**Problem Statement:**Find the centroid of the given solid region assuming a density of \( \delta(x, y, z) = 1 \).**Solid Region Definition:**\[ x^2 + y^2 + z^2 \leq 1, \quad z \geq 0 \]This represents the upper hemisphere of a unit sphere including the disk at the base.**Centroid Expression:**The coordinates of the centroid \((\bar{x}, \bar{y}, \bar{z})\) are sought, represented as:\[ \left( \bar{x}, \bar{y}, \bar{z} \right) = \left( \begin{array}{c} \text{[blank space for solution]} \end{array} \right) \]**Additional Notes:**To compute the centroid of this solid region, one would generally use integral calculus. The symmetry in terms of \(x\) and \(y\) might simplify certain calculations, particularly due to the symmetric nature of the solid around the z-axis.

Given the solid region S, x2+y2+z2≤1, z≥0 Note that the given solid is the upper hemisphere of the sphere x2+y2+z2=1 with center at the origin 0,0,0. The coordinates to determine the centroid x,y,z is given by, m=∭δ(x,y,z)dVMyz=∭xδ(x,y,z)dVMxz=∭yδ(x,y,z)dVMxy=∭zδ(x,y,z)dV Then, x=Myzm, y=Mxzm, z=MxymSince the central axis of the solid S is z-axis, the center of mass of the solid lies in the z-axis. Thus, x=0, y=0 Now, m=∭Sδ(x,y,z)dV=∫02π∫01∫01-r21r dz dr dθ=∫02π∫01rz01-r2=∫02π∫01r1-r2dr dθ=∫02π13 dθ=13θ02π=2π3Next, Mxy=∭SzdV=∫02π∫01∫01-r2zr dz dr dθ=∫02π∫01rz2201-r2dr dθ=∫02π∫01r21-r2dr dθ=∫02π18dθ=18θ02π=2π8 Then, z=Mxym=2π8×32π=38 Thus, the centroid of the given solid S lies at 0,0,38.