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Surface area of the Sphere

  1. Jul 8, 2014 #1

    Nathanael

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    I'm wondering why my method for finding the surface area of a sphere is invalid.

    Essentially I'm integrating the perimeter of the circle perpendicular to the radius along the entire radius, and then multiplying by 2 (because the radius only covers half the sphere)
    (I hope that made sense; sorry, I really don't know how to say it properly.)

    The reason I'm so thrown off as to why this is wrong is because it worked correctly for the volume. I integrated the area of the circle perpendicular to the radius along the entire radius (and multiplied by 2) and got the correct value for volume: [itex](^{R}_{0}∫2\pi (R^2-x^2)dx)=\frac{4\pi r^3}{3}[/itex]

    My mathematics goes like this:
    The radius of the circle (perpendicular to the sphere's radius) at any distance x from the center of the sphere is [itex]\sqrt{R^2-x^2}[/itex] where R is the radius of the sphere

    So I need to integrate that from x=0 to x=R and multiply each step by 2pi (so the 2pi should factor out) but then I need to multiply it by 2 because that integral only accounts for half of the surface area so I get:
    [itex]4\pi (^{R}_{0}∫\sqrt{R^2-x^2})=(\pi r)^2\neq4\pi r^2[/itex]



    Why did it work for the volume but not the surface area? Did I make a mistake with the surface area? Was it just a coincidence that it worked for the volume? Is there a fundamental difference that I'm overlooking?

    (I realize the surface area is just the derivative of the volume, but that doesn't help for the problem I'm thinking about)

    Thanks for any help

    (Sorry if I was unclear about my method, I'm not sure how to explain it very well verbally.)
     
  2. jcsd
  3. Jul 8, 2014 #2

    HallsofIvy

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    Approximating the surface area by the sides of cylinders does not converge uniformly to the surface area. Instead approximate each section by the side of a truncated cone having the two disks cut off the sphere as top and bottom.
     
  4. Jul 8, 2014 #3

    Nathanael

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    Thank you. That seems to explain why my method yielded answers slightly below expected.

    At first I was under the false intuitive assumption that the surface area (of just the side part) of a truncated cone would be the average perimeter ([itex]2\pi \frac{R+r}{2}[/itex]) multiplied by the height, whereas it is actually multiplied by the "side length" from top to bottom.

    So instead of multiplying each infintesimal step by dx we would multiply it by a slightly larger factor that represents the "secant distance" (or "side length") (which should increase as you move out along the radius)
    I won't try to explain, (because as you can tell I'm not good at articulating mathematical thoughts) but I've concluded the factor should be [itex]\frac{dx}{\sqrt{1-(\frac{x}{R})^2}}[/itex]

    Integrating it against that instead of dx you get [itex]2\pi r^2[/itex] (and then multiplying by 2 for the other half gives you the correct surface area)


    Thank you for your insight, I am very grateful. You gave me just enough for me to understand my intuition's error, yet you didn't spoil the problem for me. Thanks
     
  5. Jul 8, 2014 #4

    Nathanael

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    I still don't know what you meant by this or where I was supposed to go from there but the rest of your post was very very helpful



    EDIT:

    I have another question;

    Why is it valid to calculate the volume via cylinders?
     
    Last edited: Jul 9, 2014
  6. Jul 9, 2014 #5

    HallsofIvy

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    ?? I just said it was NOT. If you meant "Why is it invalid", that's because cylinders are not the best linear approximation to the surface, truncated cones are.
     
  7. Jul 9, 2014 #6

    Nathanael

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    For surface area I had to rewrite my equations using truncated cones (thank you btw), but for volume I essentially used cylinders and got the correct answer.

    I asked why does it work for VOLUME
     
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