Question

Please show the full solution and use integration

Determine the centre of mass of the region D enclosed by the surfaces x²+y²=1, z=0 andz=x+2 Consider using cylindrical coordinates.
Find using a triple integral, the volumeof the parallelepiped with vertices (0,0,0),(0,1,2), (0,2,1),(0,3,3),(2,1,1),(2,2,3),(2,3,2), and (2,4,4). Hint: Use the substitutions u= −x−2y+4z,v=−x+4y−2z,w=x, or equivalently x=w,y=(u+2v+3w)/6, z=(2u+v+3w)/6. Don't forget the Jacobian.

Answer 1

To determine the center of mass of the region D enclosed by the surfaces x²+y²=1, z=0 and z=x+2, we can use cylindrical coordinates and calculate the triple integral of the density function over the region.

Step-by-step explanation:

To determine the center of mass of the region D enclosed by the surfaces x²+y²=1, z=0 and z=x+2, we can use cylindrical coordinates. The equation of the surface x²+y²=1 represents a circle in the xy-plane with radius 1. We can express this circle in cylindrical coordinates as r=1 and θ ranging from 0 to 2π.

To find the center of mass, we need to calculate the triple integral of the density function over the region D. In cylindrical coordinates, the density function can be written as ρ=f(r,θ,z)=kz, where k is a constant.