## 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 = 2, and x = 1 is revolved around the line y = −2.

1. anonymous

$\pi \int\limits_{1}^{e^2} 4 - \ln(x)^2$

2. anonymous

okay so in know that, this would be the volume if it was spun around the x- axis just not sure what to do when its revolved around y = -2

3. anonymous

@ganeshie8 @Hero @jim_thompson5910 any idea ?

4. jim_thompson5910

|dw:1438914029970:dw|

5. jim_thompson5910

|dw:1438914089545:dw|

6. anonymous

yep yep i know what it looks like, i've graphed it

7. jim_thompson5910

|dw:1438914110494:dw|

8. jim_thompson5910

you agree it's this region? |dw:1438914141386:dw|

9. anonymous

yes

10. jim_thompson5910

we're going to rotate that region around the line y = -2 |dw:1438914218489:dw|

11. jim_thompson5910

|dw:1438914231869:dw|

12. jim_thompson5910

|dw:1438914302702:dw|

13. jim_thompson5910

does that help?

14. anonymous

it does tell me if this is right

15. jim_thompson5910

in a sense, you'll have 2 cylinders cylinder A with radius 4 cylinder B with radius ln(x)+2

16. jim_thompson5910

volume of washer = cylinder A - cylinder B

17. anonymous

$\pi \int\limits_{1}^{e^2}16 - (\ln(x) + 2)$

18. anonymous

is that right @jim_thompson5910

19. anonymous

oh woops forgot to square the (ln(x) +2)

20. jim_thompson5910

$\Large \pi \int\limits_{1}^{e^2}16 - (\ln(x) + 2)^2$ is correct

21. anonymous

thank's !!

22. jim_thompson5910

np