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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 xaxis.
Any and all help is greatly appreciated! :)
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 xaxis. Any and all help is greatly appreciated! :)

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anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0\[y=\frac{ 2 }{ \frac{ x }{ 2 } }\]

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0so y =\[\frac{ 4 }{ x }\]

amonoconnor
 2 years ago
Best ResponseYou've already chosen the best response.1No... 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.

AccessDenied
 2 years ago
Best ResponseYou've already chosen the best response.1We 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 xaxis. First, did you have any ideas on how you might set this one up? :)

AccessDenied
 2 years ago
Best ResponseYou've already chosen the best response.1Since the region is not connected to the xaxis, 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!

amonoconnor
 2 years ago
Best ResponseYou've already chosen the best response.1Sorry, 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 xaxis, they are xvalues of: 1, 4. R: 2 r: (2/(x^(1/2))

AccessDenied
 2 years ago
Best ResponseYou've already chosen the best response.1I agree with that information, yes. :)

amonoconnor
 2 years ago
Best ResponseYou've already chosen the best response.1After taking the antiderivative, but before plugging in the boundaries to solve, I had: "pi(4x  4(ln(x))) 14" ?

amonoconnor
 2 years ago
Best ResponseYou've already chosen the best response.1And... For a final answer, I am getting "4pi(4  ln(4))" ??

AccessDenied
 2 years ago
Best ResponseYou've already chosen the best response.1Looks like something was lost. What was the steps after you plugged in your limits of integration?

amonoconnor
 2 years ago
Best ResponseYou've already chosen the best response.1What I have on my paper: = pi(4x  4(ln(x))) 14 = [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

amonoconnor
 2 years ago
Best ResponseYou've already chosen the best response.1Oh!!! I think I se a mistake... Is it that I didn't subtract 4pi from 16pi, so to should be: 4pi(3  ln(4))

AccessDenied
 2 years ago
Best ResponseYou've already chosen the best response.1Yep, 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

amonoconnor
 2 years ago
Best ResponseYou've already chosen the best response.1Alright, sweet, so the final answer is indeed 4pi(3  ln(4)) ? :D

AccessDenied
 2 years ago
Best ResponseYou've already chosen the best response.1That's what I got as well.

amonoconnor
 2 years ago
Best ResponseYou've already chosen the best response.1Awesome! 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!

AccessDenied
 2 years ago
Best ResponseYou've already chosen the best response.1Always glad to help! :D
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