A) \[r,\theta \]
B) \[r,\phi \]
C) \[\theta ,\phi \]
D) \[r\]
Correct Answer: C
Solution :
| [c]\[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{r}^{2}}{{\sin }^{2}}\theta {{\cos }^{2}}\phi +{{r}^{2}}{{\sin }^{2}}\theta \cdot {{\sin }^{2}}\phi +{{r}^{2}}{{\cos }^{2}}\theta \] |
| \[={{r}^{2}}{{\sin }^{2}}\theta ({{\cos }^{2}}\phi +{{\sin }^{2}}\phi )+{{r}^{2}}{{\cos }^{2}}\theta \] |
| \[={{r}^{2}}({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )={{r}^{2}},\] which is independent of \[\theta ,\phi \]. |
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