## 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.

1. campbell_st

I'll use the same suggestion I used when you posted this question previously plot the information.... that is step 1

2. campbell_st

|dw:1439071497355:dw| so start by finding the y value when x = 1

3. campbell_st

to 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

4. anonymous

is it 0?

5. campbell_st

well 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...

6. campbell_st

oops limits on the integral need to be reversed... -1 to 0