To find the coordinates of the center of mass of a solid with a variable density, we use the following steps. The center of mass is given by the coordinates $( \bar{x}, \bar{y}, \bar{z} )$, where:

$\bar{x} = \frac{1}{M} \int \int \int_{R} x \rho(x, y, z) \, dV$ $\bar{y} = \frac{1}{M} \int \int \int_{R} y \rho(x, y, z) \, dV$ $\bar{z} = \frac{1}{M} \int \int \int_{R} z \rho(x, y, z) \, dV$

Here, $M$ is the total mass of the solid, which is given by:

$M = \int \int \int_{R} \rho(x, y, z) \, dV$

Step 1: Define the region $R$

The region $R$ is the solid defined by: $0 \leq x \leq 4, \quad 0 \leq y \leq 3, \quad 0 \leq z \leq 3$

Step 2: Density function $\rho(x, y, z)$

The density function is given by: $\rho(x, y, z) = 1 + \frac{x}{3}$

Step 3: Compute the total mass $M$

The total mass is: $M = \int_0^4 \int_0^3 \int_0^3 \left(1 + \frac{x}{3}\right) dz \, dy \, dx$ First, integrate with respect to $z$: $\int_0^3 1 \, dz = 3, \quad \int_0^3 \frac{x}{3} \, dz = \frac{x}{3} \times 3 = x$ Thus, the integral becomes: $M = \int_0^4 \int_0^3 (3 + x) \, dy \, dx$ Now integrate with respect to $y$: $\int_0^3 3 \, dy = 9, \quad \int_0^3 x \, dy = 3x$ So the integral is now: $M = \int_0^4 (9 + 3x) \, dx$ Now integrate with respect to $x$: $\int_0^4 9 \, dx = 36, \quad \int_0^4 3x \, dx = 3 \times \frac{x^2}{2} \Big|_0^4 = 3 \times 8 = 24$ Thus, the total mass is: $M = 36 + 24 = 60$

Step 4: Compute $\bar{x}$

$\bar{x} = \frac{1}{M} \int_0^4 \int_0^3 \int_0^3 x \left( 1 + \frac{x}{3} \right) dz \, dy \, dx$ First, compute the $z$-integral: $\int_0^3 1 \, dz = 3, \quad \int_0^3 \frac{x}{3} \, dz = x$ So the integral becomes: $\bar{x} = \frac{1}{60} \int_0^4 \int_0^3 x(3 + x) \, dy \, dx$ Now integrate with respect to $y$: $\int_0^3 x(3 + x) \, dy = 3x(3 + x)$ Thus, the integral becomes: $\bar{x} = \frac{1}{60} \int_0^4 3x(3 + x) \, dx$ Expanding: $3x(3 + x) = 9x + 3x^2$ Now integrate with respect to $x$: $\int_0^4 9x \, dx = 9 \times \frac{x^2}{2} \Big|_0^4 = 9 \times 8 = 72$ $\int_0^4 3x^2 \, dx = 3 \times \frac{x^3}{3} \Big|_0^4 = 3 \times 21.33 = 64$ Thus: $\bar{x} = \frac{1}{60} \times (72 + 64) = \frac{136}{60} = 2.267$

Step 5: Compute $\bar{y}$

Since the density function does not depend on $y$, the center of mass in the $y$-direction will be at the midpoint: $\bar{y} = \frac{1}{2} \times (0 + 3) = 1.5$

Step 6: Compute $\bar{z}$

Similarly, since the density function does not depend on $z$, the center of mass in the $z$-direction will be at the midpoint: $\bar{z} = \frac{1}{2} \times (0 + 3) = 1.5$

Final Answer

The coordinates of the center of mass are approximately: $\boxed{(2.27, 1.5, 1.5)}$

Would you like any details or further clarifications?

Related questions:

1. How would the center of mass change if the density function depended on $y$ as well?
2. What would happen to the center of mass if the region were not a rectangular prism?
3. How would the total mass change if the density function increased more steeply with $x$?
4. Can you compute the center of mass of a cylindrical region with variable density?
5. What if the density function were $\rho(x, y, z) = 1 + x + y + z$? How would that affect the center of mass?

Tip: When the density function is symmetric in some variables, the center of mass in those directions often lies at the midpoint of the region.