## amonoconnor 2 years ago Would someone be willing to check my answer for the following problem involving volumes of solids? y = 2/(x^1/2), y = 2, x = 4; about the x-axis. Any and all help is greatly appreciated! :)

1. anonymous

so basically

2. anonymous

$y=\frac{ 2 }{ \frac{ x }{ 2 } }$

3. anonymous

so y =$\frac{ 4 }{ x }$

4. amonoconnor

No... one of the equations defining the area to be revolved is y = 2/(x^(1/2)) and another serving as a limit, and boundary value for the integral, is y = 2.

5. AccessDenied

We might start with a sketch of our situation. This one is pretty tough to visualize, so a graphing calculator might help if that is available. Here is Wolfram's plot: http://www.wolframalpha.com/input/?i=plot+y%3D2%2Fx%5E%281%2F2%29%2C+y%3D2%2C+x%3D4 You want to revolve this region about the x-axis. First, did you have any ideas on how you might set this one up? :)

6. AccessDenied

Since the region is not connected to the x-axis, I think the washer method is looking good here. We need to figure out the limits, and then decide what our inner and outer radius are. If you need more help, I can return to this if you reply soon (I just saw you are offline), or just bump it and someone else can help. Good luck!

7. amonoconnor

Sorry, my iPad does show me as inactive in OS even when I'm not some times... Annoying :p Anyways, yes, I used the Washer Method. I determined... (And please tell me if any of this is incorrect) Boundaries: since we are revolving around the x-axis, they are x-values of: 1, 4. R: 2 r: (2/(x^(1/2))

8. AccessDenied

I agree with that information, yes. :)

9. amonoconnor

After taking the anti-derivative, but before plugging in the boundaries to solve, I had: "pi(4x - 4(ln(x))) 1|4" ?

10. AccessDenied

Looks good!

11. amonoconnor

Sweeeet! :)

12. amonoconnor

And... For a final answer, I am getting "4pi(4 - ln(4))" ??

13. amonoconnor

*units cubed

14. AccessDenied

Looks like something was lost. What was the steps after you plugged in your limits of integration?

15. amonoconnor

What I have on my paper: = pi(4x - 4(ln(x))) 1|4 = [pi(4(4) - 4(ln(4)))] - [pi(4(1) - 4(ln(1)))] = [pi(16 - 4(ln4))] - [pi(4 - 0)] = 16pi - 4pi(ln4) - 4pi = 4pi(4 - ln(4)) units^3

16. amonoconnor

Oh!!! I think I se a mistake... Is it that I didn't subtract 4pi from 16pi, so to should be: 4pi(3 - ln(4))

17. AccessDenied

Yep, there you go! I was thinking either that or you made the other x=1 limit all 0 (just because ln 1 = 0). But that fixes it! :D

18. amonoconnor

Alright, sweet, so the final answer is indeed 4pi(3 - ln(4)) ? :D

19. AccessDenied

That's what I got as well.

20. amonoconnor

Awesome! Thank you very much for your patience and willingness to work with me! It helped, and I fully understand what went wrong, I'm not just confused. Thank you!

21. AccessDenied