Let x = r cos(t), y = r sin(t) cos(u), z = r sin(t) sin(u). Then find the value of x² + y² + z².

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Given: x = r cos(t), y = r sin(t) cos(u), z = r sin(t) sin(u).

We need to find x² + y² + z².

Calculate each term:
- x² = (r cos(t))² = r² cos²(t)
- y² = (r sin(t) cos(u))² = r² sin²(t) cos²(u)
- z² = (r sin(t) sin(u))² = r² sin²(t) sin²(u)

Sum them up:

x² + y² + z² = r² cos²(t) + r² sin²(t) cos²(u) + r² sin²(t) sin²(u)
= r² cos²(t) + r² sin²(t) (cos²(u) + sin²(u))

Since cos²(u) + sin²(u) = 1,

x² + y² + z² = r² cos²(t) + r² sin²(t) = r² (cos²(t) + sin²(t))

Again, cos²(t) + sin²(t) = 1,

Therefore, x² + y² + z² = r².

Hence, the correct answer is r².

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