Why can't I use the disk method to compute surface area? [duplicate]

The main problem is that you can't approximate the length of a non-horizontal segment with horizontal segments; not even straight lines.

Consider a very simple case of the diagonal of a square of side length $1$; this diagonal has length $\sqrt{2}$.

You are trying to approximate the diagonal by dividing the interval $[0,1]$ into $n$ equal parts, each of length $\frac{1}{n}$, and taking the horizontal line on top of a little "bar" as an approximation for the diagonal, then adding them up.

But those "horizontal bars" on top will always add up to $1$, never more than that, even as you let $n\to\infty$. So you are not really approximating the diagonal, so your Riemann sums don't converge to the length of the diagonal.

The same thing happens with your computation above: you aren't actually approximating the surface area of the sphere, you are approximating something else, because those "disks" don't have surface area converging to the surface of the sphere, just like the little bars in the square example don't converge to the diagonal of the square:

                /|
               /||
              /=||
             /| ||
            /=| ||
           /| | ||
          /=| | ||
         /=_|_|_||

(The = are the lengths you are adding up)

There was a question that kept popping up here with a "proof" that $\pi=4$ which used this same error, of thinking that adding up lengths of straight segments would yield that length of the limit was the limit of the lengths. See here.

What you want to look at is the error in the approximation. In order for the integral to give you the correct value, the errors have to go to $0$ as $dx$ goes to $0$. In my example above with the diagonal, the error between approximating with the horizontal segment and the true length of the diagonal segment is $$\mathrm{Error} = \mathrm{Actual\ value}-\mathrm{Estimated\ value} = \sqrt{\frac{1}{n^2}+\frac{1}{n^2}} - \frac{1}{n} = \frac{\sqrt{2}-1}{n}.$$ The total error over the entire interval is therefore $\frac{n(\sqrt{2}-1)}{n}=\sqrt{2}-1$. But as $n\to\infty$ (that is, as $dx\to 0$), this error does not decrease: it remains at $\sqrt{2}-1$ throughout. So in the end, when I perform my final computation with the integral, my answer is going to be $\sqrt{2}-1$ away from the true answer. Indeed, using the above computations I would be led to compute $$\int_0^1\,dx = 1,$$ whereas the right answer is $\sqrt{2}$.

By contrast, if I were computing the area this issue would not occur: if you compute the area using the rectangles, then the error per interval would be $$\begin{align*} \text{Error per interval} &= \text{Actual area} - \text{Computed area}\\ &= \left(\frac{x}{n} + \frac{1}{2n^2}\right) - \frac{x}{n} = \frac{1}{2n^2}. \end{align*}$$ So the total error in the computation would be $\frac{1}{2n}$; hence, the error goes to $0$ as $n\to\infty$, so "at the limit" we should get the right answer.

The same thing happens when you try to use disks with the sphere: if you are looking at the volume, you are fine because the error does goes to $0$; but if you are looking at surface area, the error in the estimation does not got to $0$, so the Riemann integral that you compute at the end is not computing the actual surface area of the sphere, but rather something else.

In order to actually approximate the length of the limiting curve you need to consider the secant line, not the line segment parallel to the $x$-axis.