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

Correct — C. Square each coordinate, then group the terms that share sin²(t).

x² = r²cos²(t)
y² = r²sin²(t)cos²(u)
z² = r²sin²(t)sin²(u)

y² + z² = r²sin²(t)[cos²(u) + sin²(u)] = r²sin²(t)

x² + y² + z² = r²cos²(t) + r²sin²(t)
= r²[cos²(t) + sin²(t)] = r² → option (c)

Why the others are wrong

• (a)
  2r — 2r has the wrong power. Every term of x² + y² + z² carries r², so the sum must be r² times something.
• (b)
  r² cos (u) — No u can survive: cos²(u) + sin²(u) = 1 removes u as soon as y² and z² are grouped.
• (d)
  r² sin (t) — No sin(t) can survive: after grouping, the t-terms are cos²(t) + sin²(t) = 1. A leftover sin(t) would need an unsquared term, and there is none.

The formulas place a point at distance r from the origin — spherical coordinates with the x-axis as the axis. x² + y² + z² is the squared distance, so it has to be r².

The algebra is sin²θ + cos²θ = 1 used twice: first on u to merge y² and z², then on t to merge the result with x².

Key facts

• sin²θ + cos²θ = 1 for every angle θ.
• Here y² + z² = r²sin²(t), because the u-terms add to 1.
• x = r cos t, y = r sin t cos u, z = r sin t sin u always satisfy x² + y² + z² = r².

Common traps

• Expanding all three squares and hunting for cancellations instead of grouping y² and z² first
• Forgetting to square r, which leads towards r or 2r
• Stopping at r²cos²(t) + r²sin²(t) and keeping just one of the two terms

No directly related past PYQ was found.