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I should calculate this mixed product, where $\overline r=\overrightarrow{OP}$,

$$\overline r\cdot (\overline r_{\theta}\times\overline r_{\varphi}),$$ to get with the determinant,

$$\overline r\cdot (\overline r_{\theta}\times\overline r_{\varphi})=\begin{vmatrix}x & y & z \\ x_{\theta} & y_{\theta} & z_{\theta} \\ x_{\varphi} & y_{\varphi} & z_{\varphi} \end{vmatrix}=\color{red}{r^3\sin \theta}.\tag 1$$

If

$$\begin{cases} x=r(\theta,\varphi)\sin \theta\cos \varphi\\ y=r(\theta,\varphi)\sin \theta\sin \varphi\\ z=r(\theta,\varphi)\cos \theta \end{cases}$$

why are they true these systems?

$$\begin{cases} x_\theta(\theta,\varphi)=r_\theta\sin \theta\cos\varphi+r\cos \theta\cos \varphi\\ x_\varphi(\theta,\varphi)=r_\varphi\sin \theta\cos\varphi-r\sin\theta\sin \varphi \end{cases} \tag 2$$

$$\begin{cases} y_\theta(\theta,\varphi)=r_\theta\sin \theta\sin\varphi+r\cos \theta\sin \varphi\\ y_\varphi(\theta,\varphi)=r_\varphi\sin \theta\sin\varphi+r\sin\theta\cos \varphi \end{cases} \tag 3$$

$$\begin{cases} z_\theta(\theta,\varphi)=r_\theta\cos\theta-r\sin \theta\\ z_\varphi(\theta,\varphi)=r_\varphi\cos\theta \end{cases} \tag 4$$

With $(2), (3), (4)$ I have not get $r^3\sin \theta$ with the determinant $(1)$. Please, can I have your attention and help?

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A strategy suggestion to calculate by hand this determinant:

First, I'll use condensed notations: $r$ instead of $r(\theta,\varphi)$. Next, observe each column is the sum of two column vectors $$\begin{bmatrix}x\\x_\theta\\x_\varphi\end{bmatrix}=\underbrace{\begin{bmatrix}r\sin\theta\cos\varphi\\r_\theta\sin\theta\cos\varphi\\r_\varphi\sin\theta\cos\varphi\end{bmatrix}}_{\textstyle C_x}+\underbrace{\begin{bmatrix}0\\r\cos\theta\cos\varphi\\-r\sin\theta\sin\varphi\end{bmatrix}}_{\textstyle D_x}=\sin\theta\cos\varphi\begin{bmatrix}r\\r_\theta\\r_\varphi\end{bmatrix}+r\begin{bmatrix}0\\\cos\theta\cos\varphi\\-\sin\theta\sin\varphi\end{bmatrix}$$ and similarly for the other columns.

Now the determinant, with these notations becomes, by the multilinearity properties of determinants \begin{align}\begin{vmatrix}x & y & z \\ x_{\theta} & y_{\theta} & z_{\theta} \\ x_{\varphi} & y_{\varphi} & z_{\varphi} \end{vmatrix}&=\begin{vmatrix}C_x{+}D_x & C_y {+}D_y& C_z{+}D_z\\ \end{vmatrix}\\&=\bigl|C_x\enspace C_y\enspace C_z \bigr|+\bigl|D_x\enspace C_y\enspace C_z \bigr|+\bigl|C_x\enspace C_y\enspace D_z \bigr|+\bigl| D_x\enspace C_y\enspace D_z \bigr|\\[1ex] &\quad+\bigl| C_x\enspace D_y\enspace C_z \bigr|+\bigl|D_x\enspace D_y\enspace C_z \bigr|+\bigl| C_x\enspace D_y\enspace D_z \bigr|+\bigl| D_x\enspace D_y\enspace D_z \bigr| \end{align} Each of $C_x, C_y, C_z$ is collinear to the column vector $\;\begin{bmatrix}r\\r_\theta\\r_\varphi\end{bmatrix}$, so any of the determinants in the sum which comprises two of these column vectors is $0$, further $\bigl| D_x\enspace D_y\enspace D_z \bigr|$ has a row of $0$s. Therefore there remains $$\begin{vmatrix}x & y & z \\ x_{\theta} & y_{\theta} & z_{\theta} \\ x_{\varphi} & y_{\varphi} & z_{\varphi} \end{vmatrix} =\bigl| C_x\enspace D_y\enspace D_z \bigr|+\bigl|D_x\enspace C_y\enspace D_z \bigr|+\bigl| D_x\enspace D_y\enspace C_z \bigr|, $$ which can easily be computed by hand.

Hope this will help!

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  • $\begingroup$ Thank you very much. Surely you and other users you have helped me. Last help into my question. Please why $(1)\implies (2), (3)$ and $(4)$? $\endgroup$ – Sebastiano Dec 14 '19 at 9:58
  • $\begingroup$ @Sebastiano Shouldn't it be the other way around? Note that (2),(3), and (4) are consequences from the definition of $x,y,z$ by taking partial derivatives. You then use the relations (2),(3) and (4) for computing the determinant given in (1). The result is $r^3\sin\theta$ as I sketched in my answer. $\endgroup$ – Pythagoras Dec 14 '19 at 20:41
  • $\begingroup$ @Pythagoras I have taken the partial derivatives. Infact being $x=f\cdot g \cdot h$ using the partial derivates (supposing only for $\theta$) I should have 3 terms and not 2 terms. $\endgroup$ – Sebastiano Dec 14 '19 at 20:47
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    $\begingroup$ Using your example, the last factor $h$ does not depend on $\theta$, so the derivative is zero. They are treating $\theta,\phi$ as independent variables. So a function of $\phi$ is like a constant with respect to the variable $\theta$. $\endgroup$ – Pythagoras Dec 14 '19 at 20:54
  • $\begingroup$ @Pythagoras I was and actually in tilt for a lot of work. I have not considered that: $x_\theta=f_\theta \cdot g \cdot h+f \cdot g_\theta \cdot h + f \cdot g \cdot h_\theta$ where $\phi=h_\theta=0$. You're right and thank you very much again. $\endgroup$ – Sebastiano Dec 15 '19 at 21:50
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Hint: write out the determinant in the matrix form.

Step 1: Multiply the first row by $-r_{\theta}/r$ and add it to the second row. Do similar thing with respect to the third row. You will see that all the terms involving $r_{\theta}$ and $r_{\phi}$ will be canceled.

Step 2: Extract the factors $r,r,r\sin\theta$ outside from the first, second and third row, respectively. This explains the factor $r^3\sin\theta$.

Step 3: Work on the remaining determinant by expanding with respect to the third row. It can be easily checked that the result is $\sin^2\phi+\cos^2\phi=1$, hence the result

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Equations (2), (3) and (4) are true because of the product rule for derivatives.

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