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    I'm looking at a proof of the formula for the area of a surface of revolution:

    The points P (x,y) and Q (x+dx, y+dy) are close together. The surface area generated by an arc between P and Q (length ds), rotated about the x-axis, is dS.

    dS = surface area of a cylinder = circumference x height
    dS = 2 \pi y \times ds

S=\int 2 \pi y\ ds

    with some more work,

    S = 2 \pi \int y \sqrt {1+ (\frac{dy}{ds})^2}\ dx

    What I want to know is this: if P and Q are so close together that ds is a straight horizontal line, then ds must be the same as dx, so surely dx could also be used as the height of the cylinder, giving S=\int 2 \pi y\ dx. Why is this not possible?
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    (Original post by bzzz)
    with some more work,

    S = 2 \pi \int y \sqrt {1+ (\frac{dy}{ds})^2}\ dx
    I think you mean:

    S = 2 \pi \int y \sqrt {1+ (\frac{dy}{dx})^2}\ dx

    What I want to know is this: if P and Q are so close together that ds is a straight horizontal line, then ds must be the same as dx, so surely dx could also be used as the height of the cylinder, giving S=\int 2 \pi y\ dx. Why is this not possible?
    ds is not horizontal; it's tangential to the curve.
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    (Original post by ghostwalker)
    I think you mean:

    S = 2 \pi \int y \sqrt {1+ (\frac{dy}{dx})^2}\ dx



    ds is not horizontal; it's tangential to the curve.
    Surely if P and Q were so close that they were practically on top of each other, then ds would be horizontal?
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    (Original post by bzzz)
    Surely if P and Q were so close that they were practically on top of each other, then ds would be horizontal?
    Why would it be horizontal, in your mind? Think about the definition of the derivative (it uses the same principles). If this were the case and all lines between two infinitely close points were horizontal, then if we were to consider the derivative of any function at any point, the answer would always be zero. Which is obviously not true.

    The line ds has the same gradient as the curve does at the point (x,y).
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    (Original post by Farhan.Hanif93)
    Why would it be horizontal, in your mind? Think about the definition of the derivative (it uses the same principles). If this were the case and all lines between two infinitely close points were horizontal, then if we were to consider the derivative of any function at any point, the answer would always be zero. Which is obviously not true.

    The line ds has the same gradient as the curve does at the point (x,y).
    Alright. Does that mean the proof isn't correct?
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    (Original post by bzzz)
    Alright. Does that mean the proof isn't correct?
    The result at the end is correct (if you correct the mistake/typo that Ghostwalker pointed out) and the work leading up to it is fine, I think. I'm not too sure what would constitute a "proof" in this case as I'm not too sure how much rigour is required. You've certainly shown it to be true, with an assumption or two along the way, at the very least.

    The statement that ds is horizontal is definitely not correct.
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    (Original post by Farhan.Hanif93)
    The result at the end is correct (if you correct the mistake/typo that Ghostwalker pointed out) and the work leading up to it is fine, I think. I'm not too sure what would constitute a "proof" in this case as I'm not too sure how much rigour is required. You've certainly shown it to be true, with an assumption or two along the way, at the very least.

    The statement that ds is horizontal is definitely not correct.
    If ds is not horizontal, then wouldn't the use of a cylinder to calculate S be incorrect?
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    (Original post by bzzz)
    If ds is not horizontal, then wouldn't the use of a cylinder to calculate S be incorrect?
    Hmm, so maybe you shouldn't use cylinders and rather use frustrums? It's easy to determine the general formula for the slant-face surface areas of them.

    I think it would be best to wait for someone with a better knowledge of this than me to come and help out.
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    (Original post by Farhan.Hanif93)
    Hmm, so maybe you shouldn't use cylinders and rather use frustrums? It's easy to determine the general formula for the slant-face surface areas of them.

    I think it would be best to wait for someone with a better knowledge of this than me to come and help out.
    Yeah, all the other proofs of this I've seen use frustrums rather than cylinders, which makes more sense.
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    (Original post by bzzz)
    If ds is not horizontal, then wouldn't the use of a cylinder to calculate S be incorrect?
    As Farhan.Hanif93 said, you actually have a frustrum of a cone when considering a small change in s.

    I'm not going to type out all the maths, but you basically end up with.

    \displaystyle\frac{\delta S}{\delta s}= 2\pi y + \delta s c_s


    Where c_s is a constant dependent on s.

    Then taking the limit as \delta s goes to zero, you end up with the original formula you had.

    \displaystyle\frac{dS}{ds}= 2\pi y

    Which is what you generated from considering a cylinder rather than a frustrum of a cone.
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    (Original post by ghostwalker)
    As Farhan.Hanif93 said, you actually have a frustrum of a cone when considering a small change in s.

    I'm not going to type out all the maths, but you basically end up with.

    \displaystyle\frac{\delta S}{\delta s}= 2\pi y + \delta s c_s


    Where c_s is a constant dependent on s.

    Then taking the limit as \delta s goes to zero, you end up with the original formula you had.

    \displaystyle\frac{dS}{ds}= 2\pi y

    Which is what you generated from considering a cylinder rather than a frustrum of a cone.
    Thanks
 
 
 
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