Coordinate transformation from spherical to rectangular

First use the definitions of unit vectors
$$
\hat e_r = \frac{\frac{\partial \vec r}{\partial r}}{ \left | \frac{\partial \vec r}{\partial r} \right |}, \hat e_\theta = \frac{\frac{\partial \vec r}{\partial \theta}}{ \left | \frac{\partial \vec r}{\partial \theta} \right |}, \;\; \hat e_\phi = \frac{\frac{\partial \vec r}{\partial \phi}}{ \left | \frac{\partial \vec r}{\partial \phi} \right |}
$$
to arrive at a system of linear equations relating the unit vectors in spherical and in Cartesian coordinates. Then define an arbitrary vector ##\mathbf{A}## written in spherical coordinate, ##\mathbf{A} = A_r \hat{e}_r + A_\theta \hat{e}_\theta + A_\phi \hat{e}_\phi##. Substitute the spherical coordinate unit vectors in terms of the Cartesian unit vectors using the relations you just derived.

 

##\mathbf{r}## is a position vector, thus ##\mathbf{r} = x\hat{e}_x + y\hat{e}_y + z\hat{e}_z##. Then express ##x##, ##y##, and ##z## in terms of ##r##, ##\theta##, and ##\phi##.

 

Which relations can you not reach? If you mean the definitions of unit vectors in post #2, as I hinted before, ##\mathbf{r}## is a position vector. Taking derivative like ##
\frac{\partial \mathbf{r}}{\partial r}## is done by simply substituting the component-expanded expression of ##\mathbf{r}## as given in post #4 into the derivative. Thus
$$
\frac{\partial \mathbf{r}}{\partial r} = \frac{\partial}{\partial r} \left( x\hat{e}_x + y\hat{e}_y + z\hat{e}_z \right)
$$
First try to calculate ##\frac{\partial \mathbf{r}}{\partial r}## and let's see what you will get.

 

Check again whether there should be sine or cosine.

After incorporating the correction I suggested above, you should get a simple answer for the magnitude.

Yes.
Then group the resulting terms according to the Cartesian unit vectors they contain.

 

Please check again the terms marked as red.

 

If you follow the derivation I suggested above, you will find something different. Could you possibly confuse the matrix transformation for the unit vectors with that of the components in that book? The two transformation matrices indeed look alike except for the third column.

 

Sorry, my bad. I was switching to a different definition of unit vectors without realizing that it is different than the one I was previously using for writing the equations in post #2. So, you are correct.

If you fix two coordinates among ##(r,\theta,\phi)## and let the last one varies, you will get the so-called "space curve" or the more technical term "3D curve". For example fix ##r## and ##\theta##, then you will get a space curve ##\mathbf{r}(\phi)## which forms a circle of radius ##r\sin \theta## "floating" horizontally above the ##xy## plane at a height of ##r\cos\theta##. Now, define a vector ##\mathbf{u} = \frac{d \mathbf{r}}{d s}##, where ##ds = r\sin\theta d\phi## is an element of arc length along the circle. Vector ##\mathbf{u}## is tangent to the circle, and its explicit expression is
$$
\mathbf{u} = \frac{d \mathbf{r}}{d s} = \frac{1}{r\sin\theta} \frac{d \mathbf{r}}{d\phi }
$$
But ##r\sin\theta## is the length of the vector ##\frac{d \mathbf{r}}{d\phi }## itself. Therefore one can write
$$
\mathbf{u} = \frac{\frac{d\mathbf{r}}{d\phi}}{ \left | \frac{d\mathbf{r}}{d\phi} \right |}
$$
Because ##\mathbf{u}## is tangent to the circle at any given point and it has unit length, this vector must be the unit vector in the direction of increasing ##\phi##.

Such a task is of course doable but tedious. I suggest that you read "Mathematical Methods for Physics and Engineering" by K. F. Riley, Hobson, and Bence subchapter 10.10.