To find the mass $M$ of the rectangular volume with the given dimensions and density, we start with the formula for mass in terms of density:

$M = \int_V \rho \, dV$

Where:

• $\rho = p_0(1 + \frac{xy}{ab})$ is the density function,
• $V$ is the volume of the rectangular region with $x$ ranging from 0 to $a$, $y$ ranging from 0 to $b$, and $z$ ranging from 0 to $c$.

Step 1: Set up the integral for the mass

The mass is given by the triple integral over the volume of the rectangular region:

$M = \int_0^a \int_0^b \int_0^c \rho(x, y, z) \, dz \, dy \, dx$

Substitute the density $\rho = p_0 \left( 1 + \frac{xy}{ab} \right)$ into the integral:

$M = \int_0^a \int_0^b \int_0^c p_0 \left( 1 + \frac{xy}{ab} \right) \, dz \, dy \, dx$

Step 2: Simplify the integral

Now, perform the integration step by step.

Integrate with respect to $z$:

Since the integrand does not depend on $z$, we have:

$\int_0^c \, dz = c$

Thus, the mass becomes:

$M = p_0 \int_0^a \int_0^b \left( 1 + \frac{xy}{ab} \right) c \, dy \, dx$

This simplifies to:

$M = p_0 c \int_0^a \int_0^b \left( 1 + \frac{xy}{ab} \right) \, dy \, dx$

Step 3: Break the integral into two parts

We now split the integrand into two terms:

$M = p_0 c \left( \int_0^a \int_0^b 1 \, dy \, dx + \int_0^a \int_0^b \frac{xy}{ab} \, dy \, dx \right)$

First term: $\int_0^a \int_0^b 1 \, dy \, dx$

This is a straightforward integral:

$\int_0^a \int_0^b 1 \, dy \, dx = a \times b$

Second term: $\int_0^a \int_0^b \frac{xy}{ab} \, dy \, dx$

We can factor out the constants:

$\frac{1}{ab} \int_0^a \int_0^b xy \, dy \, dx$

First, integrate with respect to $y$:

$\int_0^b xy \, dy = x \left[ \frac{y^2}{2} \right]_0^b = x \frac{b^2}{2}$

Now, integrate with respect to $x$:

$\int_0^a x \frac{b^2}{2} \, dx = \frac{b^2}{2} \left[ \frac{x^2}{2} \right]_0^a = \frac{b^2}{2} \cdot \frac{a^2}{2} = \frac{a^2 b^2}{4}$

Thus, the second term becomes:

$\frac{1}{ab} \cdot \frac{a^2 b^2}{4} = \frac{ab}{4}$

Step 4: Combine the results

Now combine the two integrals:

$M = p_0 c \left( ab + \frac{ab}{4} \right)$

Factor out $ab$:

$M = p_0 c \cdot ab \left( 1 + \frac{1}{4} \right) = p_0 c \cdot ab \cdot \frac{5}{4}$

Thus, the mass is:

$M = \frac{5}{4} p_0 abc$

Final Answer:

The mass is $M = \frac{5}{4} p_0 abc$.

Would you like more details on any part of the process or have any other questions? Here are a few related questions:

1. How would the mass change if the density had a different functional form?
2. What if the volume was a cylinder instead of a rectangular box—how would the integral change?
3. How can we calculate the center of mass for this system?
4. What would happen if the density were zero at certain points in the volume?
5. How do you interpret the physical meaning of the term $\frac{xy}{ab}$ in the density function?

Tip: When working with triple integrals in geometry, always check if symmetry or separability can simplify your calculations.