By integrating the constant 1 with respect to z from 0 to 2, the inner integral evaluates to 2. This leaves a double integral ∫03∫042dydx, 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 dzdydx.
By integrating the constant 1 with respect to z from 0 to 2, the inner integral evaluates to 2. This leaves a double integral ∫03∫042dydx, 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 dzdydx.
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 ∫∫∫1dV 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 ∫∫∫1dV 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.
The final volume of the rectangular box is 24 cubic units. This is obtained by evaluating the iterated triple integral ∫03∫04∫021dzdydx from the inside out.
Conditions: The region is a rectangular box with bounds 0≤x≤3, 0≤y≤4, and 0≤z≤2.; The integrand is 1.
The final volume of the rectangular box is 24 cubic units. This is obtained by evaluating the iterated triple integral ∫03∫04∫021dzdydx from the inside out.
Conditions: The region is a rectangular box with bounds 0≤x≤3, 0≤y≤4, and 0≤z≤2.; The integrand is 1.