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

OH..about x axis. sorry

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

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

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

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

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

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

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

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

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

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

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

why 3pi(1/16)^2?

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

like logically figuring shape;etc.

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

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

|dw:1365102773341:dw|

i just orientated it in a more normal looking fashion

x goes from 1 to 4..?

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

h = 1 to 1/sqrt(y)

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

why did you shift up?

r = y

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

notice that we still have an empty core

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

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

dbl chk my limits on y tho

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