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