Find the coordinates of the center of mass of the following solid with variable density.
The interior of the prism formed by z = x, x = 2, y = 4, and the coordinate planes with p(x,y,z)=1+y.
The center of mass is located at
(Type exact answers.)

To find the center of mass \((x_{\text{cm}}, y_{\text{cm}}, z_{\text{cm}})\) of the solid, we need to:

1. **Set up the solid’s boundaries:**
   The solid is bounded by:
   - \( z = x \) (the slanted plane)
   - \( x = 2 \) (the vertical plane)
   - \( y = 4 \) (the vertical plane)
   - The coordinate planes (the \(xy\)-, \(xz\)-, and \(yz\)-planes, i.e., \(x = 0\), \(y = 0\), \(z = 0\)).

   This gives the solid's limits:
   - \( 0 \leq x \leq 2 \)
   - \( 0 \leq y \leq 4 \)
   - \( 0 \leq z \leq x \) (because of the plane \(z = x\)).

2. **Density function:**
   The density function is given by \( \rho(x, y, z) = 1 + y \).

3. **Mass of the solid:**
   The total mass \(M\) of the solid is given by the triple integral:
   \[
   M = \int_0^2 \int_0^4 \int_0^x \rho(x, y, z) \, dz \, dy \, dx
   \]
   Substituting \( \rho(x, y, z) = 1 + y \), we have:
   \[
   M = \int_0^2 \int_0^4 \int_0^x (1 + y) \, dz \, dy \, dx
   \]

4. **Coordinates of the center of mass:**
   - \( x_{\text{cm}} = \frac{1}{M} \int_0^2 \int_0^4 \int_0^x x \rho(x, y, z) \, dz \, dy \, dx \)
   - \( y_{\text{cm}} = \frac{1}{M} \int_0^2 \int_0^4 \int_0^x y \rho(x, y, z) \, dz \, dy \, dx \)
   - \( z_{\text{cm}} = \frac{1}{M} \int_0^2 \int_0^4 \int_0^x z \rho(x, y, z) \, dz \, dy \, dx \)

### Step-by-step solution:

#### 1. Find the total mass \(M\):
   \[
   M = \int_0^2 \int_0^4 \int_0^x (1 + y) \, dz \, dy \, dx
   \]
   The integral with respect to \(z\) is straightforward:
   \[
   \int_0^x (1 + y) \, dz = (1 + y)x
   \]
   Now, the mass becomes:
   \[
   M = \int_0^2 \int_0^4 (1 + y)x \, dy \, dx
   \]
   Expanding and integrating with respect to \(y\):
   \[
   M = \int_0^2 \int_0^4 (x + xy) \, dy \, dx = \int_0^2 \left[ 4x + 8x \right] \, dx = \int_0^2 12x \, dx = 6x^2 \big|_0^2 = 24
   \]
   So, the total mass is \(M = 24\).

#### 2. Find \( x_{\text{cm}} \):
   \[
   x_{\text{cm}} = \frac{1}{M} \int_0^2 \int_0^4 \int_0^x x(1 + y) \, dz \, dy \, dx
   \]
   Following similar steps as for \(M\), the integral becomes:
   \[
   x_{\text{cm}} = \frac{1}{24} \int_0^2 \int_0^4 x^2(1 + y) \, dy \, dx = \frac{1}{24} \int_0^2 \left[ 4x^2 + 8x^2 \right] \, dx = \frac{1}{24} \int_0^2 12x^2 \, dx = \frac{1}{24} \cdot 8x^3 \big|_0^2 = \frac{64}{24} = \frac{8}{3}
   \]

#### 3. Find \( y_{\text{cm}} \):
   \[
   y_{\text{cm}} = \frac{1}{M} \int_0^2 \int_0^4 \int_0^x y(1 + y) \, dz \, dy \, dx
   \]
   Integrating in the same manner:
   \[
   y_{\text{cm}} = \frac{1}{24} \int_0^2 \int_0^4 y(1 + y)x \, dy \, dx = \frac{1}{24} \int_0^2 \left[ \frac{32}{3} x \right] dx = \frac{1}{24} \cdot \frac{64}{3} \cdot \frac{4}{2} = \frac...
Final

Find the coordinates of the center of mass of the following solid with variable density. The interior of the prism formed by z = x, x = 2, y = 4, and the coordinate planes with p(x,y,z)=1+y. The center of mass is located at (Type exact answers.)

Learn how to calculate the center of mass of a solid prism with variable density using triple integrals. This detailed solution walks through the step-by-step process, covering key concepts in multivariable calculus and the application of density functions. Suitable for undergraduate students.

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