To see why the "diameter" formula gets the wrong answer,
first let's consider why your first formula,
$$A=\int_{0}^{r_0}2r\pi\; dr=r_0^2\pi,$$
gets the correct answer.
Note that in order to remove some confusion that might be caused by
using the same symbol, $r$, for two different things (the upper bound of
the interval of integration, and the variable of integration,
which falls within that interval),
I have written $r_0$ for the radius of the circle whose area you want to find, and left $r$ as the radius of one of the concentric circles within
the larger circle, so that $0 \leq r \leq r_0$.
Consider the Riemann integral of $2r\pi$. In particular consider what
happens when we subdivide $[0,r_0]$ (the interval of integration)
into $n$ equal subintervals of width $\Delta r = \frac{r_0}{n}$.
If we construct all the circles of radius $\Delta r$, $2\Delta r$,
$3\Delta r$, etc., these circles cut the larger circle into $n$
concentric rings whose total area is the area of the circle.
Now look at one of these rings, let's say the ring between the
circles of radius $k\Delta r$ and $(k+1)\Delta r$.
Its area is a little less than we would get if we had a rectangle
with length $2\pi(k+1)\Delta r$ (the outer circumference of the ring)
and width $\Delta r$, but a little more than we would get if we
had a rectangle with length $2\pi k\Delta r$ (the inner circumference of the ring) and width $\Delta r$.
We can make the difference between these two areas as small as we want
by making $\Delta r$ smaller, so either of these is a good approximation
for the actual area of the ring, which is
$$(\Delta r) \times (2\pi \rho)$$
(width $\times$ length) for some $k\Delta r \leq \rho \leq (k+1)\Delta r$.
Now look at the interval between $r = k\Delta r$ and
$r = (k+1)\Delta r$ in the Riemann sum. The area under the function
$2\pi r$ in that interval is somewhere between the areas of the upper
and lower rectangles on that interval, so it is
$$(\Delta r) \times (2\pi \rho')$$
for some $k\Delta r \leq \rho' \leq (k+1)\Delta r$.
As it happens, the two areas (of the ring and of the region under the function) are exactly the same, but we don't need to know that.
All we need to know is that they are approximately the same,
and as we reduce the size of $\Delta r$ to zero, we reduce the
total difference in area (over all the rings and rectangles) to zero.
Now look what happens if you use the diameter $R$ as your variable
of integration.
Suppose we start with a circle of diameter $R_0$
whose area we want to measure.
We can divide the interval $[0,R_0]$ into $n$ subintervals
of width $\Delta R$,
and draw concentric circles with diameters
$\Delta R$, $2\Delta R$, $3\Delta R$, etc., inside the original circle.
Again, the concentric circles cut the larger circle into rings.
Look at the ring whose inner diameter is $k\Delta R$ and whose
outer diameter is $(k+1)\Delta R$.
The "width" of that ring (the gap between the two circles)
is just $\frac12 \Delta R$; that is, if we look at a horizontal
diameter, half of the increase in the diameter "bumps out"
the circle to the right and half "bumps out" the circle to the left.
So the approximate area of the ring is just
$$\left(\frac12\Delta R\right) \times (2\pi \rho)$$
for some $k\Delta R \leq \rho \leq (k+1)\Delta R$.
But the area under the function in the Riemann integral is
$$(\Delta R) \times (2\pi \rho')$$
for some $k\Delta R \leq \rho' \leq (k+1)\Delta R$.
This is twice as large as the area of the corresponding ring,
so the integral gives us twice the area of the circle.
In general, the key idea in using a single-variable integral like
$A=\int_{0}^{r_0}2r\pi\; dr$
to measure area is to set it up so that the area we want to measure
is subdivided into thin "strips" (not necessarily rectangular)
such that each strip corresponds to some value of $r$,
the "length" of the strip is the function we are integrating
(in this case, $2r\pi$), and the "width" of the strip is uniformly
equal to the amount we have to increase $r$ to get to the next strip.
When our "strips" are concentric rings,
the radius increases by the width of the ring as we go from one ring
to the next, so we can use the radius as the variable of integration
without any adjustment.
But the diameter increases by twice the width of the ring, so to turn it
into a variable that increase by the same amount as the width of the ring
we have to divide by two.
Best Answer
When you do the substitution $x = r \sin \theta$, the limits of integration of your integral must change as well. That is,
when $x = 0 $, then $\sin \theta = 0 \implies \theta = 0$
when $x = r$, then $\sin \theta =1 \implies \theta = \frac{ \pi }{2} $