How would you draw a cylinder for the shell method for the region that is enclosed by x = pi/2, x = -pi/2 and y = cosx?
This is rotated about y = -1.

- anonymous

- schrodinger

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- anonymous

@FoolForMath
@myininaya
@experimentX
@shivam_bhalla
@amistre64
@Zarkon
@Hero
Anyone?

- anonymous

@jim_thompson5910

- anonymous

the definite integrand of f(x)=cosx from -pi/2 to pi/2
sinx from -pi/2 to pi/2 = 1-(-1)=2

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## More answers

- anonymous

I do not want the answer. @akamushi

- anonymous

@FoolForMath
@myininaya
@experimentX
@shivam_bhalla
@amistre64
@Zarkon
@Hero
Anyone?

- anonymous

@FoolForMath
@myininaya
@experimentX
@shivam_bhalla
@amistre64
@Zarkon
@Hero
Anyone?

- lgbasallote

@lgbasallote lol jk...idk shell using trig functions

- anonymous

|dw:1337557715702:dw|

- anonymous

The height would be 2cosx, and the radius would be cosx - (-1)?

- anonymous

ie.
h = 2y
r = y - (-1)
?

- anonymous

|dw:1337558085638:dw|

- anonymous

The height (which are the horizontal lines of the cylinder) would be given by
h = 2cosx?
And the radius would be given by r = cos(x) - (-1)?

- anonymous

Wait, I think the height would be:
h = 2x, due to symmetry

- anonymous

|dw:1337558202025:dw|

- anonymous

So then, radius is r = cos(x) - (-1) for sure, right?

- anonymous

And I can restate that cos(x) as y
So then, r = y +1, h = 2x = 2arccos(y), I believe.

- anonymous

now you are on the right track.

- anonymous

if you are familiar with winplot, you can make a nice 3d representation.

- anonymous

@shivam_bhalla! Yes you finally came.

- anonymous

|dw:1337558589994:dw|

- anonymous

I have one question for either of you:
I drew a vertical line from the x-axis to the function cos(x). Can that generically be represented as y?

- anonymous

@QRAwarrior , try this http://openstudy.com/updates/4fb44286e4b05565342a2a03
You will find it a lot more helpful :)
The answer to your question you asked before is yes

- amistre64

x = pi/2, x = -pi/2 and y = cosx|dw:1337561871090:dw|

- amistre64

|dw:1337562804283:dw|

- anonymous

Ok so it is:
h = x
y = cosx - (-1)

- amistre64

we only need half of this, then double the results
height is a measure of arccos(y)
radius is from from 1 to 2 by perspective
\[2*2pi\int_{0}^{1}(y+1)*\cos^{-1}(y)\ dy\]

- anonymous

I think I came up with this a few posts back

- amistre64

height is a function of x;
y=cos(x)
arccos(y) = x = height

- anonymous

--------------------------------------------------------
Ok consider this question:
Find the length of y^2 = x^3 from (0,0) to (4,8)
I know the formula for arclength, but I end up getting this:
|dw:1337563218324:dw|

- anonymous

|dw:1337563244465:dw|

- anonymous

How the heck can you resolve the radical when you have:
|dw:1337563293983:dw|
As your argument?

- amistre64

have you tried looking at it form both forms?
\[\sqrt{1+(x')^2}\ or\ \sqrt{1+(y')^2}\]

- anonymous

No. I think I will try the other form, BUT the problem is that I am not sure whether to pick the POSITIVE root or the NEGATIVE root..

- amistre64

for the most part, i think you stick with the positive root

- anonymous

Why do you say "most part"

- amistre64

distance tends to be a measure in absolute value, so the positive root seems most apt

- anonymous

WOW! I just realized how EASY it is doing it wrt x.

- anonymous

|dw:1337563549037:dw|

- anonymous

Easy U-sub. Thanks a lot amistre64. What level of education is your mathematics?

- anonymous

I was just curious

- amistre64

i dunno what level it is, im guessing its at least on a 3rd grade level ;)

- anonymous

Ok anyways, thanks for your help

- amistre64

youre welcome. Levelwise is difficult since im self taught for the most part of this stuff

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