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anonymous
 one year ago
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 = −1.
anonymous
 one year ago
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 = −1.

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campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1I'll use the same suggestion I used when you posted this question previously plot the information.... that is step 1

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1dw:1439071497355:dw so start by finding the y value when x = 1

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1to find the volume you will need to integrate with respect to y so change the equation so that x is the subject, which means finding the inverse you will need an upper and lower value to integrate between you know the lower is y = 1 what is the y value for the upper limit... when x = 1

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1well when its volume around the y axis I use \[V = \pi \left \int\limits_{0}^{1} x^2~dy \right\] so if y = ln(x) then \[x = e^y\] and \[x^2 = (e^y)^2 ~~or~~~x^2 = e^{2y}\] make that substitution and away you go...

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1oops limits on the integral need to be reversed... 1 to 0
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