Finding Volume of Solid Revolved About Line y=4

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SUMMARY

The discussion focuses on calculating the volume of a solid generated by revolving the area between the functions f(x) = e^x and g(x) = ln(x) around the line y = 4, specifically between x = 1/2 and x = 1. The integral setup for volume is V = π * ∫ ((4 - e^x)² - (4 - ln(x))²) dx. A participant questions whether the correct expression should be (4 - e^x) or (e^x - 4) and reports an incorrect volume calculation of approximately -23. Additionally, the discussion includes a separate problem involving the rates of people entering and leaving an amusement park, modeled by E(t) and L(t), respectively.

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Homework Statement



f(x) =e^x and g(x)= ln(x)

Find the volume of the solid generated when the region enclosed by the graphs of f and g between x=1/2 and x=1 is revolved about the line y=4

Homework Equations



v= pi* integral( f(x)^2 - g(x)^2 dx)

The Attempt at a Solution



SO for the part about the volume

I set up my integral as V= pi* integral ( (4-e^x)^2 - (4- ln(x))^2 dx) from 1/2 to 1 and get around -23. Am I right ? should it be 4-e^x or e^x-4 ?
 
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Also another problem if you guys don't mind

The rate at which people enter an amusement park on a given day is modeled by the function E defined by E(t)= 15600/(t^2-24t+160)

The rate at which people leave is L(t)= 9890/(t^2-38t+370)

Both E(t) and L(t) are measured in people/hour an d time t is measured in hours after midnight. These functions are valid for 9<= x<= 23. At t=9, no one is in the park.

Let H(t)= integral( E(x)-L(x)) from 9 to t for 9<=t<=23. At what time t, for 9<=t<=23, does the model predict the number of people in the park is maximum.

My approach:

H(x) max when H'(x) min ( I am doubting this connection. AM I right ?). Then I find H''(x) and set it =0 and solve for t then plug in t back to the model. Is my approach right or should I do something else since I cannot t ( I keep getting negative numbers)
 

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