In the text and problems so far, we have found the e vectors for Question: Using the results of Problem 1, express the vector in Problem 4 in spherical coordinates.

Spherical coordinates are given below.

$$\begin{gathered} x=r\cos \varphi \sin \theta \\ y=r\sin \varphi \sin 0 \\ z=r\cos \theta \end{gathered}$$

The coordinate system primarily utilized in three-dimensional systems is the cylindrical coordinates of the system. The cylindrical coordinate system is used to find the surface area in three-dimensional space.

Assume the following quantity.

$$\begin{gathered} C=\left(i,j,k\right) \\ S=\left(e_r,e_0,e_{\varphi }\right) \\ {e}_r=i\cos \varphi \sin \theta +j\sin \varphi \sin \theta +k\cos \theta \\ {e}_{\theta }=i\cos \varphi \sin \theta +j\sin \varphi \sin \theta k\cos \theta \\ {e}_{\varphi }=-i\cos \varphi +j\sin \varphi \end{gathered}$$

T is the transformation operator defined by $T{c}_j=s_j$.

$$\begin{gathered} T\left(C\right)=\left(\begin{array}{ccc}\cos \varphi \sin \theta & \cos \varphi \sin \theta & -\sin \varphi \\ \cos \varphi \sin \theta & \cos \varphi \sin \theta & \cos \varphi \\ \cos \theta & -\sin \theta & 0\end{array}\right) \\ T{\left(C\right)}^{-1}=\left(\begin{array}{ccc}\cos \varphi \sin \theta & \cos \varphi \sin \theta & -\sin \varphi \\ \cos \varphi \sin \theta & \cos \varphi \sin \theta & \cos \varphi \\ -\sin \varphi & \cos \varphi & 0\end{array}\right) \end{gathered}$$.

Formula states that $V\left(S\right)=T{\left(C\right)}^{-1}V\left(C\right)$

$$\begin{gathered} T\left(C\right)=\left(\begin{array}{ccc}\cos \varphi \sin \theta & \sin \varphi \sin \theta & \cos \theta \\ \cos \varphi \cos \theta & \cos \varphi \sin \theta & \sin \theta \\ -\sin \varphi & \cos \varphi & 0\end{array}\right) \\ T{\left(C\right)}^{-1}=\left(\begin{array}{c}\cos \theta \\ -\sin \theta \\ -r\sin \theta \end{array}\right) \\ V\left(S\right)=r_r\cos \theta -e_{\theta }-e_{\varphi }r\sin \theta \end{gathered}$$

Hence, the vector is $V\left(S\right)=e_r\cos \theta -e_{\theta }\sin \theta -e_{\varphi }r\sin \theta$.