How do you convert \[{x^2} + {y^2} = z\] into spherical and cylindrical form?

Hint: Here in this question we have to convert the given cartesian form of an equation to the spherical and cylindrical form. While converting the cartesian form to cylindrical form, we use \[x = r\cos \theta ,\,y = r\sin \theta ,z = z\] and for the spherical form we use \[x = \rho \sin \phi \cos \theta ,\,y = \rho \cos \phi \sin \theta ,z = \rho \cos \phi \].Then on simplification we obtain the answer.

Complete step by step answer:
Cartesian form: A function or relation is written using the \[(x,y)\] or \[(x,y,z)\] coordinates.
Cylindrical form: A function or relation is written using the \[(r\cos \theta ,r\sin \theta ,z)\] coordinates.
Spherical form: A function or a relation is written using the \[(\rho \sin \phi \cos \theta ,\rho \cos \varphi \sin \theta ,\rho \cos \phi )\] coordinates.
Now we have to convert the cartesian form into spherical and cylindrical form.
When we are converting the cartesian form to cylindrical form. We consider
\[x = r\cos \theta ,\,y = r\sin \theta ,z = z\]--------(1)
Now consider the given question.
\[{x^2} + {y^2} = z\]--------(2)

On substituting the equation (1) to the equation (2) we have
\[ \Rightarrow {(r\cos \theta )^2} + {(r\sin \theta )^2} = z\]
\[ \Rightarrow {r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta = z\]
Take \[{r^2}\] as a common in the LHS we have
\[ \Rightarrow {r^2}({\cos ^2}\theta + {\sin ^2}\theta ) = z\]
By the trigonometric identity we have \[{\cos ^2}\theta + {\sin ^2}\theta = 1\], so we have
\[ \Rightarrow {r^2} = z\]
Therefore the above function represents the cylindrical form.
Now we convert the cartesian form into spherical form.
When we are converting the cartesian form to spherical form. We consider
\[x = \rho \sin \phi \cos \theta ,\,y = \rho \cos \phi \sin \theta ,z = \rho \cos \phi \]--------(3)
Now consider the given question.
\[{x^2} + {y^2} = z\]--------(2)
To the equation (2) add \[{z^2}\] on both sides
\[ \Rightarrow {x^2} + {y^2} + {z^2} = z + {z^2}\] ---- (4)
As we know that the \[{x^2} + {y^2} + {z^2} = {\rho ^2}\]--- (5)

On substituting the equation (5) and equation (3) to the equation (2) we get
\[ \Rightarrow {\rho ^2} = \rho \cos \phi + {\rho ^2}{\cos ^2}\phi \]
Take \[{\rho ^2}{\cos ^2}\phi \] to LHS
\[ \Rightarrow {\rho ^2} - {\rho ^2}{\cos ^2}\phi = \rho \cos \phi \]
Take \[\rho \] as a common in the LHS
\[ \Rightarrow {\rho ^2}(1 - {\cos ^2}\phi ) = \rho \cos \phi \]
On cancelling the \[\rho \] both in LHS and RHS we get
\[ \Rightarrow \rho (1 - {\cos ^2}\phi ) = \cos \phi \]
By the trigonometric identity we have \[(1 - {\cos ^2}\phi ) = {\sin ^2}\phi \], on substituting we have
\[ \Rightarrow \rho {\sin ^2}\phi = \cos \phi \]
On dividing both sides by \[{\sin ^2}\phi \]
\[ \therefore \rho = \cot \phi \csc \phi \]
Hence this is the spherical form.

Note: We can convert the equation from cartesian form to spherical form, cylindrical form and vice versa, the only thing we have to remember is the values. Since both cylindrical and spherical forms involve the trigonometric ratios we should know about the trigonometric identities and it is implemented whenever it is used.