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Answers for “长方体体积的三重积分如何简化为二重积分?”

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By integrating the constant 1 with respect to zz from 0 to 2, the inner integral evaluates to 2. This leaves a double integral ∫03∫042 dy dx\int_{0}^{3}\int_{0}^{4} 2 \, dy \, dx, which represents the volume under the constant surface z(x,y)=2z(x,y)=2 over the xyxy-region.

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

Understand why

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When computing volume, the integrand is implicitly 1 because the goal is to sum the infinitesimal volume elements dVdV over the entire region. The triple integral ∫∫∫1 dV\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.

Meet the concept

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The final volume of the rectangular box is 24 cubic units. This is obtained by evaluating the iterated triple integral ∫03∫04∫021 dz dy dx\int_{0}^{3}\int_{0}^{4}\int_{0}^{2} 1 \, dz \, dy \, dx from the inside out.

Conditions: The region is a rectangular box with bounds 0≤x≤30 \le x \le 3, 0≤y≤40 \le y \le 4, and 0≤z≤20 \le z \le 2.; The integrand is 1.