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Math Help - The Relationship between the Surface Area of a Curve and A Circle's Circumference

  1. #1
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    The Relationship between the Surface Area of a Curve and A Circle's Circumference

    In my BC Calc class, we are learning about the formula for finding the surface area of curves of revolution. I don't like just memorizing formulas without understanding why they work, so I have set out to discover how the formula for the surface area of a curve was derived.

    I'm concerned as to why the formula does not represent the sum of the circumference of the infinite number of circles formed by the curve, rather than the sum of surface area of an inifinite number of cylinders. When I solve problems using my hypothetical method I get a different answer than I would using the formula I have been taught.

    Can anyone please help me to understand what is wrong with the formula I made up?
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by JewelsofHearts View Post
    In my BC Calc class, we are learning about the formula for finding the surface area of curves of revolution. I don't like just memorizing formulas without understanding why they work, so I have set out to discover how the formula for the surface area of a curve was derived.

    I'm concerned as to why the formula does not represent the sum of the circumference of the infinite number of circles formed by the curve, rather than the sum of surface area of an inifinite number of cylinders. When I solve problems using my hypothetical method I get a different answer than I would using the formula I have been taught.

    Can anyone please help me to understand what is wrong with the formula I made up?
    We derive the formula for surface area in much the same way that we derived the formula for arc length. So, take a curve in the plane y=f(x), and for simplicity assume that the curve exists in the first quadrant over the interval [a,b]. Begin by slicing the interval into subinterval of length \Delta{x} and on each subinterval we will approximate the lentgh of the function by summing the straight lines. Then Rotate! That's it. The rotation genterates a bunch of surface areas of frustrums and the formula for the surface of a frustrum is A_f=2\pi{r}l where r is the average radius of the left hand radius and the right hand radius, or r=\frac{1}{2}(r_1+r_2).

    For the frustrum on the interval [x_{i-1},x_i] we have r_1=f(x_i) and r_2=f(x_{i-1}).

    Now the length of the ith line segment is given by \sqrt{1+[f'(x)]^2}\Delta{x}. This can be derived and I can do it if you like. So, we have

    A_i=2\pi\left(\frac{f(x_i)+f(x_{i-1})}{2}\right)\sqrt{1+[f'(x)]^2}\Delta{x}. But if delta x is small, we can say that A_i=2\pi{f(c_i)}\sqrt{1+[f'(c_i)]^2}\Delta{x} for some c_i\in[x_{i-1},x_i]

    Summing and taking the limit we get

    \displaystyle{A}=\lim_{n\to\infty}\sum_{i=1}^\inft  y2\pi{f(c_i)}\sqrt{1+[f'(c_i)]^2}\Delta{x}=\int_a^b2\pi{f(x)}\sqrt{1+[f'(x)]^2}dx
    Last edited by VonNemo19; February 9th 2011 at 10:45 AM.
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  3. #3
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    Incidentally, JewelsofHearts, you might want to make your language more precise. All space curves have zero surface area. (I don't think you were taking about space curves - perhaps surfaces?)
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Ah, what the heck. I'll go ahead and derive the other part for ya.

    OK. You know that if I break a smooth curve up into a bunch of little straight lines, then from two consecutive points on the curve, P_{i-1} and P_i a line can be drawn. The distance between these two points will be given by the distance formula like so

    d_{P_{i-1}P_i}=\sqrt{[f(x_{i})-f(x_{i-1})]^2+(x_i-x_{i-1})^2}.

    Now, changing notation for convenience, we let f(x_{i})-f(x_{i-1})=\Delta{y} and x_i-x_{i-1}=\Delta{x} sucht that

    d_{P_{i-1}P_i}=\sqrt{\Delta{y}^2+\Delta{x}^2}

    =\sqrt{\Delta{x}^2(1+\frac{\Delta{y}^2}{\Delta{x}^  2})}=\Delta{x}\sqrt{1+(\frac{\Delta{y}}{\Delta{x}}  )^2}.

    But, \frac{\Delta{y}}{\Delta{x}}=\frac{f(x_i)-f(x_{i-1})}{x_i-x_{i-1}}=f'(c_i) for c_i\in[x_{i-1},x_1] by the mean value theorem.

    So therefore d=\Delta{x}\sqrt{1+[f'(c_i)]^2}
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  5. #5
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    Wow, VonNemo19! That was a lengthy explanation. Thanks for taking the time to do that.

    What I was really wondering, though, was if the surface area of a curve of revolution could be represented by the sum of the circumferences of an inifinite number of circles on a given interval.
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