Deriving the formula of the surface of a sphere using triangles.
$begingroup$
My friend tried to find the formula of the surface of a sphere using the following reasoning, but we can't see the mistake:
Let's first take half a sphere and divide the sphere into infinitely small triangles like this:
Bad drawing, but I hope you understand the idea.
Then, we can unwrap it and arrange the hemisphere into half a rectangle: 
The height will be $frac{pi r}{2}$ because it is a quarter of the length of a circumference and the base will be $2 pi r$.
We can now insert the other half rectangle of the other hemisphere divided in infinitely small triangles in this way: 
It will nicely create a rectangle (if we rearrange the side triangles) and to get the area of the rectangle, just multiply the base times height, ending with $pi^2 r^2$.
We know that that the correct answer is $4 pi r^2$, but we can't figure out the error.
geometry area spheres
edited Dec 8 '18 at 17:25
Alex Vong
1,284819
asked Dec 8 '18 at 16:34
Matemagia D13G0Matemagia D13G0
132
$endgroup$
add a comment |
$begingroup$
My friend tried to find the formula of the surface of a sphere using the following reasoning, but we can't see the mistake:
Let's first take half a sphere and divide the sphere into infinitely small triangles like this:
Bad drawing, but I hope you understand the idea.
Then, we can unwrap it and arrange the hemisphere into half a rectangle: The height will be $frac{pi r}{2}$ because it is a quarter of the length of a circumference and the base will be $2 pi r$.
We can now insert the other half rectangle of the other hemisphere divided in infinitely small triangles in this way:
It will nicely create a rectangle (if we rearrange the side triangles) and to get the area of the rectangle, just multiply the base times height, ending with $pi^2 r^2$.
We know that that the correct answer is $4 pi r^2$, but we can't figure out the error.
geometry area spheres
edited Dec 8 '18 at 17:25
Alex Vong
1,284819
asked Dec 8 '18 at 16:34
Matemagia D13G0Matemagia D13G0
132
$endgroup$
add a comment |
$begingroup$
My friend tried to find the formula of the surface of a sphere using the following reasoning, but we can't see the mistake:
Let's first take half a sphere and divide the sphere into infinitely small triangles like this:
Bad drawing, but I hope you understand the idea.
Then, we can unwrap it and arrange the hemisphere into half a rectangle: The height will be $frac{pi r}{2}$ because it is a quarter of the length of a circumference and the base will be $2 pi r$.
We can now insert the other half rectangle of the other hemisphere divided in infinitely small triangles in this way:
It will nicely create a rectangle (if we rearrange the side triangles) and to get the area of the rectangle, just multiply the base times height, ending with $pi^2 r^2$.
We know that that the correct answer is $4 pi r^2$, but we can't figure out the error.
geometry area spheres
edited Dec 8 '18 at 17:25
Alex Vong
1,284819
asked Dec 8 '18 at 16:34
Matemagia D13G0Matemagia D13G0
132
$endgroup$
My friend tried to find the formula of the surface of a sphere using the following reasoning, but we can't see the mistake:
Let's first take half a sphere and divide the sphere into infinitely small triangles like this:
Bad drawing, but I hope you understand the idea.
Then, we can unwrap it and arrange the hemisphere into half a rectangle: The height will be $frac{pi r}{2}$ because it is a quarter of the length of a circumference and the base will be $2 pi r$.
We can now insert the other half rectangle of the other hemisphere divided in infinitely small triangles in this way:
It will nicely create a rectangle (if we rearrange the side triangles) and to get the area of the rectangle, just multiply the base times height, ending with $pi^2 r^2$.
We know that that the correct answer is $4 pi r^2$, but we can't figure out the error.
geometry area spheres
geometry area spheres
edited Dec 8 '18 at 17:25
Alex Vong
1,284819
asked Dec 8 '18 at 16:34
Matemagia D13G0Matemagia D13G0
132
edited Dec 8 '18 at 17:25
Alex Vong
1,284819
asked Dec 8 '18 at 16:34
Matemagia D13G0Matemagia D13G0
132
edited Dec 8 '18 at 17:25
Alex Vong
1,284819
edited Dec 8 '18 at 17:25
Alex Vong
1,284819
edited Dec 8 '18 at 17:25
Alex Vong
1,284819
1,284819
asked Dec 8 '18 at 16:34
Matemagia D13G0Matemagia D13G0
132
asked Dec 8 '18 at 16:34
Matemagia D13G0Matemagia D13G0
132
asked Dec 8 '18 at 16:34
Matemagia D13G0Matemagia D13G0
132
132
add a comment |
add a comment |
1 Answer
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At the most abstract level, the discrepancy is due to the sphere not being a developable surface: it cannot be projected onto a plane while preserving areas and angles at once. Thus, it is not correct to say that the triangle on the unit sphere with angles $fracpi2$, $fracpi2$ and $dx$ – the "infinitely small triangles" in the argument – has the same area as the planar right triangle of leg lengths $dx$ and $fracpi2$.
answered Dec 8 '18 at 17:00
Parcly TaxelParcly Taxel
41.8k1372101
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1 Answer
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1 Answer
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active
oldest
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active
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active
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votes
$begingroup$
At the most abstract level, the discrepancy is due to the sphere not being a developable surface: it cannot be projected onto a plane while preserving areas and angles at once. Thus, it is not correct to say that the triangle on the unit sphere with angles $fracpi2$, $fracpi2$ and $dx$ – the "infinitely small triangles" in the argument – has the same area as the planar right triangle of leg lengths $dx$ and $fracpi2$.
answered Dec 8 '18 at 17:00
Parcly TaxelParcly Taxel
41.8k1372101
$endgroup$
add a comment |
$begingroup$
At the most abstract level, the discrepancy is due to the sphere not being a developable surface: it cannot be projected onto a plane while preserving areas and angles at once. Thus, it is not correct to say that the triangle on the unit sphere with angles $fracpi2$, $fracpi2$ and $dx$ – the "infinitely small triangles" in the argument – has the same area as the planar right triangle of leg lengths $dx$ and $fracpi2$.
answered Dec 8 '18 at 17:00
Parcly TaxelParcly Taxel
41.8k1372101
$endgroup$
add a comment |
$begingroup$
At the most abstract level, the discrepancy is due to the sphere not being a developable surface: it cannot be projected onto a plane while preserving areas and angles at once. Thus, it is not correct to say that the triangle on the unit sphere with angles $fracpi2$, $fracpi2$ and $dx$ – the "infinitely small triangles" in the argument – has the same area as the planar right triangle of leg lengths $dx$ and $fracpi2$.
answered Dec 8 '18 at 17:00
Parcly TaxelParcly Taxel
41.8k1372101
$endgroup$
At the most abstract level, the discrepancy is due to the sphere not being a developable surface: it cannot be projected onto a plane while preserving areas and angles at once. Thus, it is not correct to say that the triangle on the unit sphere with angles $fracpi2$, $fracpi2$ and $dx$ – the "infinitely small triangles" in the argument – has the same area as the planar right triangle of leg lengths $dx$ and $fracpi2$.
answered Dec 8 '18 at 17:00
Parcly TaxelParcly Taxel
41.8k1372101
answered Dec 8 '18 at 17:00
Parcly TaxelParcly Taxel
41.8k1372101
answered Dec 8 '18 at 17:00
Parcly TaxelParcly Taxel
41.8k1372101
answered Dec 8 '18 at 17:00
Parcly TaxelParcly Taxel
41.8k1372101
41.8k1372101
add a comment |
add a comment |
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