Why is the integrand implicitly 1 when computing the volume of a rectangular box with a triple integral?

Conditions

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

Reasoning, step by step

1. Identify the goal of the integral: computing volume.
2. Recognize that volume is the sum of infinitesimal volume elements $dV$.
3. Understand that multiplying $dV$ by 1 does not change its value.
4. Conclude that the implicit integrand is 1.

Example

The speaker points out that no function is written inside the integral, then explains that this means the integrand is implicitly 1. A red 1 is added before dz, making the integral $\int_{0}^{3}\int_{0}^{4}\int_{0}^{2} 1\,dz\,dy\,dx$.

Common misconceptions

• Assuming a blank integrand means zero.
• Thinking that a triple integral always requires a complex function as the integrand.

Watch the explanation

Explore next

• How does the triple integral reduce to a double integral for this box?
• What is the final volume computed for the box?