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Different proof for surface area of a sphere?
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Take a sphere, and look at its circumference. We then take that circumference of length $(2pi r)$, and move it around the sphere, so that it covers the entire surface area of the sphere. How much area did it cover? Well it would be it’s length, multiplied by the height (diameter) of the sphere. That would be $(2pi r)(2r) = (4pi r^2)$. Is this a correct proof?
geometry
edited Jul 23 at 20:59
Strants
5,09921636
asked Jul 23 at 20:25
Jon due
163
add a comment |Â
up vote
2
down vote
favorite
Take a sphere, and look at its circumference. We then take that circumference of length $(2pi r)$, and move it around the sphere, so that it covers the entire surface area of the sphere. How much area did it cover? Well it would be it’s length, multiplied by the height (diameter) of the sphere. That would be $(2pi r)(2r) = (4pi r^2)$. Is this a correct proof?
geometry
edited Jul 23 at 20:59
Strants
5,09921636
asked Jul 23 at 20:25
Jon due
163
But that is the surface area of a right circular cylinder of height $2r$ and radius $r$
– ab123
Jul 23 at 20:31
Do you mean the lateral surface?
– Jon due
Jul 23 at 20:32
yes, the lateral surface area
– ab123
Jul 23 at 20:33
@Jondue Try your approach with a cylinder. Does it yield the true surface area of the cylinder?
– user76284
Jul 23 at 20:33
2
No, it's not a correct proof. You have stumbled on an important coincidence that dates back to Archimedes: en.wikipedia.org/wiki/On_the_Sphere_and_Cylinder
– Ethan Bolker
Jul 23 at 20:39
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Take a sphere, and look at its circumference. We then take that circumference of length $(2pi r)$, and move it around the sphere, so that it covers the entire surface area of the sphere. How much area did it cover? Well it would be it’s length, multiplied by the height (diameter) of the sphere. That would be $(2pi r)(2r) = (4pi r^2)$. Is this a correct proof?
geometry
edited Jul 23 at 20:59
Strants
5,09921636
asked Jul 23 at 20:25
Jon due
163
Take a sphere, and look at its circumference. We then take that circumference of length $(2pi r)$, and move it around the sphere, so that it covers the entire surface area of the sphere. How much area did it cover? Well it would be it’s length, multiplied by the height (diameter) of the sphere. That would be $(2pi r)(2r) = (4pi r^2)$. Is this a correct proof?
geometry
edited Jul 23 at 20:59
Strants
5,09921636
asked Jul 23 at 20:25
Jon due
163
edited Jul 23 at 20:59
Strants
5,09921636
edited Jul 23 at 20:59
Strants
5,09921636
edited Jul 23 at 20:59
Strants
5,09921636
5,09921636
asked Jul 23 at 20:25
Jon due
163
asked Jul 23 at 20:25
Jon due
163
asked Jul 23 at 20:25
Jon due
163
163
But that is the surface area of a right circular cylinder of height $2r$ and radius $r$
– ab123
Jul 23 at 20:31
Do you mean the lateral surface?
– Jon due
Jul 23 at 20:32
yes, the lateral surface area
– ab123
Jul 23 at 20:33
@Jondue Try your approach with a cylinder. Does it yield the true surface area of the cylinder?
– user76284
Jul 23 at 20:33
2
No, it's not a correct proof. You have stumbled on an important coincidence that dates back to Archimedes: en.wikipedia.org/wiki/On_the_Sphere_and_Cylinder
– Ethan Bolker
Jul 23 at 20:39
add a comment |Â
But that is the surface area of a right circular cylinder of height $2r$ and radius $r$
– ab123
Jul 23 at 20:31
Do you mean the lateral surface?
– Jon due
Jul 23 at 20:32
yes, the lateral surface area
– ab123
Jul 23 at 20:33
@Jondue Try your approach with a cylinder. Does it yield the true surface area of the cylinder?
– user76284
Jul 23 at 20:33
2
No, it's not a correct proof. You have stumbled on an important coincidence that dates back to Archimedes: en.wikipedia.org/wiki/On_the_Sphere_and_Cylinder
– Ethan Bolker
Jul 23 at 20:39
But that is the surface area of a right circular cylinder of height $2r$ and radius $r$
– ab123
Jul 23 at 20:31
But that is the surface area of a right circular cylinder of height $2r$ and radius $r$
– ab123
Jul 23 at 20:31
Do you mean the lateral surface?
– Jon due
Jul 23 at 20:32
Do you mean the lateral surface?
– Jon due
Jul 23 at 20:32
yes, the lateral surface area
– ab123
Jul 23 at 20:33
yes, the lateral surface area
– ab123
Jul 23 at 20:33
@Jondue Try your approach with a cylinder. Does it yield the true surface area of the cylinder?
– user76284
Jul 23 at 20:33
@Jondue Try your approach with a cylinder. Does it yield the true surface area of the cylinder?
– user76284
Jul 23 at 20:33
2
2
No, it's not a correct proof. You have stumbled on an important coincidence that dates back to Archimedes: en.wikipedia.org/wiki/On_the_Sphere_and_Cylinder
– Ethan Bolker
Jul 23 at 20:39
No, it's not a correct proof. You have stumbled on an important coincidence that dates back to Archimedes: en.wikipedia.org/wiki/On_the_Sphere_and_Cylinder
– Ethan Bolker
Jul 23 at 20:39
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
It's incorrect proof. The circumference strip fits the sphere only when it lies on a plane that passes through its center. But that's not the problem tho. Even if you get half circumference strip and "rotate it around" so that it sweeps the sphere and use your reasoning, you will get a wrong formula ($pi r * 2 pi r$).
Generally you can't find the surface area by moving around one dimensional curves that do not have surface area (they only have length). Intuitively in order to calculate the surface area you must approximate the surface with elements that have easily computable surface area (rectangles) such that the approximation gets arbitrary close to the surface the more you tessellate it take the limit as the tessellation gets infinitely fine.
Now in order to do this legitimately and rigorously in a system where the word proof actually makes sense you must first define what a "surface" actually means and then define what "area" means which it turns out is not easy at all. The definition of the word "surface" (at least for kinds of surfaces relevant to this context) usually involves the notion of smooth manifold which is essential object in differential geometry. The definition of the word "area" usually involves some form of integration (can be "unsigned" measure theoretic, or some form of signed integration, such as integrating differential forms on manifolds).
Of course this can be done even with basic calculus using Riemann integration and some heuristics in $mathbb R^3$ without properly defining what area or surface are. And people have done it thousands of years before calculus was invented with heuristics and hand waving. It depends on what counts as proof according to you.
answered Jul 23 at 21:01
gcc-6.0
58017
add a comment |Â
up vote
0
down vote
Your idea sounds like Pappus's Centroid Theorem.
If you want to find the surface are of the sphere by revolving a curve about an axis, you need to multiply the distance traveled by the centroid of the curve, by the length of the curve.
In your case the curve is a semi circle revolved about the x-axis.
If the centroid of the semi-circle is d units apart from the x-axis, then the surface area of the sphere is $$2pi d times pi R$$ where R is radius of the sphere.
Apparently $d=frac 2Rpi$ so you get the correct answer for the surface area.
answered Jul 23 at 21:25
Mohammad Riazi-Kermani
27.5k41852
add a comment |Â
up vote
0
down vote
Let's see--we have a proposed method for measuring the area of a surface
when there is a one-dimensional shape that can "sweep" the entire surface.
In particular we have a circle that can be "moved around" to cover the entire surface of a sphere, and when we multiply the length of the circle times the diameter of the sphere we get $4pi r^2.$
We happen to know this is the correct area of the sphere. Success!
It seems a shame to waste this method on just one use, doesn't it?
Surely it is a very useful method for solving many problems.
Let's apply the method to a cylinder of diameter $d$ and height $d,$
with closed ends. The area of this cylinder by conventional methods
is $pi d times d + 2(frac14pi d^2) = frac32 pi d^2.$
There's a circle around the cylinder that we can use to sweep the curved surface, but that doesn't cover the closed ends of the cylinder.
So instead let's take a square of side $d$ and fit it around the cylinder
so there is an edge of the square going diametrically across each end of the cylinder.
Then we can move the square around to cover not just the curved part of the surface but also the closed ends.
The perimeter of the square is $4d,$ and the diameter of the cylinder is $d,$
so we expect the surface area to be $4d^2.$
How well did we do?
In case it is not clear, I did not really mean for anyone to believe everything I wrote in the first two paragraphs.
I did mean for you to believe everything I wrote in the third paragraph, however.
I believe the third paragraph is a fair application of the method used in the question, but with slightly less hand-waving.
answered Jul 23 at 21:46
David K
48.2k340107
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
It's incorrect proof. The circumference strip fits the sphere only when it lies on a plane that passes through its center. But that's not the problem tho. Even if you get half circumference strip and "rotate it around" so that it sweeps the sphere and use your reasoning, you will get a wrong formula ($pi r * 2 pi r$).
Generally you can't find the surface area by moving around one dimensional curves that do not have surface area (they only have length). Intuitively in order to calculate the surface area you must approximate the surface with elements that have easily computable surface area (rectangles) such that the approximation gets arbitrary close to the surface the more you tessellate it take the limit as the tessellation gets infinitely fine.
Now in order to do this legitimately and rigorously in a system where the word proof actually makes sense you must first define what a "surface" actually means and then define what "area" means which it turns out is not easy at all. The definition of the word "surface" (at least for kinds of surfaces relevant to this context) usually involves the notion of smooth manifold which is essential object in differential geometry. The definition of the word "area" usually involves some form of integration (can be "unsigned" measure theoretic, or some form of signed integration, such as integrating differential forms on manifolds).
Of course this can be done even with basic calculus using Riemann integration and some heuristics in $mathbb R^3$ without properly defining what area or surface are. And people have done it thousands of years before calculus was invented with heuristics and hand waving. It depends on what counts as proof according to you.
answered Jul 23 at 21:01
gcc-6.0
58017
add a comment |Â
up vote
1
down vote
It's incorrect proof. The circumference strip fits the sphere only when it lies on a plane that passes through its center. But that's not the problem tho. Even if you get half circumference strip and "rotate it around" so that it sweeps the sphere and use your reasoning, you will get a wrong formula ($pi r * 2 pi r$).
Generally you can't find the surface area by moving around one dimensional curves that do not have surface area (they only have length). Intuitively in order to calculate the surface area you must approximate the surface with elements that have easily computable surface area (rectangles) such that the approximation gets arbitrary close to the surface the more you tessellate it take the limit as the tessellation gets infinitely fine.
Now in order to do this legitimately and rigorously in a system where the word proof actually makes sense you must first define what a "surface" actually means and then define what "area" means which it turns out is not easy at all. The definition of the word "surface" (at least for kinds of surfaces relevant to this context) usually involves the notion of smooth manifold which is essential object in differential geometry. The definition of the word "area" usually involves some form of integration (can be "unsigned" measure theoretic, or some form of signed integration, such as integrating differential forms on manifolds).
Of course this can be done even with basic calculus using Riemann integration and some heuristics in $mathbb R^3$ without properly defining what area or surface are. And people have done it thousands of years before calculus was invented with heuristics and hand waving. It depends on what counts as proof according to you.
answered Jul 23 at 21:01
gcc-6.0
58017
add a comment |Â
up vote
1
down vote
up vote
1
down vote
It's incorrect proof. The circumference strip fits the sphere only when it lies on a plane that passes through its center. But that's not the problem tho. Even if you get half circumference strip and "rotate it around" so that it sweeps the sphere and use your reasoning, you will get a wrong formula ($pi r * 2 pi r$).
Generally you can't find the surface area by moving around one dimensional curves that do not have surface area (they only have length). Intuitively in order to calculate the surface area you must approximate the surface with elements that have easily computable surface area (rectangles) such that the approximation gets arbitrary close to the surface the more you tessellate it take the limit as the tessellation gets infinitely fine.
Now in order to do this legitimately and rigorously in a system where the word proof actually makes sense you must first define what a "surface" actually means and then define what "area" means which it turns out is not easy at all. The definition of the word "surface" (at least for kinds of surfaces relevant to this context) usually involves the notion of smooth manifold which is essential object in differential geometry. The definition of the word "area" usually involves some form of integration (can be "unsigned" measure theoretic, or some form of signed integration, such as integrating differential forms on manifolds).
Of course this can be done even with basic calculus using Riemann integration and some heuristics in $mathbb R^3$ without properly defining what area or surface are. And people have done it thousands of years before calculus was invented with heuristics and hand waving. It depends on what counts as proof according to you.
answered Jul 23 at 21:01
gcc-6.0
58017
It's incorrect proof. The circumference strip fits the sphere only when it lies on a plane that passes through its center. But that's not the problem tho. Even if you get half circumference strip and "rotate it around" so that it sweeps the sphere and use your reasoning, you will get a wrong formula ($pi r * 2 pi r$).
Generally you can't find the surface area by moving around one dimensional curves that do not have surface area (they only have length). Intuitively in order to calculate the surface area you must approximate the surface with elements that have easily computable surface area (rectangles) such that the approximation gets arbitrary close to the surface the more you tessellate it take the limit as the tessellation gets infinitely fine.
Now in order to do this legitimately and rigorously in a system where the word proof actually makes sense you must first define what a "surface" actually means and then define what "area" means which it turns out is not easy at all. The definition of the word "surface" (at least for kinds of surfaces relevant to this context) usually involves the notion of smooth manifold which is essential object in differential geometry. The definition of the word "area" usually involves some form of integration (can be "unsigned" measure theoretic, or some form of signed integration, such as integrating differential forms on manifolds).
Of course this can be done even with basic calculus using Riemann integration and some heuristics in $mathbb R^3$ without properly defining what area or surface are. And people have done it thousands of years before calculus was invented with heuristics and hand waving. It depends on what counts as proof according to you.
answered Jul 23 at 21:01
gcc-6.0
58017
answered Jul 23 at 21:01
gcc-6.0
58017
answered Jul 23 at 21:01
gcc-6.0
58017
answered Jul 23 at 21:01
gcc-6.0
58017
58017
add a comment |Â
add a comment |Â
up vote
0
down vote
Your idea sounds like Pappus's Centroid Theorem.
If you want to find the surface are of the sphere by revolving a curve about an axis, you need to multiply the distance traveled by the centroid of the curve, by the length of the curve.
In your case the curve is a semi circle revolved about the x-axis.
If the centroid of the semi-circle is d units apart from the x-axis, then the surface area of the sphere is $$2pi d times pi R$$ where R is radius of the sphere.
Apparently $d=frac 2Rpi$ so you get the correct answer for the surface area.
answered Jul 23 at 21:25
Mohammad Riazi-Kermani
27.5k41852
add a comment |Â
up vote
0
down vote
Your idea sounds like Pappus's Centroid Theorem.
If you want to find the surface are of the sphere by revolving a curve about an axis, you need to multiply the distance traveled by the centroid of the curve, by the length of the curve.
In your case the curve is a semi circle revolved about the x-axis.
If the centroid of the semi-circle is d units apart from the x-axis, then the surface area of the sphere is $$2pi d times pi R$$ where R is radius of the sphere.
Apparently $d=frac 2Rpi$ so you get the correct answer for the surface area.
answered Jul 23 at 21:25
Mohammad Riazi-Kermani
27.5k41852
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Your idea sounds like Pappus's Centroid Theorem.
If you want to find the surface are of the sphere by revolving a curve about an axis, you need to multiply the distance traveled by the centroid of the curve, by the length of the curve.
In your case the curve is a semi circle revolved about the x-axis.
If the centroid of the semi-circle is d units apart from the x-axis, then the surface area of the sphere is $$2pi d times pi R$$ where R is radius of the sphere.
Apparently $d=frac 2Rpi$ so you get the correct answer for the surface area.
answered Jul 23 at 21:25
Mohammad Riazi-Kermani
27.5k41852
Your idea sounds like Pappus's Centroid Theorem.
If you want to find the surface are of the sphere by revolving a curve about an axis, you need to multiply the distance traveled by the centroid of the curve, by the length of the curve.
In your case the curve is a semi circle revolved about the x-axis.
If the centroid of the semi-circle is d units apart from the x-axis, then the surface area of the sphere is $$2pi d times pi R$$ where R is radius of the sphere.
Apparently $d=frac 2Rpi$ so you get the correct answer for the surface area.
answered Jul 23 at 21:25
Mohammad Riazi-Kermani
27.5k41852
answered Jul 23 at 21:25
Mohammad Riazi-Kermani
27.5k41852
answered Jul 23 at 21:25
Mohammad Riazi-Kermani
27.5k41852
answered Jul 23 at 21:25
Mohammad Riazi-Kermani
27.5k41852
27.5k41852
add a comment |Â
add a comment |Â
up vote
0
down vote
Let's see--we have a proposed method for measuring the area of a surface
when there is a one-dimensional shape that can "sweep" the entire surface.
In particular we have a circle that can be "moved around" to cover the entire surface of a sphere, and when we multiply the length of the circle times the diameter of the sphere we get $4pi r^2.$
We happen to know this is the correct area of the sphere. Success!
It seems a shame to waste this method on just one use, doesn't it?
Surely it is a very useful method for solving many problems.
Let's apply the method to a cylinder of diameter $d$ and height $d,$
with closed ends. The area of this cylinder by conventional methods
is $pi d times d + 2(frac14pi d^2) = frac32 pi d^2.$
There's a circle around the cylinder that we can use to sweep the curved surface, but that doesn't cover the closed ends of the cylinder.
So instead let's take a square of side $d$ and fit it around the cylinder
so there is an edge of the square going diametrically across each end of the cylinder.
Then we can move the square around to cover not just the curved part of the surface but also the closed ends.
The perimeter of the square is $4d,$ and the diameter of the cylinder is $d,$
so we expect the surface area to be $4d^2.$
How well did we do?
In case it is not clear, I did not really mean for anyone to believe everything I wrote in the first two paragraphs.
I did mean for you to believe everything I wrote in the third paragraph, however.
I believe the third paragraph is a fair application of the method used in the question, but with slightly less hand-waving.
answered Jul 23 at 21:46
David K
48.2k340107
add a comment |Â
up vote
0
down vote
Let's see--we have a proposed method for measuring the area of a surface
when there is a one-dimensional shape that can "sweep" the entire surface.
In particular we have a circle that can be "moved around" to cover the entire surface of a sphere, and when we multiply the length of the circle times the diameter of the sphere we get $4pi r^2.$
We happen to know this is the correct area of the sphere. Success!
It seems a shame to waste this method on just one use, doesn't it?
Surely it is a very useful method for solving many problems.
Let's apply the method to a cylinder of diameter $d$ and height $d,$
with closed ends. The area of this cylinder by conventional methods
is $pi d times d + 2(frac14pi d^2) = frac32 pi d^2.$
There's a circle around the cylinder that we can use to sweep the curved surface, but that doesn't cover the closed ends of the cylinder.
So instead let's take a square of side $d$ and fit it around the cylinder
so there is an edge of the square going diametrically across each end of the cylinder.
Then we can move the square around to cover not just the curved part of the surface but also the closed ends.
The perimeter of the square is $4d,$ and the diameter of the cylinder is $d,$
so we expect the surface area to be $4d^2.$
How well did we do?
In case it is not clear, I did not really mean for anyone to believe everything I wrote in the first two paragraphs.
I did mean for you to believe everything I wrote in the third paragraph, however.
I believe the third paragraph is a fair application of the method used in the question, but with slightly less hand-waving.
answered Jul 23 at 21:46
David K
48.2k340107
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let's see--we have a proposed method for measuring the area of a surface
when there is a one-dimensional shape that can "sweep" the entire surface.
In particular we have a circle that can be "moved around" to cover the entire surface of a sphere, and when we multiply the length of the circle times the diameter of the sphere we get $4pi r^2.$
We happen to know this is the correct area of the sphere. Success!
It seems a shame to waste this method on just one use, doesn't it?
Surely it is a very useful method for solving many problems.
Let's apply the method to a cylinder of diameter $d$ and height $d,$
with closed ends. The area of this cylinder by conventional methods
is $pi d times d + 2(frac14pi d^2) = frac32 pi d^2.$
There's a circle around the cylinder that we can use to sweep the curved surface, but that doesn't cover the closed ends of the cylinder.
So instead let's take a square of side $d$ and fit it around the cylinder
so there is an edge of the square going diametrically across each end of the cylinder.
Then we can move the square around to cover not just the curved part of the surface but also the closed ends.
The perimeter of the square is $4d,$ and the diameter of the cylinder is $d,$
so we expect the surface area to be $4d^2.$
How well did we do?
In case it is not clear, I did not really mean for anyone to believe everything I wrote in the first two paragraphs.
I did mean for you to believe everything I wrote in the third paragraph, however.
I believe the third paragraph is a fair application of the method used in the question, but with slightly less hand-waving.
answered Jul 23 at 21:46
David K
48.2k340107
Let's see--we have a proposed method for measuring the area of a surface
when there is a one-dimensional shape that can "sweep" the entire surface.
In particular we have a circle that can be "moved around" to cover the entire surface of a sphere, and when we multiply the length of the circle times the diameter of the sphere we get $4pi r^2.$
We happen to know this is the correct area of the sphere. Success!
It seems a shame to waste this method on just one use, doesn't it?
Surely it is a very useful method for solving many problems.
Let's apply the method to a cylinder of diameter $d$ and height $d,$
with closed ends. The area of this cylinder by conventional methods
is $pi d times d + 2(frac14pi d^2) = frac32 pi d^2.$
There's a circle around the cylinder that we can use to sweep the curved surface, but that doesn't cover the closed ends of the cylinder.
So instead let's take a square of side $d$ and fit it around the cylinder
so there is an edge of the square going diametrically across each end of the cylinder.
Then we can move the square around to cover not just the curved part of the surface but also the closed ends.
The perimeter of the square is $4d,$ and the diameter of the cylinder is $d,$
so we expect the surface area to be $4d^2.$
How well did we do?
In case it is not clear, I did not really mean for anyone to believe everything I wrote in the first two paragraphs.
I did mean for you to believe everything I wrote in the third paragraph, however.
I believe the third paragraph is a fair application of the method used in the question, but with slightly less hand-waving.
answered Jul 23 at 21:46
David K
48.2k340107
answered Jul 23 at 21:46
David K
48.2k340107
answered Jul 23 at 21:46
David K
48.2k340107
answered Jul 23 at 21:46
David K
48.2k340107
48.2k340107
add a comment |Â
add a comment |Â
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But that is the surface area of a right circular cylinder of height $2r$ and radius $r$
– ab123
Jul 23 at 20:31
Do you mean the lateral surface?
– Jon due
Jul 23 at 20:32
yes, the lateral surface area
– ab123
Jul 23 at 20:33
@Jondue Try your approach with a cylinder. Does it yield the true surface area of the cylinder?
– user76284
Jul 23 at 20:33
2
No, it's not a correct proof. You have stumbled on an important coincidence that dates back to Archimedes: en.wikipedia.org/wiki/On_the_Sphere_and_Cylinder
– Ethan Bolker
Jul 23 at 20:39