Question

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! :)

Answer 1

so basically

Answer 2

\[y=\frac{ 2 }{ \frac{ x }{ 2 } }\]

Answer 3

so y =\[\frac{ 4 }{ x }\]

Answer 4

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.

Answer 5

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? :)

Answer 6

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!

Answer 7

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))

Answer 8

I agree with that information, yes. :)

Answer 9

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

Answer 10

Looks good!

Answer 11

Sweeeet! :)

Answer 12

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

Answer 13

*units cubed

Answer 14

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

Answer 15

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

Answer 16

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))

Answer 17

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

Answer 18

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

Answer 19

That's what I got as well.

Answer 20

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!

Answer 21

Always glad to help! :D