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MathSofiya
Group Title
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?
 2 years ago
 2 years ago
MathSofiya Group Title
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?
 2 years ago
 2 years ago

This Question is Closed

Libniz Group TitleBest ResponseYou've already chosen the best response.0
what the devil is this?
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
Area of a Surface of Revolution
 2 years ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
surface area ?
 2 years ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
let me make some drawing
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
@TuringTest
 2 years ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
dw:1341858624540:dw
 2 years ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
we are taking circumference of each plate , they have height(radius) of 'y' 2 Pi r= 2 Pi y
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
what if we're rotating about the y axis...would it still be 2 pi y
 2 years ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
it would be much more complicated
 2 years ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
not that simple
 2 years ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
first ,you gotta define function in term of x
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
as in x= .....y....
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
or x=g(y)
 2 years ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
it is too complicated unlike finding volume , we just use x axis
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
dw:1341859119434:dw
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
\[S=\int 2\pi x ds\]
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
but the book still has (dy/dx)^2
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
@helder_edwin please help
 2 years ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
it is 3 dimensional
 2 years ago

helder_edwin Group TitleBest ResponseYou've already chosen the best response.1
give a second to check my books it's been a long time
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
Everyone abandoned me :'(
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
I didn't get your ping... weird I'll have to ask administration about that... ok, what do we ave here, let me read...
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
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...
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
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?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
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 viceversa, so each of these formulas is potentially two depending on how we look at our ds
 2 years ago

helder_edwin Group TitleBest ResponseYou've already chosen the best response.1
can you read spanish?
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
sorry I can't
 2 years ago

SmoothMath Group TitleBest ResponseYou've already chosen the best response.0
lol. That guy.
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
here is a nice full explanation if you care to dig deeper http://tutorial.math.lamar.edu/Classes/CalcII/SurfaceArea.aspx
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
I will , thanks @TuringTest .
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
welcome :)
 2 years ago

SmoothMath Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

helder_edwin Group TitleBest ResponseYou've already chosen the best response.1
sorry i wanted to send you something but it's in spanish. but i goes much in the same way as what @TuringTest did
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.1
oh ok....thanks everyone Thanks @SmoothMath !!!
 2 years ago

SmoothMath Group TitleBest ResponseYou've already chosen the best response.0
My pleasure =D
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
@helder_edwin me lo mandas por favor? quiero aprender mas la terminologia en espanol
 2 years ago

helder_edwin Group TitleBest ResponseYou've already chosen the best response.1
claro!
 2 years ago

helder_edwin Group TitleBest ResponseYou've already chosen the best response.1
@TuringTest recibiste el pdf?
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
no, donde lo pusiste? podrias mandarme un ("link"?) o url ? ya puesto que estoy tu "fan" es posible mandar mensajes privadas
 2 years ago
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