By integrating the constant 1 with respect to from 0 to 2, the inner integral evaluates to 2. This leaves a double integral , which represents the volume under the constant surface over the -region.
Conditions: The region is a rectangular box with constant top surface .; The integral is evaluated in the order .
By integrating the constant 1 with respect to from 0 to 2, the inner integral evaluates to 2. This leaves a double integral , which represents the volume under the constant surface over the -region.
Conditions: The region is a rectangular box with constant top surface .; The integral is evaluated in the order .