GATE CS 2023 Q21 — Calculus: Limits, Continuity, Differentiability

We can evaluate the given triple integral by splitting it into two parts using the linearity of integration:

Since the limits of integration are independent constants, we can separate the variables for each part:

1. First part ($I_1$): $$I_1 = \left( \int_{-3}^{3} 4x^2 \, \mathrm{d}x \right) \left( \int_{-2}^{2} y \, \mathrm{d}y \right) \left( \int_{-1}^{1} 1 \, \mathrm{d}z \right)$$ Since $y$ is an odd function and the interval $[-2, 2]$ is symmetric about $0$, we have: $$\int_{-2}^{2} y \, \mathrm{d}y = 0 \implies I_1 = 0$$
2. Second part ($I_2$): $$I_2 = \left( \int_{-3}^{3} 1 \, \mathrm{d}x \right) \left( \int_{-2}^{2} 1 \, \mathrm{d}y \right) \left( \int_{-1}^{1} z^3 \, \mathrm{d}z \right)$$ Similarly, since $z^3$ is an odd function and the interval $[-1, 1]$ is symmetric about $0$, we have: $$\int_{-1}^{1} z^3 \, \mathrm{d}z = 0 \implies I_2 = 0$$

Combining both parts, we get: