Gabriel's Horn Volume is $pi$ but the Circle Area at $x = 1$ is $pi$
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We know that Gabriel's Horn has the volume $pi $ when rotating $f(x) = frac1x$ around the $x$-axis for $x in [1,infty)$. From disk method, we calculate the volume of this object by summing an infinite sum of a disk with the radius $f(x)$ and a really small height of the disk $dx$.
But we know that the circle(disk with infinitesimal $dx$) area at the entrance of the Gabriel's Horn is $pi*frac11^2 = pi$; thus the volume of Gabriel's Horn should be bigger than $pi$ but that, of course, contradicts the integration calculation.
Where could I possibly go wrong? Is there any conceptual understanding that I miss or misunderstood?
calculus integration proof-verification proof-explanation
asked Sep 8 at 7:23
Rayyan
112
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up vote
1
down vote
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We know that Gabriel's Horn has the volume $pi $ when rotating $f(x) = frac1x$ around the $x$-axis for $x in [1,infty)$. From disk method, we calculate the volume of this object by summing an infinite sum of a disk with the radius $f(x)$ and a really small height of the disk $dx$.
But we know that the circle(disk with infinitesimal $dx$) area at the entrance of the Gabriel's Horn is $pi*frac11^2 = pi$; thus the volume of Gabriel's Horn should be bigger than $pi$ but that, of course, contradicts the integration calculation.
Where could I possibly go wrong? Is there any conceptual understanding that I miss or misunderstood?
calculus integration proof-verification proof-explanation
asked Sep 8 at 7:23
Rayyan
112
2
Why on earth do you expect the volume to be more than $pi$?
– Lord Shark the Unknown
Sep 8 at 7:30
1
As dangerous as entering this kind of logic is, the volume of the "infinitesimal cylinder" would be $pi dx$, which is another "infinitesimal", not $pi$.
– Saucy O'Path
Sep 8 at 7:42
If there were units involved in the calculation, it would be clear that the area and the volume had different units and that you couldn't compare them. Just because the calculation is done without units doesn't mean you get to compare apples to oranges.
– Henrik
Sep 8 at 7:43
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
We know that Gabriel's Horn has the volume $pi $ when rotating $f(x) = frac1x$ around the $x$-axis for $x in [1,infty)$. From disk method, we calculate the volume of this object by summing an infinite sum of a disk with the radius $f(x)$ and a really small height of the disk $dx$.
But we know that the circle(disk with infinitesimal $dx$) area at the entrance of the Gabriel's Horn is $pi*frac11^2 = pi$; thus the volume of Gabriel's Horn should be bigger than $pi$ but that, of course, contradicts the integration calculation.
Where could I possibly go wrong? Is there any conceptual understanding that I miss or misunderstood?
calculus integration proof-verification proof-explanation
asked Sep 8 at 7:23
Rayyan
112
We know that Gabriel's Horn has the volume $pi $ when rotating $f(x) = frac1x$ around the $x$-axis for $x in [1,infty)$. From disk method, we calculate the volume of this object by summing an infinite sum of a disk with the radius $f(x)$ and a really small height of the disk $dx$.
But we know that the circle(disk with infinitesimal $dx$) area at the entrance of the Gabriel's Horn is $pi*frac11^2 = pi$; thus the volume of Gabriel's Horn should be bigger than $pi$ but that, of course, contradicts the integration calculation.
Where could I possibly go wrong? Is there any conceptual understanding that I miss or misunderstood?
calculus integration proof-verification proof-explanation
calculus integration proof-verification proof-explanation
asked Sep 8 at 7:23
Rayyan
112
asked Sep 8 at 7:23
Rayyan
112
asked Sep 8 at 7:23
Rayyan
112
asked Sep 8 at 7:23
Rayyan
112
asked Sep 8 at 7:23
Rayyan
112
112
2
Why on earth do you expect the volume to be more than $pi$?
– Lord Shark the Unknown
Sep 8 at 7:30
1
As dangerous as entering this kind of logic is, the volume of the "infinitesimal cylinder" would be $pi dx$, which is another "infinitesimal", not $pi$.
– Saucy O'Path
Sep 8 at 7:42
If there were units involved in the calculation, it would be clear that the area and the volume had different units and that you couldn't compare them. Just because the calculation is done without units doesn't mean you get to compare apples to oranges.
– Henrik
Sep 8 at 7:43
add a comment |Â
2
Why on earth do you expect the volume to be more than $pi$?
– Lord Shark the Unknown
Sep 8 at 7:30
1
As dangerous as entering this kind of logic is, the volume of the "infinitesimal cylinder" would be $pi dx$, which is another "infinitesimal", not $pi$.
– Saucy O'Path
Sep 8 at 7:42
If there were units involved in the calculation, it would be clear that the area and the volume had different units and that you couldn't compare them. Just because the calculation is done without units doesn't mean you get to compare apples to oranges.
– Henrik
Sep 8 at 7:43
2
2
Why on earth do you expect the volume to be more than $pi$?
– Lord Shark the Unknown
Sep 8 at 7:30
Why on earth do you expect the volume to be more than $pi$?
– Lord Shark the Unknown
Sep 8 at 7:30
1
1
As dangerous as entering this kind of logic is, the volume of the "infinitesimal cylinder" would be $pi dx$, which is another "infinitesimal", not $pi$.
– Saucy O'Path
Sep 8 at 7:42
As dangerous as entering this kind of logic is, the volume of the "infinitesimal cylinder" would be $pi dx$, which is another "infinitesimal", not $pi$.
– Saucy O'Path
Sep 8 at 7:42
If there were units involved in the calculation, it would be clear that the area and the volume had different units and that you couldn't compare them. Just because the calculation is done without units doesn't mean you get to compare apples to oranges.
– Henrik
Sep 8 at 7:43
If there were units involved in the calculation, it would be clear that the area and the volume had different units and that you couldn't compare them. Just because the calculation is done without units doesn't mean you get to compare apples to oranges.
– Henrik
Sep 8 at 7:43
add a comment |Â
2 Answers
2
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up vote
2
down vote
accepted
$À$ represents the volume of Gabriel's Horn, and the volume of its cross section at $x = 1$ contributes very little to the total volume of the horn itself, and is far smaller in value than the area of the circle you speak of.
I think this problem lies in a misunderstanding of the relationship between volume and area. The volume of an entire object can be less than the area of one of its sides. Take a right rectangular prism (a box) that has measurements of $4l space * 8w space *0.25h$. Its largest face has an area of $32$ units$^2$, whereas its entire volume is $8$ units$^3$.
This is exactly the case in Gabriel's Horn. While the area of its left-most circular face is indeed $À$, we do not calculate the total volume by summing up the areas of each cross section's circular face, instead we sum up the each cross section's infinitesimally small volume, which we calculate as $À$ using the disk method.
answered Sep 8 at 7:57
Michael Lilley
828
add a comment |Â
up vote
2
down vote
What you're missing here is that the disc in question has no volume at all: instead, you must take all the infinite number of infinitely thin sheets in order to create the whole volume. To see this in a more immediately graspable situation, let's consider the square pyramid of base length 1 and height 1; the true volume is 1/3 (in cube units) and the area of the base is 1 (in square units). The cross section as you go up the pyramid decreases quickly enough that it doesn't fill the unit cube at all. The same is true of Gabriel's Horn.
answered Sep 8 at 7:33
Dan Uznanski
6,25021327
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
$À$ represents the volume of Gabriel's Horn, and the volume of its cross section at $x = 1$ contributes very little to the total volume of the horn itself, and is far smaller in value than the area of the circle you speak of.
I think this problem lies in a misunderstanding of the relationship between volume and area. The volume of an entire object can be less than the area of one of its sides. Take a right rectangular prism (a box) that has measurements of $4l space * 8w space *0.25h$. Its largest face has an area of $32$ units$^2$, whereas its entire volume is $8$ units$^3$.
This is exactly the case in Gabriel's Horn. While the area of its left-most circular face is indeed $À$, we do not calculate the total volume by summing up the areas of each cross section's circular face, instead we sum up the each cross section's infinitesimally small volume, which we calculate as $À$ using the disk method.
answered Sep 8 at 7:57
Michael Lilley
828
add a comment |Â
up vote
2
down vote
accepted
$À$ represents the volume of Gabriel's Horn, and the volume of its cross section at $x = 1$ contributes very little to the total volume of the horn itself, and is far smaller in value than the area of the circle you speak of.
I think this problem lies in a misunderstanding of the relationship between volume and area. The volume of an entire object can be less than the area of one of its sides. Take a right rectangular prism (a box) that has measurements of $4l space * 8w space *0.25h$. Its largest face has an area of $32$ units$^2$, whereas its entire volume is $8$ units$^3$.
This is exactly the case in Gabriel's Horn. While the area of its left-most circular face is indeed $À$, we do not calculate the total volume by summing up the areas of each cross section's circular face, instead we sum up the each cross section's infinitesimally small volume, which we calculate as $À$ using the disk method.
answered Sep 8 at 7:57
Michael Lilley
828
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
$À$ represents the volume of Gabriel's Horn, and the volume of its cross section at $x = 1$ contributes very little to the total volume of the horn itself, and is far smaller in value than the area of the circle you speak of.
I think this problem lies in a misunderstanding of the relationship between volume and area. The volume of an entire object can be less than the area of one of its sides. Take a right rectangular prism (a box) that has measurements of $4l space * 8w space *0.25h$. Its largest face has an area of $32$ units$^2$, whereas its entire volume is $8$ units$^3$.
This is exactly the case in Gabriel's Horn. While the area of its left-most circular face is indeed $À$, we do not calculate the total volume by summing up the areas of each cross section's circular face, instead we sum up the each cross section's infinitesimally small volume, which we calculate as $À$ using the disk method.
answered Sep 8 at 7:57
Michael Lilley
828
$À$ represents the volume of Gabriel's Horn, and the volume of its cross section at $x = 1$ contributes very little to the total volume of the horn itself, and is far smaller in value than the area of the circle you speak of.
I think this problem lies in a misunderstanding of the relationship between volume and area. The volume of an entire object can be less than the area of one of its sides. Take a right rectangular prism (a box) that has measurements of $4l space * 8w space *0.25h$. Its largest face has an area of $32$ units$^2$, whereas its entire volume is $8$ units$^3$.
This is exactly the case in Gabriel's Horn. While the area of its left-most circular face is indeed $À$, we do not calculate the total volume by summing up the areas of each cross section's circular face, instead we sum up the each cross section's infinitesimally small volume, which we calculate as $À$ using the disk method.
answered Sep 8 at 7:57
Michael Lilley
828
edited Sep 8 at 8:04
answered Sep 8 at 7:57
Michael Lilley
828
answered Sep 8 at 7:57
Michael Lilley
828
answered Sep 8 at 7:57
Michael Lilley
828
828
add a comment |Â
add a comment |Â
up vote
2
down vote
What you're missing here is that the disc in question has no volume at all: instead, you must take all the infinite number of infinitely thin sheets in order to create the whole volume. To see this in a more immediately graspable situation, let's consider the square pyramid of base length 1 and height 1; the true volume is 1/3 (in cube units) and the area of the base is 1 (in square units). The cross section as you go up the pyramid decreases quickly enough that it doesn't fill the unit cube at all. The same is true of Gabriel's Horn.
answered Sep 8 at 7:33
Dan Uznanski
6,25021327
add a comment |Â
up vote
2
down vote
What you're missing here is that the disc in question has no volume at all: instead, you must take all the infinite number of infinitely thin sheets in order to create the whole volume. To see this in a more immediately graspable situation, let's consider the square pyramid of base length 1 and height 1; the true volume is 1/3 (in cube units) and the area of the base is 1 (in square units). The cross section as you go up the pyramid decreases quickly enough that it doesn't fill the unit cube at all. The same is true of Gabriel's Horn.
answered Sep 8 at 7:33
Dan Uznanski
6,25021327
add a comment |Â
up vote
2
down vote
up vote
2
down vote
What you're missing here is that the disc in question has no volume at all: instead, you must take all the infinite number of infinitely thin sheets in order to create the whole volume. To see this in a more immediately graspable situation, let's consider the square pyramid of base length 1 and height 1; the true volume is 1/3 (in cube units) and the area of the base is 1 (in square units). The cross section as you go up the pyramid decreases quickly enough that it doesn't fill the unit cube at all. The same is true of Gabriel's Horn.
answered Sep 8 at 7:33
Dan Uznanski
6,25021327
What you're missing here is that the disc in question has no volume at all: instead, you must take all the infinite number of infinitely thin sheets in order to create the whole volume. To see this in a more immediately graspable situation, let's consider the square pyramid of base length 1 and height 1; the true volume is 1/3 (in cube units) and the area of the base is 1 (in square units). The cross section as you go up the pyramid decreases quickly enough that it doesn't fill the unit cube at all. The same is true of Gabriel's Horn.
answered Sep 8 at 7:33
Dan Uznanski
6,25021327
answered Sep 8 at 7:33
Dan Uznanski
6,25021327
answered Sep 8 at 7:33
Dan Uznanski
6,25021327
answered Sep 8 at 7:33
Dan Uznanski
6,25021327
6,25021327
add a comment |Â
add a comment |Â
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2
Why on earth do you expect the volume to be more than $pi$?
– Lord Shark the Unknown
Sep 8 at 7:30
1
As dangerous as entering this kind of logic is, the volume of the "infinitesimal cylinder" would be $pi dx$, which is another "infinitesimal", not $pi$.
– Saucy O'Path
Sep 8 at 7:42
If there were units involved in the calculation, it would be clear that the area and the volume had different units and that you couldn't compare them. Just because the calculation is done without units doesn't mean you get to compare apples to oranges.
– Henrik
Sep 8 at 7:43