Question

Convert the vector P to Cartesian coordinates where P = r ar + cos⁡θ aφ.

(a) \(\frac{1}{\sqrt{x^2+y^2+z^2}} [(\frac{x}{\sqrt{x^2+y^2+z^2}}-\frac{xyz}{\sqrt{x^2+y^2 }})az + (\frac{y}{\sqrt{x^2+y^2+z^2}}+\frac{xyz}{\sqrt{x^2+y^2}})ay+ \frac{z}{\sqrt{x^2+y^2+z^2}} ax] \)

(b) \(\frac{1}{\sqrt{x^2+y^2+z^2}} [(\frac{x}{\sqrt{x^2+y^2+z^2}}-\frac{yz}{\sqrt{x^2+y^2 }})ax + (\frac{y}{\sqrt{x^2+y^2+z^2}}+\frac{xz}{\sqrt{x^2+y^2}})ay+ \frac{z}{\sqrt{x^2+y^2+z^2}} az] \)

(c) \(\frac{1}{\sqrt{x^2+y^2+z^2}} [(\frac{x}{\sqrt{x^2+y^2+z^2}}-\frac{y}{\sqrt{x^2+y^2 }})ax + (\frac{y}{\sqrt{x^2+y^2+z^2}}+\frac{z}{\sqrt{x^2+y^2}})ay+ \frac{z}{\sqrt{x^2+y^2+z^2}} az] \)

(d) \(\frac{1}{\sqrt{x^2+y^2+z^2}} [(\frac{x}{\sqrt{x^2+y^2+z^2}}-\frac{y}{\sqrt{x^2+y^2 }})ax + (\frac{y}{\sqrt{x^2+y^2+z^2}}+\frac{x}{\sqrt{x^2+y^2}})ay+ \frac{z}{\sqrt{x^2+y^2+z^2}} az] \)

This question was addressed to me by my college professor while I was bunking the class.

Question is taken from Conversion from Cartesian, Cylindrical and Spherical Coordinates topic in portion Vector Differential Calculus of Engineering Mathematics

Answer 1

Correct answer is (b) \(\frac{1}{\sqrt{x^2+y^2+z^2}} [(\frac{x}{\sqrt{x^2+y^2+z^2}}-\frac{yz}{\sqrt{x^2+y^2 }})ax + (\frac{y}{\sqrt{x^2+y^2+z^2}}+\frac{xz}{\sqrt{x^2+y^2}})ay+ \frac{z}{\sqrt{x^2+y^2+z^2}} az] \)

The best explanation: The formula to convert any vector from Spherical coordinates to Cartesian coordinates is given by \(\begin{bmatrix}

Px\\

Py\\

Pz\\

\end{bmatrix} \)

\(= \begin{bmatrix}

sin\theta cos⁡φ & cos\theta cos⁡φ & -sinφ\\

sin\theta sin⁡φ & cos⁡\theta sinφ & cosφ\\

cos\theta & -sin\theta & 0\\

\end{bmatrix} \)

\(= \begin{bmatrix}

r\\

0\\

cos\theta\\

\end{bmatrix} \)

after substituting the values of the vector. Now, solving the matrix we get the answer

\(\frac{1}{\sqrt{x^2+y^2+z^2}} [(\frac{x}{\sqrt{x^2+y^2+z^2}}-\frac{yz}{\sqrt{x^2+y^2 }})ax + (\frac{y}{\sqrt{x^2+y^2+z^2}}+\frac{xz}{\sqrt{x^2+y^2}})ay+ \frac{z}{\sqrt{x^2+y^2+z^2}} az] \).