountering.
Best Answer
Let's first try to imagine the quarter of the solid sphere:
It has a curved surface and $2$ mutually perpendicular plane surfaces
$$\text{T.S.A} = \underbrace{\frac{1}{4} 4\pi r^2}_{\text{curved surface area}} + \overbrace{\frac{1}{2}\pi r^2 + \frac{1}{2}\pi r^2}^{\text{Plane surface surface}} = 2\pi r^2$$