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

step-by-step explanation:

you can add the two numbers and divide the result by 2. or you can find the gap between the two numbers and add half of this to the smaller number.

step-by-step explanation:

de-gf

hj=3x+20

ji=15x-64