Set up (do not evaluate) the integral which gives the volume when region bounded by curves y=Ln(x), y=2, and x=1 is revolved around the line y=-2. Thanks in advance!

Step-by-step explanation:

Graph the region: desmos.com/calculator/rbe6rq61a2

When the region is rotated about y=-2, the resulting shape is a horizontal, hollow cylinder.  The volume can be found with either washer method or shell method.

To use washer method, cut a thin vertical slice of the region.  Rotated around y=-2, this slice becomes a washer.  The width of this washer is dx.  The outer radius is 2 − (-2) = 4.  The inner radius is y − (-2) = y + 2.  The volume of the washer is:

dV = π (4² − (y + 2)²) dx

dV = π (4² − (ln x + 2)²) dx

The total volume is the sum of the washers from x=1 to x=e².

V = ∫ dV

V = ∫₁ᵉ² π (4² − (ln x + 2)²) dx

To instead use shell method, cut a thin horizontal slice of the region.  Rotated around y=-2, this slice becomes a cylindrical shell.  The thickness of the shell is dy.  The radius is y − (-2) = y + 2.  The width is x − 1.  The volume of the shell is:

dV = 2π (y + 2) (x − 1) dy

dV = 2π (y + 2) (eʸ − 1) dy

The total volume is the sum of the shells from y=0 to y=2.

V = ∫ dV

V = ∫₀² 2π (y + 2) (eʸ − 1) dy