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amonoconnor
 11 months 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
 11 months 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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faisalalif1999
 11 months ago
Best ResponseYou've already chosen the best response.0\[y=\frac{ 2 }{ \frac{ x }{ 2 } }\]

faisalalif1999
 11 months ago
Best ResponseYou've already chosen the best response.0so y =\[\frac{ 4 }{ x }\]

amonoconnor
 11 months 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
 11 months 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
 11 months 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
 11 months 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
 11 months ago
Best ResponseYou've already chosen the best response.1I agree with that information, yes. :)

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

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

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

amonoconnor
 11 months 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
 11 months ago
Best ResponseYou've already chosen the best response.1Always glad to help! :D
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