Formalization of an Intuitive Proof of the Surface Area of a Sphere
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Consider the following proof that the surface area of a sphere is $4 pi r^2$.
First let's try to squash an arbitrary sphere into its net. If we cut the sphere in half then open it up, we get two tangent bowl shaped figures. Now we attempt to squash them to flatten them int a 2D net. This 'rips' them at the top and bottom and results in a new, thinner bowl shaped figure. We repeat this process over and over. Note that since there is always some finite height of the entire net, it is never 2D. So, if we perform this operation $n$ times, then the limit of the shape of the net as $n$ tends to $infty$ is a rectangle. Below is a diagram of an intermediate step in the squashification of the sphere.
The rectangle's length is the circumference of the great circle of the sphere which is $2 pi r$ and the height of the rectangle in the height of the sphere which is $2r$. Thus the surface area of the sphere is the area of the rectangle which is $2 pi r cdot 2 r = 4 pi r^2$.
Isn't it kind of sketchy how I said the limit of the entire figure as the number of 'rips' tends to infinity is a rectangle? How could one formalize this proof?
proof-writing spheres proof-theory
asked Jan 19 at 3:49
Shrey JoshiShrey Joshi
31413
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add a comment |
$begingroup$
Consider the following proof that the surface area of a sphere is $4 pi r^2$.
First let's try to squash an arbitrary sphere into its net. If we cut the sphere in half then open it up, we get two tangent bowl shaped figures. Now we attempt to squash them to flatten them int a 2D net. This 'rips' them at the top and bottom and results in a new, thinner bowl shaped figure. We repeat this process over and over. Note that since there is always some finite height of the entire net, it is never 2D. So, if we perform this operation $n$ times, then the limit of the shape of the net as $n$ tends to $infty$ is a rectangle. Below is a diagram of an intermediate step in the squashification of the sphere.
The rectangle's length is the circumference of the great circle of the sphere which is $2 pi r$ and the height of the rectangle in the height of the sphere which is $2r$. Thus the surface area of the sphere is the area of the rectangle which is $2 pi r cdot 2 r = 4 pi r^2$.
Isn't it kind of sketchy how I said the limit of the entire figure as the number of 'rips' tends to infinity is a rectangle? How could one formalize this proof?
proof-writing spheres proof-theory
asked Jan 19 at 3:49
Shrey JoshiShrey Joshi
31413
$endgroup$
$begingroup$
Your attempt to link a picture didn't work for me; I don't think that's a valid URL.
$endgroup$
– jmerry
Jan 19 at 3:57
$begingroup$
It's fixed now, thanks!
$endgroup$
– Shrey Joshi
Jan 19 at 5:07
$begingroup$
3Blue1Brown has a very intuitive approach to showing this: youtube.com/watch?v=GNcFjFmqEc8
$endgroup$
– Aniruddh Venkatesan
Jan 19 at 5:09
add a comment |
$begingroup$
Consider the following proof that the surface area of a sphere is $4 pi r^2$.
First let's try to squash an arbitrary sphere into its net. If we cut the sphere in half then open it up, we get two tangent bowl shaped figures. Now we attempt to squash them to flatten them int a 2D net. This 'rips' them at the top and bottom and results in a new, thinner bowl shaped figure. We repeat this process over and over. Note that since there is always some finite height of the entire net, it is never 2D. So, if we perform this operation $n$ times, then the limit of the shape of the net as $n$ tends to $infty$ is a rectangle. Below is a diagram of an intermediate step in the squashification of the sphere.
The rectangle's length is the circumference of the great circle of the sphere which is $2 pi r$ and the height of the rectangle in the height of the sphere which is $2r$. Thus the surface area of the sphere is the area of the rectangle which is $2 pi r cdot 2 r = 4 pi r^2$.
Isn't it kind of sketchy how I said the limit of the entire figure as the number of 'rips' tends to infinity is a rectangle? How could one formalize this proof?
proof-writing spheres proof-theory
asked Jan 19 at 3:49
Shrey JoshiShrey Joshi
31413
$endgroup$
Consider the following proof that the surface area of a sphere is $4 pi r^2$.
First let's try to squash an arbitrary sphere into its net. If we cut the sphere in half then open it up, we get two tangent bowl shaped figures. Now we attempt to squash them to flatten them int a 2D net. This 'rips' them at the top and bottom and results in a new, thinner bowl shaped figure. We repeat this process over and over. Note that since there is always some finite height of the entire net, it is never 2D. So, if we perform this operation $n$ times, then the limit of the shape of the net as $n$ tends to $infty$ is a rectangle. Below is a diagram of an intermediate step in the squashification of the sphere.
The rectangle's length is the circumference of the great circle of the sphere which is $2 pi r$ and the height of the rectangle in the height of the sphere which is $2r$. Thus the surface area of the sphere is the area of the rectangle which is $2 pi r cdot 2 r = 4 pi r^2$.
Isn't it kind of sketchy how I said the limit of the entire figure as the number of 'rips' tends to infinity is a rectangle? How could one formalize this proof?
proof-writing spheres proof-theory
proof-writing spheres proof-theory
asked Jan 19 at 3:49
Shrey JoshiShrey Joshi
31413
asked Jan 19 at 3:49
Shrey JoshiShrey Joshi
31413
edited Jan 19 at 5:05
Shrey Joshi
asked Jan 19 at 3:49
Shrey JoshiShrey Joshi
31413
asked Jan 19 at 3:49
Shrey JoshiShrey Joshi
31413
asked Jan 19 at 3:49
Shrey JoshiShrey Joshi
31413
31413
$begingroup$
Your attempt to link a picture didn't work for me; I don't think that's a valid URL.
$endgroup$
– jmerry
Jan 19 at 3:57
$begingroup$
It's fixed now, thanks!
$endgroup$
– Shrey Joshi
Jan 19 at 5:07
$begingroup$
3Blue1Brown has a very intuitive approach to showing this: youtube.com/watch?v=GNcFjFmqEc8
$endgroup$
– Aniruddh Venkatesan
Jan 19 at 5:09
add a comment |
$begingroup$
Your attempt to link a picture didn't work for me; I don't think that's a valid URL.
$endgroup$
– jmerry
Jan 19 at 3:57
$begingroup$
It's fixed now, thanks!
$endgroup$
– Shrey Joshi
Jan 19 at 5:07
$begingroup$
3Blue1Brown has a very intuitive approach to showing this: youtube.com/watch?v=GNcFjFmqEc8
$endgroup$
– Aniruddh Venkatesan
Jan 19 at 5:09
$begingroup$
Your attempt to link a picture didn't work for me; I don't think that's a valid URL.
$endgroup$
– jmerry
Jan 19 at 3:57
$begingroup$
Your attempt to link a picture didn't work for me; I don't think that's a valid URL.
$endgroup$
– jmerry
Jan 19 at 3:57
$begingroup$
It's fixed now, thanks!
$endgroup$
– Shrey Joshi
Jan 19 at 5:07
$begingroup$
It's fixed now, thanks!
$endgroup$
– Shrey Joshi
Jan 19 at 5:07
$begingroup$
3Blue1Brown has a very intuitive approach to showing this: youtube.com/watch?v=GNcFjFmqEc8
$endgroup$
– Aniruddh Venkatesan
Jan 19 at 5:09
$begingroup$
3Blue1Brown has a very intuitive approach to showing this: youtube.com/watch?v=GNcFjFmqEc8
$endgroup$
– Aniruddh Venkatesan
Jan 19 at 5:09
add a comment |
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$begingroup$
Your attempt to link a picture didn't work for me; I don't think that's a valid URL.
$endgroup$
– jmerry
Jan 19 at 3:57
$begingroup$
It's fixed now, thanks!
$endgroup$
– Shrey Joshi
Jan 19 at 5:07
$begingroup$
3Blue1Brown has a very intuitive approach to showing this: youtube.com/watch?v=GNcFjFmqEc8
$endgroup$
– Aniruddh Venkatesan
Jan 19 at 5:09