- #1
mmmboh
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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?
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?