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QRAwarrior
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.
@FoolForMath @myininaya @experimentX @shivam_bhalla @amistre64 @Zarkon @Hero Anyone?
the definite integrand of f(x)=cosx from -pi/2 to pi/2 sinx from -pi/2 to pi/2 = 1-(-1)=2
I do not want the answer. @akamushi
@FoolForMath @myininaya @experimentX @shivam_bhalla @amistre64 @Zarkon @Hero Anyone?
@FoolForMath @myininaya @experimentX @shivam_bhalla @amistre64 @Zarkon @Hero Anyone?
@lgbasallote lol jk...idk shell using trig functions
The height would be 2cosx, and the radius would be cosx - (-1)?
ie. h = 2y r = y - (-1) ?
|dw:1337558085638:dw|
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)?
Wait, I think the height would be: h = 2x, due to symmetry
|dw:1337558202025:dw|
So then, radius is r = cos(x) - (-1) for sure, right?
And I can restate that cos(x) as y So then, r = y +1, h = 2x = 2arccos(y), I believe.
now you are on the right track.
if you are familiar with winplot, you can make a nice 3d representation.
@shivam_bhalla! Yes you finally came.
|dw:1337558589994:dw|
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?
@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
x = pi/2, x = -pi/2 and y = cosx|dw:1337561871090:dw|
|dw:1337562804283:dw|
Ok so it is: h = x y = cosx - (-1)
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 \[2*2pi\int_{0}^{1}(y+1)*\cos^{-1}(y)\ dy\]
I think I came up with this a few posts back
height is a function of x; y=cos(x) arccos(y) = x = height
-------------------------------------------------------- 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|
|dw:1337563244465:dw|
How the heck can you resolve the radical when you have: |dw:1337563293983:dw| As your argument?
have you tried looking at it form both forms? \[\sqrt{1+(x')^2}\ or\ \sqrt{1+(y')^2}\]
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..
for the most part, i think you stick with the positive root
Why do you say "most part"
distance tends to be a measure in absolute value, so the positive root seems most apt
WOW! I just realized how EASY it is doing it wrt x.
|dw:1337563549037:dw|
Easy U-sub. Thanks a lot amistre64. What level of education is your mathematics?
i dunno what level it is, im guessing its at least on a 3rd grade level ;)
Ok anyways, thanks for your help
youre welcome. Levelwise is difficult since im self taught for the most part of this stuff