anonymous
  • anonymous
just making sure I understand it.... \[S=\int_{a}^{b} 2\pi y \sqrt{1+{(\frac{dy}{dx})}^2} dx\] \[S=\int_{c}^{d} 2\pi y \sqrt{1+{(\frac{dx}{dy})}^2} dy\] \[S=\int_{a}^{b} 2\pi x \sqrt{1+{(\frac{dy}{dx})}^2} dx\] One is for revolution about the x axis...the other for revolution about the y- axis...and then we have one more equation....why?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
what the devil is this?
anonymous
  • anonymous
Area of a Surface of Revolution
anonymous
  • anonymous
surface area ?

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anonymous
  • anonymous
yes
anonymous
  • anonymous
let me make some drawing
anonymous
  • anonymous
@TuringTest
anonymous
  • anonymous
|dw:1341858624540:dw|
anonymous
  • anonymous
we are taking circumference of each plate , they have height(radius) of 'y' 2 Pi r= 2 Pi y
anonymous
  • anonymous
what if we're rotating about the y axis...would it still be 2 pi y
anonymous
  • anonymous
no
anonymous
  • anonymous
it would be much more complicated
anonymous
  • anonymous
2 pi x
anonymous
  • anonymous
not that simple
anonymous
  • anonymous
first ,you gotta define function in term of x
anonymous
  • anonymous
ok
anonymous
  • anonymous
as in x= .....y....
anonymous
  • anonymous
or x=g(y)
anonymous
  • anonymous
it is too complicated unlike finding volume , we just use x axis
anonymous
  • anonymous
ok
anonymous
  • anonymous
|dw:1341859119434:dw|
anonymous
  • anonymous
\[S=\int 2\pi x ds\]
anonymous
  • anonymous
but the book still has (dy/dx)^2
anonymous
  • anonymous
@helder_edwin please help
anonymous
  • anonymous
it is 3 dimensional
helder_edwin
  • helder_edwin
give a second to check my books it's been a long time
anonymous
  • anonymous
ok
anonymous
  • anonymous
Everyone abandoned me :'(
helder_edwin
  • helder_edwin
no
TuringTest
  • TuringTest
I didn't get your ping... weird I'll have to ask administration about that... ok, what do we ave here, let me read...
TuringTest
  • TuringTest
so you want to know why we have 4 formulas, right? my answer is that there are really only two, but each one can be seen from two different perspectives...
anonymous
  • anonymous
ok
TuringTest
  • TuringTest
first consider the arc length formula:\[ds=\sqrt{1+[f'(x)]^2}dx\]now this is the formula for arc length taken from the perspective of y being a funcion of x but arc length is the same regardless of whether you look at the function as f(x) or g(y) since the arc itself will still have the same length. so we can also write\[ds=\sqrt{1+[g'(y)]^2}dy\] and as long as we are talking about the same curve they should be equal, since the arc can only have one length. make sense so far?
anonymous
  • anonymous
Yes
TuringTest
  • TuringTest
now for a revolution, the formula is\[A=\int2\pi yds\]or\[A=\int2\pi xds\]depending on which axis we are going around but as I just explained above, ds (the arc length differential) can always be written two ways depending on whether we consider y a function of x or vice-versa, so each of these formulas is potentially two depending on how we look at our ds
helder_edwin
  • helder_edwin
can you read spanish?
TuringTest
  • TuringTest
yo si
anonymous
  • anonymous
sorry I can't
anonymous
  • anonymous
lol. That guy.
TuringTest
  • TuringTest
here is a nice full explanation if you care to dig deeper http://tutorial.math.lamar.edu/Classes/CalcII/SurfaceArea.aspx
anonymous
  • anonymous
I will , thanks @TuringTest .
TuringTest
  • TuringTest
welcome :)
anonymous
  • anonymous
Sofiya, try to break down each integral like this: First, look at the end part, the variable you're integrating with respect to. Then, look at the limits, so you say to yourself, "Okay, travelling along x from a to b." or something like that. Then, look at the function inside and try to break that down, and what that means at each particular x. For these particular integrals, the insides have 2 basic parts. The first part has the form 2pi*something, where that something is the radius. The second part is the arclength formula. So it's calculating the arclength, and then it's multiplying that in a circle.
helder_edwin
  • helder_edwin
sorry i wanted to send you something but it's in spanish. but i goes much in the same way as what @TuringTest did
anonymous
  • anonymous
oh ok....thanks everyone Thanks @SmoothMath !!!
anonymous
  • anonymous
My pleasure =D
TuringTest
  • TuringTest
@helder_edwin me lo mandas por favor? quiero aprender mas la terminologia en espanol
helder_edwin
  • helder_edwin
claro!
helder_edwin
  • helder_edwin
@TuringTest recibiste el pdf?
TuringTest
  • TuringTest
no, donde lo pusiste? podrias mandarme un ("link"?) o url ? ya puesto que estoy tu "fan" es posible mandar mensajes privadas

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