Filling a hemisphere with cylinders as so: https://i.stack.imgur.com/n5VRS.png
Now, if the radius of the sphere is R and radius of the disk(r) at any given height y is:
$$r = \sqrt{R^2 - y^2}$$
Now the sum of all the areas on the side of the cylinders i.e the surface area of a hemisphere must be:
$$\int_0^R 2\pi rdy$$ or $$\int_0^R 2\pi \sqrt{R^2 - y^2}dy$$
this solves to $$\frac{\pi^2 R^2}{2} $$ which for the sphere is $$\pi^2 R^2$$
This is a big problem that beginners run into while calculating areas and such integrals where it is correct to take your given element in hindsight but is actually incorrect (useful for maybe mathematics or even in physics).
So the problem lies in the fact that to find the surface area of an object with curved surface, the element must cover all of the surface and while integrating, nothing from the object's surface is left. Let us take an example of the surface area of a cone (because that is easier to draw in powerpoint)
What I mean by this is : look at this image (consider it to be 3d). According to your method, the surface area of the disk $ = \frac{2\pi Ry\;dy}{H}$ and then you would integrate this to get the surface area of the cone. This is clearly incorrect as you may now see because we are forgetting the fact that the cone has a slanted surface and to really take the surface area at a height y nd spanning a width dy, we must have the disk slanted. What I mean is this. So if we now see the surface area of the disk, it will not be $ = \frac{2\pi Ry\;dy}{H}$. Since the disk spans a very small width, so we can assume the slanted height to be the vertical height of a new disk (radius is same because change in radius is negligible in this height change) (so that we can write the surface area of that disk = surface area of a new disk $= 2\pi (radius)(vertical\;height)$). After calculating we get the correct surface area of the disk $ = \frac{2\pi Ry\;dl}{H} = \frac{2\pi Ry\;dy \sqrt{1 + \frac{R^2}{H^2}}}{H}$. Integrate this and you will get the correct surface area of the cone.
Similiarly with the suface area of the sphere (let radius of sphere is r), the disk you took is not "slanted" and so you would have to do this. After calculating the correct surface area of the disk and integrating the expression $ = 2\pi R\;dl = 2\pi r\sin (\theta)\;\frac{dy}{\sin (\theta)} = 2\pi r dy$ (y goes from -r to +r), we get te correct surface area of the sphere $= 4\pi r^2$ (for hemisphere, y goes from 0 to r so you can just halve the area of the sphere)
