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Paradox about the volume of a cylinder
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Trying to apply Cavalieri's method of indivisibles to calculate the volume of a cylinder with radius $R$ and height $h$, I get the following paradoxical argument.
A cylinder with radius $R$ and height $h$ can be seen as a solid obtained by rotating a rectangle with height $h$ and base $R$ about its height. Therefore, the volume of the cylinder can be thought as made out of an infinity of areas of such rectangles of infinitesimal thickness rotated for $360^circ$; hence, the volume $V$ of the cylinder should the area of the rectangle $A_textrect = R cdot h$ multiplied by the circumference of the rotation circle $C_textcirc = 2pi R$:
beginalign
V = A_textrect cdot C_textcirc = 2 pi R^2 cdot h
endalign
Of course, the right volume of a cylinder with radius $R$ and height $h$ is
beginalign
V = A_textcirc cdot h = pi R^2 cdot h
endalign
where $A_textcirc = pi R^2$ is the area of the base circle of the cylinder.
Question: Where is the error in my previous argument based on infinitesimals?
calculus integration proof-verification volume infinitesimals
edited 16 mins ago
Taroccoesbrocco
4,31461535
asked 57 mins ago
Ruggiero Rilievi
335
add a comment |Â
up vote
2
down vote
favorite
Trying to apply Cavalieri's method of indivisibles to calculate the volume of a cylinder with radius $R$ and height $h$, I get the following paradoxical argument.
A cylinder with radius $R$ and height $h$ can be seen as a solid obtained by rotating a rectangle with height $h$ and base $R$ about its height. Therefore, the volume of the cylinder can be thought as made out of an infinity of areas of such rectangles of infinitesimal thickness rotated for $360^circ$; hence, the volume $V$ of the cylinder should the area of the rectangle $A_textrect = R cdot h$ multiplied by the circumference of the rotation circle $C_textcirc = 2pi R$:
beginalign
V = A_textrect cdot C_textcirc = 2 pi R^2 cdot h
endalign
Of course, the right volume of a cylinder with radius $R$ and height $h$ is
beginalign
V = A_textcirc cdot h = pi R^2 cdot h
endalign
where $A_textcirc = pi R^2$ is the area of the base circle of the cylinder.
Question: Where is the error in my previous argument based on infinitesimals?
calculus integration proof-verification volume infinitesimals
edited 16 mins ago
Taroccoesbrocco
4,31461535
asked 57 mins ago
Ruggiero Rilievi
335
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Trying to apply Cavalieri's method of indivisibles to calculate the volume of a cylinder with radius $R$ and height $h$, I get the following paradoxical argument.
A cylinder with radius $R$ and height $h$ can be seen as a solid obtained by rotating a rectangle with height $h$ and base $R$ about its height. Therefore, the volume of the cylinder can be thought as made out of an infinity of areas of such rectangles of infinitesimal thickness rotated for $360^circ$; hence, the volume $V$ of the cylinder should the area of the rectangle $A_textrect = R cdot h$ multiplied by the circumference of the rotation circle $C_textcirc = 2pi R$:
beginalign
V = A_textrect cdot C_textcirc = 2 pi R^2 cdot h
endalign
Of course, the right volume of a cylinder with radius $R$ and height $h$ is
beginalign
V = A_textcirc cdot h = pi R^2 cdot h
endalign
where $A_textcirc = pi R^2$ is the area of the base circle of the cylinder.
Question: Where is the error in my previous argument based on infinitesimals?
calculus integration proof-verification volume infinitesimals
edited 16 mins ago
Taroccoesbrocco
4,31461535
asked 57 mins ago
Ruggiero Rilievi
335
Trying to apply Cavalieri's method of indivisibles to calculate the volume of a cylinder with radius $R$ and height $h$, I get the following paradoxical argument.
A cylinder with radius $R$ and height $h$ can be seen as a solid obtained by rotating a rectangle with height $h$ and base $R$ about its height. Therefore, the volume of the cylinder can be thought as made out of an infinity of areas of such rectangles of infinitesimal thickness rotated for $360^circ$; hence, the volume $V$ of the cylinder should the area of the rectangle $A_textrect = R cdot h$ multiplied by the circumference of the rotation circle $C_textcirc = 2pi R$:
beginalign
V = A_textrect cdot C_textcirc = 2 pi R^2 cdot h
endalign
Of course, the right volume of a cylinder with radius $R$ and height $h$ is
beginalign
V = A_textcirc cdot h = pi R^2 cdot h
endalign
where $A_textcirc = pi R^2$ is the area of the base circle of the cylinder.
Question: Where is the error in my previous argument based on infinitesimals?
calculus integration proof-verification volume infinitesimals
calculus integration proof-verification volume infinitesimals
edited 16 mins ago
Taroccoesbrocco
4,31461535
asked 57 mins ago
Ruggiero Rilievi
335
edited 16 mins ago
Taroccoesbrocco
4,31461535
asked 57 mins ago
Ruggiero Rilievi
335
edited 16 mins ago
Taroccoesbrocco
4,31461535
edited 16 mins ago
Taroccoesbrocco
4,31461535
edited 16 mins ago
Taroccoesbrocco
4,31461535
4,31461535
asked 57 mins ago
Ruggiero Rilievi
335
asked 57 mins ago
Ruggiero Rilievi
335
asked 57 mins ago
Ruggiero Rilievi
335
335
add a comment |Â
add a comment |Â
1 Answer
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up vote
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The rotation contributes to the volume by $2pi R$ only for the side of the rectangle that is opposite to its rotation axis. The circumference covered by each point of the rectangle depends on its distance from the rotation axis.
Keeping an infinitesimal approach, think of the rectangle as made out of an infinity of vertical lines of infinitesimal thickness $dr$ at a distance $r$ (with $0 leq r leq R$) from the rotation axis. Every vertical line (or infinitesimal rectangle) rotates for $2pi r$, so its contribution to the volume of the cylinder is $dV = h cdot dr cdot 2 pi r$. Therefore, the volume of the cylinder is the infinite sum of such infinitesimal volumes, i.e.
beginalign
V = int dV = 2 pi h int_0^R r dr = 2 pi h left[fracr^22 right]^R_0 = pi R^2 h.
endalign
answered 23 mins ago
Taroccoesbrocco
4,31461535
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
The rotation contributes to the volume by $2pi R$ only for the side of the rectangle that is opposite to its rotation axis. The circumference covered by each point of the rectangle depends on its distance from the rotation axis.
Keeping an infinitesimal approach, think of the rectangle as made out of an infinity of vertical lines of infinitesimal thickness $dr$ at a distance $r$ (with $0 leq r leq R$) from the rotation axis. Every vertical line (or infinitesimal rectangle) rotates for $2pi r$, so its contribution to the volume of the cylinder is $dV = h cdot dr cdot 2 pi r$. Therefore, the volume of the cylinder is the infinite sum of such infinitesimal volumes, i.e.
beginalign
V = int dV = 2 pi h int_0^R r dr = 2 pi h left[fracr^22 right]^R_0 = pi R^2 h.
endalign
answered 23 mins ago
Taroccoesbrocco
4,31461535
add a comment |Â
up vote
4
down vote
The rotation contributes to the volume by $2pi R$ only for the side of the rectangle that is opposite to its rotation axis. The circumference covered by each point of the rectangle depends on its distance from the rotation axis.
Keeping an infinitesimal approach, think of the rectangle as made out of an infinity of vertical lines of infinitesimal thickness $dr$ at a distance $r$ (with $0 leq r leq R$) from the rotation axis. Every vertical line (or infinitesimal rectangle) rotates for $2pi r$, so its contribution to the volume of the cylinder is $dV = h cdot dr cdot 2 pi r$. Therefore, the volume of the cylinder is the infinite sum of such infinitesimal volumes, i.e.
beginalign
V = int dV = 2 pi h int_0^R r dr = 2 pi h left[fracr^22 right]^R_0 = pi R^2 h.
endalign
answered 23 mins ago
Taroccoesbrocco
4,31461535
add a comment |Â
up vote
4
down vote
up vote
4
down vote
The rotation contributes to the volume by $2pi R$ only for the side of the rectangle that is opposite to its rotation axis. The circumference covered by each point of the rectangle depends on its distance from the rotation axis.
Keeping an infinitesimal approach, think of the rectangle as made out of an infinity of vertical lines of infinitesimal thickness $dr$ at a distance $r$ (with $0 leq r leq R$) from the rotation axis. Every vertical line (or infinitesimal rectangle) rotates for $2pi r$, so its contribution to the volume of the cylinder is $dV = h cdot dr cdot 2 pi r$. Therefore, the volume of the cylinder is the infinite sum of such infinitesimal volumes, i.e.
beginalign
V = int dV = 2 pi h int_0^R r dr = 2 pi h left[fracr^22 right]^R_0 = pi R^2 h.
endalign
answered 23 mins ago
Taroccoesbrocco
4,31461535
The rotation contributes to the volume by $2pi R$ only for the side of the rectangle that is opposite to its rotation axis. The circumference covered by each point of the rectangle depends on its distance from the rotation axis.
Keeping an infinitesimal approach, think of the rectangle as made out of an infinity of vertical lines of infinitesimal thickness $dr$ at a distance $r$ (with $0 leq r leq R$) from the rotation axis. Every vertical line (or infinitesimal rectangle) rotates for $2pi r$, so its contribution to the volume of the cylinder is $dV = h cdot dr cdot 2 pi r$. Therefore, the volume of the cylinder is the infinite sum of such infinitesimal volumes, i.e.
beginalign
V = int dV = 2 pi h int_0^R r dr = 2 pi h left[fracr^22 right]^R_0 = pi R^2 h.
endalign
answered 23 mins ago
Taroccoesbrocco
4,31461535
edited 18 mins ago
answered 23 mins ago
Taroccoesbrocco
4,31461535
answered 23 mins ago
Taroccoesbrocco
4,31461535
answered 23 mins ago
Taroccoesbrocco
4,31461535
4,31461535
add a comment |Â
add a comment |Â
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