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?

- anonymous

- katieb

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- anonymous

what the devil is this?

- anonymous

Area of a Surface of Revolution

- anonymous

surface area ?

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## More answers

- anonymous

yes

- anonymous

let me make some drawing

- anonymous

- anonymous

|dw:1341858624540:dw|

- anonymous

we are taking circumference of each plate , they have height(radius) of 'y'
2 Pi r= 2 Pi y

- anonymous

what if we're rotating about the y axis...would it still be 2 pi y

- anonymous

no

- anonymous

it would be much more complicated

- anonymous

2 pi x

- anonymous

not that simple

- anonymous

first ,you gotta define function in term of x

- anonymous

ok

- anonymous

as in x= .....y....

- anonymous

or x=g(y)

- anonymous

it is too complicated unlike finding volume , we just use x axis

- anonymous

ok

- anonymous

|dw:1341859119434:dw|

- anonymous

\[S=\int 2\pi x ds\]

- anonymous

but the book still has (dy/dx)^2

- anonymous

@helder_edwin please help

- anonymous

it is 3 dimensional

- helder_edwin

give a second to check my books
it's been a long time

- anonymous

ok

- anonymous

Everyone abandoned me :'(

- helder_edwin

no

- 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

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

ok

- 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

Yes

- 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

can you read spanish?

- TuringTest

yo si

- anonymous

sorry I can't

- anonymous

lol. That guy.

- TuringTest

here is a nice full explanation if you care to dig deeper
http://tutorial.math.lamar.edu/Classes/CalcII/SurfaceArea.aspx

- anonymous

I will , thanks @TuringTest .

- TuringTest

welcome :)

- 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

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

oh ok....thanks everyone Thanks @SmoothMath !!!

- anonymous

My pleasure =D

- TuringTest

@helder_edwin me lo mandas por favor? quiero aprender mas la terminologia en espanol

- helder_edwin

claro!

- helder_edwin

@TuringTest recibiste el pdf?

- 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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