How does the triple integral for the volume of a rectangular box reduce to a double integral?

Conditions

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

Reasoning, step by step

1. Start with the triple integral $\int_{0}^{3}\int_{0}^{4}\int_{0}^{2} 1 \, dz \, dy \, dx$.
2. Evaluate the innermost integral with respect to $z$: $\int_{0}^{2} 1 \, dz = [z]_0^2 = 2$.
3. Substitute the result back into the expression, yielding $\int_{0}^{3}\int_{0}^{4} 2 \, dy \, dx$.
4. Recognize this as a double integral representing the volume under the surface $z(x,y)=2$.

Example

The speaker says that after integrating with respect to z, the result is a double integral and that this is the exact integral from previous videos on the double integral. The board shows $\int_{0}^{3}\int_{0}^{4} 2 \, dy \, dx$.

Common misconceptions

• Thinking a triple integral introduces a wholly new computational process unrelated to double integrals.
• Assuming the reduction only works for constant integrands.

Watch the explanation

Explore next

• What is the final volume computed for the box?
• Why is the integrand implicitly 1 when computing the volume of a rectangular box with a triple integral?