Find the coordinates of the center of mass of the following solid with variable density.
RequalsStartSet left parenthesis x comma y comma z right parenthesis : 0 less than or equals x less than or equals 4 comma 0 less than or equals y less than or equals 3 comma 0 less than or equals z less than or equals 3 EndSet​; rho​(x,y,z)equals1 plus StartFraction x Over 3 EndFraction

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.

Find the coordinates of the center of mass of the following solid with variable density.
R = {(x, y, z) : 0 <= x <= 4, 0 <= y <= 3, 0 <= z <= 3}; ρ(x, y, z) = 1 + (x / 3)

Learn how to compute the center of mass of a solid with a variable density function ρ(x, y, z) = 1 + (x/3). This problem involves triple integration over a rectangular region, with detailed steps to find the total mass and center of mass coordinates. Suitable for university students studying multivariable calculus.

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