Question 1085933
<font color="black" face="times" size="3">Graph:
<img src = "https://i.imgur.com/l9zkCxN.png">
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
<img src = "https://i.imgur.com/aQvEQ0W.png">
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


*[Tex \Large V = \pi*\int_{a}^{b}\left(\left(R\right)^2 - \left(r\right)^2\right)dx]


*[Tex \Large V = \pi*\int_{1}^{e}\left(\left(4\right)^2 - \left(Ln(x) + 3\right)^2\right)dx]


*[Tex \Large V = \pi*\int_{1}^{e}\left(16 - \left(Ln(x) + 3\right)^2\right)dx]
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