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When computing volume, the integrand is implicitly 1 because the goal is to sum the infinitesimal volume elements $dV$ over the entire region. The triple integral $\int\int\int 1 \, dV$ accumulates these small volumes to find the total volume.

Conditions: The integral is being used to compute volume.; The differential element is written without an additional function factor.

When computing volume, the integrand is implicitly 1 because the goal is to sum the infinitesimal volume elements $dV$ over the entire region. The triple integral $\int\int\int 1 \, dV$ accumulates these small volumes to find the total volume.

Conditions: The integral is being used to compute volume.; The differential element is written without an additional function factor.

By integrating the constant 1 with respect to $z$ from 0 to 2, the inner integral evaluates to 2. This leaves a double integral $\int_{0}^{3}\int_{0}^{4} 2 \, dy \, dx$, which represents the volume under the constant surface $z(x,y)=2$ over the $xy$-region.

Conditions: The region is a rectangular box with constant top surface $z=2$.; The integral is evaluated in the order $dz \, dy \, dx$.

By integrating the constant 1 with respect to $z$ from 0 to 2, the inner integral evaluates to 2. This leaves a double integral $\int_{0}^{3}\int_{0}^{4} 2 \, dy \, dx$, which represents the volume under the constant surface $z(x,y)=2$ over the $xy$-region.

Conditions: The region is a rectangular box with constant top surface $z=2$.; The integral is evaluated in the order $dz \, dy \, dx$.