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:
\[v=2\pi \int\limits_{0}^{1 }(\sqrt{y}/y)(y-1) dy\]

- anonymous

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- anonymous

answer with disc is 21/64 pi...which is correct..but I cannot get it with shell. Help please!

- anonymous

OH..about x axis. sorry

- amistre64

shell is simple enough, so long as you remember how to find the area of a rectangle ...

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## More answers

- amistre64

y=1/x^2; y=0; x=1; x=4
|dw:1365100001202:dw|

- amistre64

you might wanna go ahead and shift the graph by x-a to get it looking somewhat normal

- anonymous

ok, I had the r and h the right direction. I guess I am having trouble finding the lengths.

- amistre64

h is from a to b, radius is from f(a) to f(b)

- anonymous

since 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

\[\int_{a}^{b}2\pi~r ~h\]
\[\int_{a}^{b}2\pi~f(x) ~x~dx\]

- amistre64

one issue i see that i alluded to but dint take care of

- amistre64

the height at a is not x=a, its x-a
\[\int_{a}^{b}2\pi~f(x) ~(x-a)~dx\]

- anonymous

you have dx...thought it had to be dy for x axis rotation for shell method?

- amistre64

no, 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

Volume of a cylindar is: pi r^2 h
pi f(b)^2 (b-a)

- amistre64

we could integrate it with respect to the radius .... but that really doesnt seem necessary to me looking at the picture

- anonymous

the 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

forget 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

lets 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

i 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

or, it can be framed in terms of radius and height in terms of y, thereby creating an inegration of dy

- amistre64

we can see that the area of the shell at h=(x-1) , 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

- anonymous

i'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

that statement is a general rule of thumb but is contingent on how you are moving and what is variable along the way

- amistre64

if 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

now, one way or the other may be simpler to work out .... so knowing both ways to address it is good

- anonymous

ok; 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) ?

- anonymous

I 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

i 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=(x-1), along the interval of 1 to 4, and remember to addon the hollow core
\[2pi\int_{1}^{4}rh\]
\[2pi\int_{1}^{4}\frac{x-1}{x^2}dx\]
\[2pi\int_{1}^{4}\frac{x}{x^2}-\frac{1}{x^2}dx\]
\[2pi\int_{1}^{4}\frac{1}{x}-{x^{-2}}dx+3pi(\frac{1}{16})^2\]
this is very doable yes?

- anonymous

if 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.

- anonymous

why 3pi(1/16)^2?

- amistre64

1/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

- anonymous

oh your intuitively doing the math not by technique..is that true?

- anonymous

like logically figuring shape;etc.

- amistre64

yes, 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

- anonymous

problem 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

x = 1/y^2
y^2 = 1/x
y = +- 1/sqrt(x) for the height

- anonymous

Ha; ok...well I agree with you totally!!! wish you were the teacher :)

- anonymous

however; 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

|dw:1365102773341:dw|

- amistre64

i just orientated it in a more normal looking fashion

- anonymous

x goes from 1 to 4..?

- amistre64

x goes from 1 to 4 hasnt changed any, its an inherent part of the problem

- anonymous

so 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

h = 1 to 1/sqrt(y)

- amistre64

might have to subtract 1 again , to get a zero there

- anonymous

why did you shift up?

- amistre64

i didnt shift up, thats just the way the problem looks when orientated in this manner
\[h=0 \to (\frac{1}{\sqrt{y}}-1)\]\[y = \frac12\to1\]

- amistre64

r = y

- anonymous

oh 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

\[\int 2pi~rh\]
\[\int_{\frac12}^{1} 2pi~y(\frac{1}{\sqrt y}-1)~dy\]

- amistre64

notice that we still have an empty core

- amistre64

y/sqrty = sqrty
\[2pi~\int_{\frac12}^{1} \sqrt y-y~~dy+(core)\]

- anonymous

got it! thanks . the core adding helped. later...

- amistre64

dbl chk my limits on y tho

- amistre64

yeah, 1/16 not 1/2 .. :)

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