Why does the lecturer say a double integral would suffice for the earlier volume case but not for the mass problem?

Conditions

• The goal is volume rather than mass with variable density.
• The integrand over the volume is constant for the volume case.
• The density varies with position for the mass case.

Reasoning, step by step

1. Analyze the volume integral: the integrand is 1, so $\int \int \int 1 \, dz \, dy \, dx$ reduces to $\int \int (\int 1 \, dz) \, dy \, dx = \int \int 2 \, dy \, dx$.
2. Analyze the mass integral: the integrand is $\rho(x,y,z)=xyz$, so $\int \int \int xyz \, dz \, dy \, dx$ cannot be separated into a simple double integral because $z$ is part of the integrand.
3. Conclude that a double integral suffices for constant height/volume but not for variable density/mass.

Example

The speaker says that when you have a constant value within the volume, you could have just done a double integral, but the new goal is mass rather than volume.

Common misconceptions

• Assuming a double integral can always compute mass.
• Thinking that variable density can be factored out of the $z$-integration.

Watch the explanation

Explore next

• What does $\rho(x,y,z)=xyz$ mean in this triple-integral lesson?
• How is the differential mass element derived from a variable density function?