How do I solve the Jacobian of the transformation $y^2 = 4z\cos(k)$ & $x= 4z\sin(k)$

Question

So with this question I'm a bit confused on the fact that I don't know-how to start with y. Since it's given as y^2, do I need to take only the positive root and then take it as

$y = 2\sqrt{z\cos(k)}$ or $ y= -2\sqrt{z\cos(k)}$ ?

Answer 1

Yes, since you have a fold, you will solve for these two (nearly identical) Jacobian matrices, then take their absolute determiants. ... which, clearly are equal.

$$\begin{align}J_1&=\begin{bmatrix}\dfrac{\partial 4z \sin k}{\partial k}&\dfrac{\partial 4z\sin k}{\partial z}\\\dfrac{\partial \sqrt{4z\cos k}}{\partial k}&\dfrac{\partial \sqrt{4z\cos k}}{\partial z}\end{bmatrix}\\[2ex]J_2&=\begin{bmatrix}\dfrac{\partial 4z \sin k}{\partial k}&\dfrac{\partial 4z\sin k}{\partial z}\\-\dfrac{\partial \sqrt{4z\cos k}}{\partial k}&-\dfrac{\partial \sqrt{4z\cos k}}{\partial z}\end{bmatrix}\\[4ex]f_{\small Z,K}(z,k) &= \lVert J_1\rVert f_{\small X,Y}(4z\sin k, \sqrt{4z\cos k})+\lVert J_2\rVert f_{\small X,Y}(4z\sin k, -\sqrt{4z\cos k})\\[1ex]&=\lVert J_1\rVert \left(f_{\small X,Y}(4z\sin k, \sqrt{4z\cos k})+ f_{\small X,Y}(4z\sin k, -\sqrt{4z\cos k})\right)\end{align}$$

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Of course, if the untransformed function only supports positive $y$ values, then ...