In the spherical coordinate system
$x=r \sin \theta \cos \phi$
$y=r \sin \theta \sin \phi$
$z=r \cos \theta$
$\theta$ lies in $[0,\pi]$, while $\phi$ is in $[-\pi,\pi)$. Why do we not need for the sign of $\theta$?
Why is it sufficient to set the polar angle between 0 and 180 degrees in a spherical coordinate system
coordinate systemsspherical coordinates
Related Solutions
I can partially answer this. I believe your first matrix is not the correct general transformation matrix for cartesian to spherical coordinates because you are missing factors of $\rho$ (the radial coordinate), as well as some other incorrect pieces. So it is not clear what you are trying to show.
If you are trying to derive the general transformation matrix from spherical to cartesian, it is: $$\begin{bmatrix} A_x\\ A_y\\ A_z \end{bmatrix} = \begin{bmatrix} \sin \theta \cos \phi & \rho \cos \theta \cos \phi & -\rho \sin \theta \sin\phi\\ \sin \theta \sin \phi & \rho \cos \theta \sin \phi & \rho \sin\theta \cos\phi\\ \cos\theta & -\rho \sin\theta & 0 \end{bmatrix} \begin{bmatrix} A_r\\ A_\theta\\ A_\phi \end{bmatrix} $$ This matrix is formed from the derivative matrix in the following way: $ \sum_{(i,j)} \frac {\partial x^i} {\partial u^j} $ where $$x^1 = \rho\sin \theta \cos \phi, x^2 = \rho\sin \theta \sin \phi, x^3=\rho\cos\theta$$ and $u^1 = \rho, u^2 = \theta, u^3 = \phi $ (Note: the superscripts are indices, not exponents). If you write out all the derivatives, you get this matrix:
$$ \begin{bmatrix} A_x\\ A_y\\ A_z \end{bmatrix} = \begin{bmatrix} \frac {\partial (\rho \sin \theta \cos \phi)} {\partial \rho} & \frac {\partial (\rho \sin \theta \cos \phi)} {\partial \theta} & \frac {\partial (\rho \sin \theta \cos \phi)} {\partial \phi}\\ \frac {\partial (\rho \sin \theta \sin \phi)} {\partial \rho} & \frac {\partial (\rho \sin \theta \sin \phi)} {\partial \theta} & \frac {\partial (\rho \sin \theta \sin \phi)} {\partial \phi}\\ \frac {\partial (\rho \cos \theta)} {\partial \rho} & \frac {\partial (\rho \cos \theta)} {\partial \theta} & \frac {\partial (\rho \cos \theta)} {\partial \phi} \end{bmatrix} \begin{bmatrix} A_r\\ A_\theta\\ A_\phi \end{bmatrix} $$ ...which evaluates to the correct general transformation matrix I listed above.
$x^2+y^2+z^2 \leq 10z$ is the same as $x^2+y^2+(z-5)^2\leq 25$ so we are dealing with a solid ball of radius 5 centered at $(0,0,5)$.
Naively we can translate this inequality to $\rho^2 \leq 10\rho \cos(\phi)$ so that $\rho \leq 10\cos(\phi)$.
I have drawn a ray emanating from the origin out to the sphere [whose equation is $\rho=10\cos(\phi)$]. The angle $\phi$ should sweep from the $z$-axis down to $\phi=\pi/2$. Notice that at this point $\rho=10\cos(\pi/2)=0$ (so we should stop).
Notice that if you had allowed $\phi$ to continue to go past $\pi/2$, cosine and thus $\rho$ would become negative (which indicates something isn't quite right).
Therefore, the bounds in spherical coordinates are:
$0 \leq \rho \leq 10\cos(\phi)$, $0 \leq \phi \leq \pi/2$, and $0 \leq \theta \leq 2\pi$.
Best Answer
We have
$$x=r \sin (-\theta) \cos \phi=r (-\sin \theta) \cos \phi=r \sin \theta (-\cos \phi)=r \sin \theta \cos(\phi+\pi)$$ $$y=r \sin (-\theta) \sin \phi=r (-\sin \theta) \sin \phi=r \sin \theta (-\sin \phi)=r \sin \theta \sin (\phi+\pi)$$ $$z=r \cos (-\theta)=r \cos \theta$$
So $(r,-\theta,\phi)$ represents the same point as $(r,\theta,(\phi+\pi )\pmod {2\pi})$