If x=`rsinthetasinphi, y=rsinthetacosphi, z=rcostheta`, then `x^(2)+y^(2)+z^(2)=?`

Options

A. 1
B. r
C. `r^(2)`
D. None of these

Answer

To solve the problem, we need to find the value of \( x^2 + y^2 + z^2 \) given the definitions of \( x \), \( y \), and \( z \):

1. **Given Definitions**:
\[ x = r \sin \theta \sin \phi \]
\[ y = r \sin \theta \cos \phi \]
\[ z = r \cos \theta \]

2. **Calculate \( x^2 \)**:
\[ x^2 = (r \sin \theta \sin \phi)^2 = r^2 \sin^2 \theta \sin^2 \phi \]

3. **Calculate \( y^2 \)**:
\[ y^2 = (r \sin \theta \cos \phi)^2 = r^2 \sin^2 \theta \cos^2 \phi \]

4. **Calculate \( z^2 \)**:
\[ z^2 = (r \cos \theta)^2 = r^2 \cos^2 \theta \]

5. **Combine \( x^2 + y^2 + z^2 \)**:
\[ x^2 + y^2 + z^2 = r^2 \sin^2 \theta \sin^2 \phi + r^2 \sin^2 \theta \cos^2 \phi + r^2 \cos^2 \theta \]

6. **Factor out \( r^2 \)**:
\[ x^2 + y^2 + z^2 = r^2 (\sin^2 \theta \sin^2 \phi + \sin^2 \theta \cos^2 \phi + \cos^2 \theta) \]

7. **Use the Pythagorean identity**:
\[ \sin^2 \phi + \cos^2 \phi = 1 \]
Therefore:
\[ \sin^2 \theta (\sin^2 \phi + \cos^2 \phi) + \cos^2 \theta = \sin^2 \theta \cdot 1 + \cos^2 \theta = \sin^2 \theta + \cos^2 \theta = 1 \]

8. **Final Result**:
\[ x^2 + y^2 + z^2 = r^2 \cdot 1 = r^2 \]

Thus, the final answer is:
\[ \boxed{r^2} \]