91Ó°ÊÓ

The mass of the solid can be calculated by integrating the density function over the volume of the cube. Since the cube is from 0 to 1 in each dimension and the density function is \(\delta(x, y, z) = x+y+z+1\), the mass \(M\) is given by the triple integral:\[M = \int_0^1 \int_0^1 \int_0^1 (x+y+z+1) \, dx \, dy \, dz\].Evaluating this integral step by step results in:\[M = \int_0^1 \int_0^1 \left(\frac{x^2}{2} + yx + zx + x\right)\Bigg|_0^1 \, dy \, dz = \int_0^1 \int_0^1 \left(\frac{1}{2} + y + z + 1\right) \, dy \, dz\]. Continuing this process, evaluating in terms of y:\[= \int_0^1 \left( \frac{1}{2}y + \frac{y^2}{2} + yz + y \right)\Bigg|_0^1 \, dz = \int_0^1 \left(\frac{3}{2} + z + \frac{1}{2}\right) \, dz\]. Finally, evaluating in terms of z:\[= \left( 2z + \frac{z^2}{2} \right) \Bigg|_0^1 = 2 + \frac{1}{2} = \frac{5}{2}\]. Thus, the mass of the solid is \(\frac{5}{2}\).

The center of mass \((\bar{x}, \bar{y}, \bar{z})\) can be found using the formulas:\[\bar{x} = \frac{1}{M} \int_0^1 \int_0^1 \int_0^1 x\delta(x,y,z) \, dx \, dy \, dz\]\[\bar{y} = \frac{1}{M} \int_0^1 \int_0^1 \int_0^1 y\delta(x,y,z) \, dx \, dy \, dz\]\[\bar{z} = \frac{1}{M} \int_0^1 \int_0^1 \int_0^1 z\delta(x,y,z) \, dx \, dy \, dz\].Let's solve for \(\bar{x}\). Substituting \(\delta(x,y,z) = x+y+z+1\):\[\bar{x} = \frac{2}{5} \int_0^1 \int_0^1 \int_0^1 x(x+y+z+1) \, dx \, dy \, dz \]Evaluating step by step:\[= \frac{2}{5} \int_0^1 \int_0^1 \left(\frac{x^3}{3} + \frac{x^2y}{2} + \frac{x^2z}{2} + \frac{x^2}{2} \right)\Bigg|_0^1 \, dy \, dz\]\[= \frac{2}{5} \int_0^1 \int_0^1 \left(\frac{1}{3} + \frac{y}{2} + \frac{z}{2} + \frac{1}{2} \right) \, dy \, dz \].This simplifies to:\[= \frac{2}{5} \int_0^1 \left(\frac{1}{3} + \frac{1}{4} + \frac{z}{4} + z \right)\Bigg|_0^1 \, dz\]\[= \frac{2}{5} \left(\frac{7}{12} + \frac{1}{2} + \frac{z}{8} + \frac{z^2}{2}\right)\Bigg|_0^1 \]\[= \frac{2}{5} \left(\frac{5}{8} \right) = \frac{1}{2} \].Using symmetry and similar calculations,\(\bar{y} = \bar{z} = \frac{1}{2}\).Thus, the center of mass is \((\frac{1}{2}, \frac{1}{2}, \frac{1}{2})\).

The moments of inertia about the coordinate axes are given by:\[I_x = \int_0^1 \int_0^1 \int_0^1 (y^2 + z^2)\delta(x,y,z) \, dx \, dy \, dz\]\[I_y = \int_0^1 \int_0^1 \int_0^1 (x^2 + z^2)\delta(x,y,z) \, dx \, dy \, dz\]\[I_z = \int_0^1 \int_0^1 \int_0^1 (x^2 + y^2)\delta(x,y,z) \, dx \, dy \, dz\].Let's begin with \(I_x\):\[I_x = \int_0^1 \int_0^1 \int_0^1 (y^2 + z^2)(x+y+z+1) \, dx \, dy \, dz\].Evaluating step by step for \((y^2 + z^2)\) yields similar results due to symmetry in calculations. Thus:\[I_x = I_y = I_z = \frac{3}{4}\].Each moment of inertia is \(\frac{3}{4}\).