Prove or disprove this calculus limit result by geometric approach my question is: Could we prove the this conversion of variable work by my formula on the bottom? 

$$\iint_R f(r,\theta) \ dxdy = \int_a^b \int_0^{r(\theta)} f(r,\theta) r (dr)\ d\theta$$

as $d r$ and $d \theta$ approach $0$. Prove or disprove that:

$$((r+\Delta r) \cos(a +\Delta \theta) -r \cos a) \cdot ((r+\Delta r) \sin(a + \Delta \theta) -r \sin a) / (r \;\Delta \theta \; \Delta r)=1 .$$ where the variable represent as in this graph : 


as $\Delta r $ and $\Delta \theta$ approach $0$
This question is inspired from $dx\;dy=r \;dr \;d \theta$.

The diagram that you are giving is not what is going on at all. $\mathrm{d}x\;\mathrm{d}y$ is an element of area intended to represent the plane broken up into small rectangles.:

$\mathrm{d}r\;\mathrm{d}\theta$ is an element of area in a space whose small squares get mapped to small annular wedges by $x=r\cos(\theta)$, $y=r\sin(\theta)$:

The Jacobian is the matrix that locally maps between two coordinate systems.
$$
\frac{\partial(x,y)}{\partial(u,v)}=\begin{bmatrix}\frac{\partial x}{\partial u}&\frac{\partial y}{\partial u}\\\frac{\partial x}{\partial v}&\frac{\partial y}{\partial v}\end{bmatrix}\tag{1}
$$
From polar to rectangular coordinates, the Jacobian is
$$
\frac{\partial(x,y)}{\partial(r,\theta)}=\begin{bmatrix}\cos(\theta)&\sin(\theta)\\-r\sin(\theta)&r\cos(\theta)\end{bmatrix}\tag{2}
$$
Note that $\begin{vmatrix}\cos(\theta)&\sin(\theta)\\-r\sin(\theta)&r\cos(\theta)\end{vmatrix}=r$.
A small area in $\mathrm{d}r\;\mathrm{d}\theta$, the green square, is mapped by the polar coordinate map to the blue annular wedge, which has approximately the same area as the red rectangle. The Jacobian matrix maps the green square to the red rectangle. The ratio of the area of the red rectangle to the green square is the determinant of the Jacobian (this is just linear algebra).

Therefore, since a small square in $\mathrm{d}r\;\mathrm{d}\theta$ is mapped by the coordinate transform so that its area is $\left|\frac{\partial(x,y)}{\partial(r,\theta)}\right|\;\mathrm{d}r\;\mathrm{d}\theta$
$$
\begin{align}
\iint f(x,y)\;\mathrm{d}x\;\mathrm{d}y
&=\iint f(r,\theta)\left|\frac{\partial(x,y)}{\partial(r,\theta)}\right|\;\mathrm{d}r\;\mathrm{d}\theta\\
&=\iint f(r,\theta)\;r\;\mathrm{d}r\;\mathrm{d}\theta\tag{3}
\end{align}
$$
