SOLUTION: Set up, but do not evaluate, the integral which gives the volume when the region bounded by the curves y = Ln(x), y = 1, and x = 1 is revolved around the line y = &#8722;3.

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Question 1085933: Set up, but do not evaluate, the integral which gives the volume when the region bounded by the curves y = Ln(x), y = 1, and x = 1 is revolved around the line y = −3.

Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website!
Graph:

f(x) = Ln(x) (green curve)
g(x) = 1 (red horizontal line)
x = 1 (blue vertical line)
y = -3 (black dashed line)

Outer radius = R
R = vertical distance from the red horizontal line to the black dashed horizontal line
R = 1 - (-3)
R = 1 + 3
R = 4

Inner radius = r
r = vertical distance from the green curve to the black dashed horizontal line
r = Ln(x) - (-3)
r = Ln(x) + 3
In contrast to R, the inner radius will change based on x

Find the intersection point between the green curve and the red line
y = Ln(x)
1 = Ln(x)
e^1 = x
x = e
The intersection point is (x,y) = (e,1) where e = 2.71828 approximately

So the shaded orange region shown below

represents the region we want to revolve around y = -3 to form the solid of revolution. We're going from a = 1 to b = e

The integral to set up is therefore

SOLUTION: Set up, but do not evaluate, the integral which gives the volume when the region bounded by the curves y = Ln(x), y = 1, and x = 1 is revolved around the line y = &#8722;3.

SOLUTION: Set up, but do not evaluate, the integral which gives the volume when the region bounded by the curves y = Ln(x), y = 1, and x = 1 is revolved around the line y = −3.