Triple Integrals — Question 4

Question 4

A box $0\le x\le 2$, $0\le y\le 1$, $0\le z\le 3$ has density $\rho=1+x+z$.

Tasks

1. Find its mass.

2. Find the average density.

3. Check against the density range.

Question 4 – Solution

Strategy. Integrate density and divide by volume.

Step 1: Mass$M=\int_0^2\int_0^1\int_0^3(1+x+z)dz\,dy\,dx=\int_0^2(15/2+3x)dx=\boxed{21}$.

Step 2: AverageThe volume is $6$, so $\boxed{\rho_{\mathrm{avg}}=21/6=7/2}$.

VerificationHere $1\le\rho\le 6$; $7/2$ lies in this interval. Also the coordinate-average rule gives $1+1+3/2=7/2$.