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I'm reading the following in a calculus text book:

$$\frac{\partial(x,y)}{\partial(r,\theta)} = \left|\begin{matrix}\cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{matrix}\right| = r$$ $$\frac{\partial(y,z)}{\partial(r,\theta)} = \left|\begin{matrix}\sin\theta & r\cos\theta \\ 1 & 0 \end{matrix}\right| = -r\cos\theta$$ $$\frac{\partial(x,z)}{\partial(r,\theta)} = \left|\begin{matrix}\cos\theta & -r\sin\theta \\ 1 & 0 \end{matrix}\right| = r\sin\theta$$

These operations appear to be computing the Jacobian (the matrix of first partial derivatives), which I expect.

But then they appear to take the determinant of the Jacobian. I didn't expect that the notation $\frac{\partial(x,y)}{\partial(r,\theta)}$ implied the "determinant of the Jacobian". I thought that notation only represented the Jacobian itself.

I'm further confused by the fact that the book I'm reading doesn't actually use the term "determinant" in the text anywhere in this chapter (on surface area integrals).

Am I simply wrong in my understanding? Or perhaps I'm missing some fundamental detail here?

In my head this is true:

$$\frac{\partial(x,y)}{\partial(r,\theta)} = \begin{bmatrix}\frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{bmatrix} \ne \det \left( \begin{bmatrix}\frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{bmatrix}\right)$$

Also, can you confirm that the single bars $| \cdot |$ represent the determinant of the matrix contained within them?

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    $\begingroup$ generally in the literature the jacobian is the determinant of the jacobian matrix. $\endgroup$ – Masacroso Feb 19 '18 at 3:50
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Conventions vary. Some people use $\partial(x,y)/\partial(r,\theta)$ for the matrix, others (like in this case) for the determinant. You just have to check how the author of the text you're reading has defined the notation (or deduce it from the context).

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