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 one year ago
have to do a shell volume; can do it easy with disc but that's not the question asked. bounded by y=1/x^2; y=0; x=1; x=4.about the x axis.... I have gotten this:
\[v=2\pi \int\limits_{0}^{1 }(\sqrt{y}/y)(y1) dy\]
 one year ago
have to do a shell volume; can do it easy with disc but that's not the question asked. bounded by y=1/x^2; y=0; x=1; x=4.about the x axis.... I have gotten this: \[v=2\pi \int\limits_{0}^{1 }(\sqrt{y}/y)(y1) dy\]

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yamanaha
 one year ago
Best ResponseYou've already chosen the best response.0answer with disc is 21/64 pi...which is correct..but I cannot get it with shell. Help please!

yamanaha
 one year ago
Best ResponseYou've already chosen the best response.0OH..about x axis. sorry

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0shell is simple enough, so long as you remember how to find the area of a rectangle ...

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0y=1/x^2; y=0; x=1; x=4 dw:1365100001202:dw

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0you might wanna go ahead and shift the graph by xa to get it looking somewhat normal

yamanaha
 one year ago
Best ResponseYou've already chosen the best response.0ok, I had the r and h the right direction. I guess I am having trouble finding the lengths.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0h is from a to b, radius is from f(a) to f(b)

yamanaha
 one year ago
Best ResponseYou've already chosen the best response.0since the equation is in terms of x; I resolved in terms of y since shell with the rotation about x axis means I have to be in terms of y and use dy. correct?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0\[\int_{a}^{b}2\pi~r ~h\] \[\int_{a}^{b}2\pi~f(x) ~x~dx\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0one issue i see that i alluded to but dint take care of

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0the height at a is not x=a, its xa \[\int_{a}^{b}2\pi~f(x) ~(xa)~dx\]

yamanaha
 one year ago
Best ResponseYou've already chosen the best response.0you have dx...thought it had to be dy for x axis rotation for shell method?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0no, we are integrating aling the height, which in this case runs along the x axis. one final remark is this. We have a missing core when we get to b that is just a solid cylindar to fill in dw:1365100474873:dw

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0Volume of a cylindar is: pi r^2 h pi f(b)^2 (ba)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0we could integrate it with respect to the radius .... but that really doesnt seem necessary to me looking at the picture

yamanaha
 one year ago
Best ResponseYou've already chosen the best response.0the teacher said if you have shell and revolve area about x axis it will make r be the vertial length and h be the horizontal length like you have; but said the integrand will be y values and will be dy....because is shell method not disc....?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0forget about calling them x and y, it just throws in extra stuff that is really not related to height and radius for a shell ..... there are 2 ways we can approach this shell, drdh, or dhdr, all that changes is the order in which we run it. dw:1365100962229:dw dw:1365101040649:dw

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0lets fall f(x) ... g(y) in the bottom case \[\int_{r=0}^{r=f(a)}\int_{h=a}^{h=g(y)}~2\pi rh~dhdr\] OR \[\int_{h=a}^{h=b}\int_{r=0}^{r=f(x)}~2\pi rh~drdh\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0i prolly messed that up some putting it into double integration ... but the point is, it can be "run" along the x axis, framing the radius and height in terms of x, thereby having a dx

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0or, it can be framed in terms of radius and height in terms of y, thereby creating an inegration of dy

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0we can see that the area of the shell at h=(x1) , and radius = f(x) from x=1 to 4 gives us 2pi f(1) * 0 2pi f(3/2) * 1/2 2pi f(2) * 1 2pi f(5/2) * 3/2 2pi f(3) * 2

yamanaha
 one year ago
Best ResponseYou've already chosen the best response.0i'm looking...so confused. I thought depending on which way you revolved it had to be a set thing like if revolved about x; use dy and y value....if revolved about y; use dx and use x values for shell method. is this statement not correct?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0that statement is a general rule of thumb but is contingent on how you are moving and what is variable along the way

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0if you stick to trying to figure these things out by remembering certain formula per se, your not going to get the overall picture and be stuck on a test going .... what was the rule i should use? We know the basic set up for a shell has these components, and how you define them will depend on dx or dy \[\int2\pi~r~h\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0now, one way or the other may be simpler to work out .... so knowing both ways to address it is good

yamanaha
 one year ago
Best ResponseYou've already chosen the best response.0ok; that makes sense to not rely on formula. if I were to revolve about x axis and want my parameters on the integral to be y values; would you say the one length would be 1/(sqrt of y) ?

yamanaha
 one year ago
Best ResponseYou've already chosen the best response.0I see the x set up fine; i cannot see the y setup and need to see both as you say ...sometimes need one or the other. for some reason the y setup gets me in this case.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0i agree :) ill address the y setup in a minute, i just want to run thru this x setup .... helps me remember the things ive forgoten :) r = f(x), h=(x1), along the interval of 1 to 4, and remember to addon the hollow core \[2pi\int_{1}^{4}rh\] \[2pi\int_{1}^{4}\frac{x1}{x^2}dx\] \[2pi\int_{1}^{4}\frac{x}{x^2}\frac{1}{x^2}dx\] \[2pi\int_{1}^{4}\frac{1}{x}{x^{2}}dx+3pi(\frac{1}{16})^2\] this is very doable yes?

yamanaha
 one year ago
Best ResponseYou've already chosen the best response.0if I am spinning a shell about the x axis; the radius would be the length from the function to y=0. would I write that as (y) or as (1/sqrt y). i know the ordered pair on the function is (x,y) but y is the function. i get confused when to use y and when to use the function equation.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.01/x^2 at x=4 is 1/16 for a radius of the core volume of a cylindar is pi r^2 h: h=1 to 4 = 3 3pi (1/16)^2 to fill in the hollow core that was left over

yamanaha
 one year ago
Best ResponseYou've already chosen the best response.0oh your intuitively doing the math not by technique..is that true?

yamanaha
 one year ago
Best ResponseYou've already chosen the best response.0like logically figuring shape;etc.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0yes, im applying stuff from geometry way back when that is still applicable :) we could run a integration on it but it would be a little silly to me in terms of y, r = f(a) to f(b); h = g(y) = x = 1/y^2

yamanaha
 one year ago
Best ResponseYou've already chosen the best response.0problem is the teacher is going to grade on technique of washer method...and it has to be exactly documented as that or it's 100 percent wrong.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0x = 1/y^2 y^2 = 1/x y = + 1/sqrt(x) for the height

yamanaha
 one year ago
Best ResponseYou've already chosen the best response.0Ha; ok...well I agree with you totally!!! wish you were the teacher :)

yamanaha
 one year ago
Best ResponseYou've already chosen the best response.0however; i follow your logic...much like mine...but it won't please her to do this. that is why I am stuck. it's not like if I get the right answer it is ok...has to be her format.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0dw:1365102773341:dw

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0i just orientated it in a more normal looking fashion

yamanaha
 one year ago
Best ResponseYou've already chosen the best response.0x goes from 1 to 4..?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0x goes from 1 to 4 hasnt changed any, its an inherent part of the problem

yamanaha
 one year ago
Best ResponseYou've already chosen the best response.0so in this picture can you help me see what a vertical line from bottom to top of shape would be for length? Is it simply y or is it 1/(sqrt y) ?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0might have to subtract 1 again , to get a zero there

yamanaha
 one year ago
Best ResponseYou've already chosen the best response.0why did you shift up?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0i didnt shift up, thats just the way the problem looks when orientated in this manner \[h=0 \to (\frac{1}{\sqrt{y}}1)\]\[y = \frac12\to1\]

yamanaha
 one year ago
Best ResponseYou've already chosen the best response.0oh dear...feeling lost. I think I will take a break and look at your comments in a little. maybe it will sink in better. thanks so very much for all your help!!

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0\[\int 2pi~rh\] \[\int_{\frac12}^{1} 2pi~y(\frac{1}{\sqrt y}1)~dy\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0notice that we still have an empty core

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0y/sqrty = sqrty \[2pi~\int_{\frac12}^{1} \sqrt yy~~dy+(core)\]

yamanaha
 one year ago
Best ResponseYou've already chosen the best response.0got it! thanks . the core adding helped. later...

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0dbl chk my limits on y tho

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0yeah, 1/16 not 1/2 .. :)
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