A rhombus with a side of 8 cm and an acute angle of 60 rotates around the side. Find the surface area of the body of revolution.

The resulting surface will consist of the lateral surface of the cylinder and the lateral surfaces of two cones adjacent to the cylinder.
One cone is directed with its apex inside the cone, the other outward.
And the tops of the cones are obtained by rotating the sides of a rhombus with acute angles of 60 degrees.

S full = 2 * S side cone + S side cylinder.
Let us write down the formulas for calculating the area of ​​the lateral surface of the cone and the area of ​​the lateral surface of the cylinder.
S side.cone = π * R * L, where R is the radius of the circle formed during rotation, L is the generator, equal to 8, as it is obtained by rotating the side of the rhombus with side 8.
The angle between the generatrix and the height is 60º.
Then by the theorem of sines the radius is equal to:
R = L * sin60º = 8 * √3 / 2 = 4√3 cm.
Hence,
S side.cone = π * R * L = π * 4√3 * 8 = 32√3 * π (cm ^ 2).

The lateral surface area of ​​the cylinder is:
S side of the cylinder = 2 * π * R * h.
The height h of the cylinder is equal to the side of the rhombus or two heights of the cone, that is, 8 – Hcone + Hcone.
Hence,
S side of the cylinder = 2 * π * 4√3 * 8 = 64√3 * π (cm ^ 2).

Thus, the surface area of ​​the body of revolution is:
S full = 2 * S side of the cone + S side of the cylinder = 2 * 32√3 * π + 64√3 * π = 128√3 * π (cm ^ 2).