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Homework Help: Surface Area of a Sphere without Caluclus Question

  1. Jul 27, 2010 #1
    Hi, I was just trying to find the surface area of a sphere without calculus, just for fun, but I ran into a little problem.

    The way I wanted to do it was by imagining a tennis ball, and cutting it in half forming a hemisphere. Then I imagined flattening out this hemisphere would preserve the surface area, but would allow me to calculate the area as if it were a circle, and I concluded that the radius of the flattened hemisphere would be the arc length from the bottom to the top of the hemisphere, which is [tex]\pi[/tex]R/2. However then finding the area of that circle and multiplying it by two because there are two hemispheres, the area I get isn't at all 4[tex]\pi[/tex]R2. Something I noticed was that instead of using the arc length as the radius of the circle, if I find the length of the hypotenuse of the triangle connecting the bottom to the top, that gives me 21/2R, and then when I find the area of a circle with that radius, and multiply it by two, I get 4[tex]\pi[/tex]R2...can someone clear up what is going on?

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  2. jcsd
  3. Jul 27, 2010 #2

    Dick

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    If you had actually cut the tennis ball in half and tried to flatten it, you would have noticed you can't flatten it without stretching parts of it. Stretching it changes the area. That the hypotenuse works as the correct radius I think is just an accident.
     
  4. Jul 27, 2010 #3
    Ok, thanks! One more question I thought of, this time for finding volume. When finding the surface area of a sphere you can use the integral of 2[tex]\pi[/tex]Rsin[tex]\theta[/tex](Rd[tex]\theta[/tex]), where Rd[tex]\theta[/tex] is the thickness of the ring and 2[tex]\pi[/tex]Rsin[tex]\theta[/tex] is the circumference of the ring (R is the radius of the sphere), and you can integrate this from 0 to [tex]\pi[/tex]...but how come when finding the volume of the sphere you can't just integrate [tex]\pi[/tex](Rsin[tex]\theta[/tex])^2(Rd[tex]\theta[/tex])? since the thickness remains the same, but now we exchange circumference for area?
     
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