ISRO CS 2025 Q.58: solution and answer

ISRO 2025 · CSQ.58 · 1 mark · MCQ

The value of the triple integral $\int_{1}^{2} \int_{0}^{1} \int_{- 3}^{3} 2 x y - z^{3} d x d y d z$ is

(A)
9
(B)
18
(C)
0
(D)
None of the above

See the question in the paper

Answer and solution

ANSWER

With $x$ from $- 3$ to $3$ , $y$ from $0$ to $1$ and $z$ from $1$ to $2$ , the integral equals $- \frac{45}{2} = - 22.5$ . That value is not listed, so the answer is (d) None of the above.

Official answer: DSource: ISRO's final answer key of 15 May 2026

Match each set of limits to its variable. In $\int\int\int f d x d y d z$ the integrals are nested: the innermost sign (the one written last, just before the integrand) goes with the first differential, $d x$ , and the outermost sign goes with the last differential, $d z$ . So $x$ runs from $- 3$ to $3$ , $y$ from $0$ to $1$ and $z$ from $1$ to $2$ : $I = \int_{1}^{2} [\int_{0}^{1} [\int_{- 3}^{3} (2 x y - z^{3}) d x] d y] d z$
Innermost integral, in $x$ (here $y$ and $z$ are constants): $\int_{- 3}^{3} (2 x y - z^{3}) d x = [x^{2} y - z^{3} x]_{- 3}^{3} = (9 y - 3 z^{3}) - (9 y + 3 z^{3}) = - 6 z^{3}$ The $2 x y$ term is odd in $x$ , so it gives zero over the symmetric interval $[- 3, 3]$ . The constant $- z^{3}$ is multiplied by the interval length $6$ .
Middle integral, in $y$ : the result $- 6 z^{3}$ does not depend on $y$ , so it is multiplied by the length of $[0, 1]$ : $\int_{0}^{1} (- 6 z^{3}) d y = - 6 z^{3}$
Outer integral, in $z$ : $I = \int_{1}^{2} - 6 z^{3} d z = - 6 [\frac{z ^{4}}{4}]_{1}^{2} = - 6 \cdot \frac{16 - 1}{4} = - \frac{90}{4} = - \frac{45}{2} = - 22.5$
$- 22.5$ is not 9, 18 or 0, so the answer is (d) None of the above.
Where the other options come from: 9 comes from pairing the limits with $x, y, z$ from left to right ( $x$ from 1 to 2, $z$ from $- 3$ to $3$ ). The $2 x y$ part then gives 9 and the $z^{3}$ part vanishes by odd symmetry in $z$ . 0 comes from using the odd-function argument on the whole integrand. Neither reading follows the nesting order of the differentials.

KEY IDEA

In $\int\int\int f d x d y d z$ the innermost integral sign belongs to $d x$ and the outermost to $d z$ . Integrate from the inside out. Over $[- 3, 3]$ in $x$ the odd term $2 x y$ vanishes and the constant $- z^{3}$ is multiplied by $6$ . What is left is $\int_{1}^{2} - 6 z^{3} d z = - 22.5$ .

WATCH OUT FOR

Pairing the limits with $x, y, z$ in left-to-right order puts $x$ on $[1, 2]$ and $z$ on $[- 3, 3]$ . That makes the $z^{3}$ term vanish and gives 9, option (a). The innermost sign goes with $d x$ , so the symmetric interval $[- 3, 3]$ belongs to $x$ , not $z$ .

From the ISRO CS 2025 paper. Every question of this paper · The full paper (PDF, ISRO) · All 2 ISRO Calculus questions, with notes