Change of Variables - Polar to Cartesian A change of variables from Cartesian to Polar gives
$$\iint_{D}\,dx\,dy=\iint_{D^*}\,r\,dr\,d\theta.$$
I'm trying to change from Polar to Cartesian.   
Since 
$$r=\frac{x}{\cos\theta};\,\, r=\frac{y}{\sin\theta};\,\,\theta=\arccos(\frac{x}{r});\,\,\theta=\arcsin(\frac{y}{r}),$$
we have, 
$$\frac{\partial r}{\partial x}=\frac{1}{\cos\theta}=\frac{r}{x};\,\, \frac{\partial r}{\partial y}=\frac{1}{\sin\theta}=\frac{r}{y};\,\,\frac{\partial \theta}{\partial x}=-\frac{1}{y};\,\,\frac{\partial \theta}{\partial y}=\frac{1}{x}.$$   
So, the determinant of the Jacobian = $\frac{\partial r}{\partial x}\frac{\partial \theta}{\partial y}-\frac{\partial r}{\partial y}\frac{\partial \theta}{\partial x}=\frac{r}{x^2}+\frac{r}{y^2}.$ Then
$$\iint_{D^*}\,r\,dr\,d\theta=\iint_{D}\,r\,\left(\frac{r}{x^2}+\frac{r}{y^2}\right)\,dx\,dy.$$
I expected to get $\iint_{D}\,dx\,dy$ but I'm not. Did I mess up on the calculations or am I missing some steps?
 A: Your definitions of $r$ at the beginning are getting you into trouble.  Bad things happen at $\sin \theta = 0$ or $\cos \theta = 0$.
Try $r = \sqrt{x^2+y^2}$.  Then
$$\frac{\partial{r}}{\partial{x}} = \frac{x}{r}; \frac{\partial{r}}{\partial{y}} = \frac{y}{r}.$$
Then, using Fantini's $\theta = \tan ^{-1}(\frac{y}{x})$, we get
$$\frac{\partial \theta}{\partial x} = \frac{-\frac{y}{x^2}}{1+\frac{y^2}{x^2}} = -\frac{y}{r^2}; \frac{\partial \theta}{\partial y} = \frac{\frac{1}{x}}{1+\frac{y^2}{x^2}} = \frac{x}{r^2}.$$
Then the determinant of your Jacobian $J$ is
$$\frac{\partial r}{\partial x} \frac{\partial \theta}{\partial y} - \frac{\partial r}{\partial y} \frac{\partial \theta}{\partial x} = \frac{x}{r}\left(\frac{x}{r^2}\right) - \frac{y}{r}\left(-\frac{y}{r^2}\right) = \frac{1}{r},$$
which is what you need:
$$dA = J r dr d\theta = \frac{1}{r} r dx dy = dx dy.$$
A: You've forgotten about the chain rule in calculating the Jacobian. For instance:
$$\frac{\partial r}{\partial x} = \frac{1}{\cos\theta} + \frac{x\sin\theta}{\cos^2\!\theta}\frac{\partial \theta}{\partial x} = \frac{1}{\cos\theta}\left[ 1 + \frac{\partial}{\partial x}\left(\arccos\left(\frac{x}{\sqrt{x^2+y^2}}\right)\right)\right].$$
However, I'd highly encourage you to go the route suggested by John in his answer. Otherwise, as you've already seen, you'll end up banging your head against the wall trying to find calculation mistake after calculation mistake.
