Determinant of Jacobian of the transformation from cartesian to polar

Question

I'm reading a textbook on numerical methods and it claims with no proof that the determinant of the matrix given by

\begin{pmatrix}\frac{\partial r}{\partial x}\;\;\frac{\partial r}{\partial y}\\ \frac{\partial \theta}{\partial x}\;\;\frac{\partial \theta}{\partial y}\end{pmatrix}

is 2, (here $r=x^2+y^2$). I tried to verify it using the formulas for the derivative of arcsine and arccosine and I get the determinant as 4. Am I wrong or is the textbook wrong.

Answer 1

Recall from your trig calculus identities: $~\frac{\mathrm d \arctan u}{\mathrm d u} = (1+u^2)^{-1}$

Assuming $r=\sqrt{x^2+y^2}$ and $\theta=\arctan(y/x)$

$$ \begin{bmatrix} \frac{\partial r}{\partial x} & \frac{\partial r}{\partial y}\\ \frac{\partial\theta}{\partial x}& \frac{\partial\theta}{\partial y} \end{bmatrix} = \begin{bmatrix} x\,(x^2+y^2)^{-1/2} & y\,(x^2+y^2)^{-1/2}\\[1ex] -y\,(x^2+y^2)^{-1}& x\,(x^2+y^2)^{-1} \end{bmatrix} $$

So the determinant is $\dfrac{1}{\sqrt{x^2+y^2}}$

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After the edit:

here $r=x^2+y^2$

$$\begin{bmatrix}\frac{\partial r}{\partial x} & \frac{\partial r}{\partial y}\\ \frac{\partial\theta}{\partial x}& \frac{\partial\theta}{\partial y}\end{bmatrix}=\begin{bmatrix}2x & 2y\\[1ex] -y\,(x^2+y^2)^{-1}& x\,(x^2+y^2)^{-1}\end{bmatrix}$$

So the determinant is $2$