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Why is the formula for the surface area of a circle not (2*pi*r)^2?

  1. Sep 9, 2011 #1
    Actually I changed my mind and feel like it should be ((pi*r)(2*pi*r)) by my faulty thinking.

    Since pi*r would give you a line wrapped halfway around a sphere, I was thinking you could repeat this line in a radial pattern around the outside of a sphere (2*pi*r) times to get the surface area of the sphere.

    That kind of logic makes sense when taking the volume of a cylinder because I imagine taking the area of the circular disk base and stacking it (height) number of times to get the volume. The meters^2 of the area and meters of the height multiply to give the m^3 of volume. So I also expected the meters of the (pi*r) and the meters of the (2*pi*r) to multiply to give me the m^2 of surface area... :/




    Can someone explain in layman's terms what is wrong with this approach?
     
    Last edited: Sep 9, 2011
  2. jcsd
  3. Sep 9, 2011 #2

    Pengwuino

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    No, why would you think that?

    The explanation is simple: that's simply not the correct area for the surface area of a circle.
     
  4. Sep 9, 2011 #3
    Actually I changed my mind and feel like it should be ((pi*r)(2*pi*r)) by my faulty thinking.

    Since pi*r would give you a line wrapped halfway around a sphere, I was thinking you could repeat this line in a radial pattern around the outside of a sphere (2*pi*r) times to get the surface area of the sphere.

    That kind of logic makes sense when taking the volume of a cylinder because I imagine taking the area of the circular disk base and stacking it (height) number of times to get the volume. The meters^2 of the area and meters of the height multiply to give the m^3 of volume. So I also expected the meters of the (pi*r) and the meters of the (2*pi*r) to multiply to give me the m^2 of surface area... :/




    Can someone explain in layman's terms what is wrong with this approach?
     
    Last edited: Sep 9, 2011
  5. Sep 9, 2011 #4

    NascentOxygen

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    Where did you learn of that method? Does it work for any geometric shape, do you know?
     
  6. Sep 9, 2011 #5

    NascentOxygen

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    If you try to "flatten out" a hemisphere, it splits. Try it with half an orange after you've scooped out all the flesh. So you do end up with a circular shape, but it has splits where there is no area, so you must measure these empty areas and discount them from your computation.
     
  7. Sep 9, 2011 #6
    yes it works with many shapes for area, surface area, and volume. Like for the area of a rectangle you could take a line that is the length between opposite edges longways and lay it side by side (width) number of times and then multiplying the units together gives you the appropriate units^2 of area.

    This same approach can be extended to volume as I discussed about cylinders, but when that thinking falls apart is when I try to apply it to anything having to do with circles.
    Say I try to find the area of a circle by getting a line extending from the center of the circle to the edge (radius) and repeating that line (circumference) number of times. I would guess this doesn't work because unlike laying lines side by side in squares, rectangles, etc, when I try to do it with circles, the center is more cramped than the outside.

    So in order to keep this approach for circles I would have to find the area of triangles made up of two radiuses set a certain angle apart and with the arc length subtending that angle as the third side. Then I could multiply these triangles by (360/angle chosen) and get the area of the circle...
     
  8. Sep 9, 2011 #7

    symbolipoint

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    Look in the Circles section of the article, http://en.wikipedia.org/wiki/Area#Surface_area

    For every circle, the ratio of circumference to diameter is [itex]\pi[/itex].

    The proof for area of a circle uses a circle cut and rearranged into a parallelogram. Careful thought helps to understand that the width of the parallelogram is half of the circle's circumference. This means that [itex]\pi r[/itex] is that width.

    Since r, the radius, is the height of this parallelogram, and [itex]\pi r[/itex] is the width, the area is [itex]\pi r^2[/itex]
     
  9. Sep 9, 2011 #8

    NascentOxygen

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    That works only because (although you might not realize it), you are using a strip of [strike]1 metre width[/strike] unit width. So you are plastering the shape with strips one [strike]metre[/strike] unit wide and of length equal to the side of the rectangle.
    EDITED.

    Sure, you can do that for circles. But the moment you try to extend it to 3 dimensions, by flattening the hemisphere (however you want to go about it), you encounter the "splitting" that I pointed out. If you cut the hemisphere into sections before flattening it out so the pieces don't split, then they will still stretch and distort, so the area you determine is no longer the area of the hemisphere that you started out with.

    Getting back to 2 dimensional shapes, the arc of any "wedge" of a circle has a length equal to radius times angle, where the angle of the wedge is measured in radians. The area of that wedge is 0.5 times angle times (radius squared). This is an exact formula, not an approximation.
     
    Last edited: Sep 10, 2011
  10. Sep 9, 2011 #9

    LJW

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    Because it isn't, simple fact. With reasoning and proofs you could very quickly prove it wrong.

    Surface area, volume, etc. of circles can be evaluated with integrals. i.e. integrating the circumference with respect to r yields the area of a circle. Integrating the surface area of a sphere with respect to r yields the volume of the sphere, as does differentiating the volume leads to the surface area.
     
  11. Sep 9, 2011 #10
    thanks guys, I see the error in my ways now
     
  12. Sep 9, 2011 #11
    Ya, and I'm doing the same thing for volume of a cylinder. Is there not a technique like this for circles/ spheres?
     
  13. Sep 9, 2011 #12
    nevermind, this is a stupid thread. But hey, everybody has to start learning somewhere!
     
  14. Sep 10, 2011 #13

    HallsofIvy

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    By the way, am I the only one wondering why nickadams keeps saying "surface area of a circle" when he clearly means "surface are of a sphere"?
     
  15. Sep 10, 2011 #14
    no, HallsofIvy, I was also wondering the same thing! What an idiot OP is. He probably doesn't even realize that surface area is for 3 dimensional objects! He should be kicked off this site for trying to figure out a way of visualizing formulas that is intuitive for him!

    By the way, am I the only one wondering why after nickadams acknowledged that he made a stupid thread, HallsofIvy still felt the need to critique a minor detail?
     
  16. Sep 10, 2011 #15

    symbolipoint

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    ?

    The original question about surface area of a circle seems to have been answered.
     
  17. Sep 10, 2011 #16

    HallsofIvy

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    :biggrin::biggrin:

    Hey, I'm just a troublemaker.:devil:
     
  18. Sep 10, 2011 #17

    HallsofIvy

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    Yep, a circle doesn't have a surface so the surface area is 0.
     
  19. Sep 11, 2011 #18

    NascentOxygen

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    You may have a point, but I think subject lines can't be edited.

    Besides, I'm not one to pick people up on tiny mistakes. :rolleyes:
     
    Last edited: Sep 11, 2011
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