Mixing
Find volume of crossed cylinders without calculus.
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I found this puzzle here. (It's labeled "crossed cylinders".) Here's the description:
Two cylinders of equal radius are intersected at right angles as shown at left. Find the volume of the intersection between the two cylinders, without using calculus! A 3D picture of the intersection is shown at right.
Hint (medium hint - exactly which high school formulae you need): 1) Area of circle = pi * radius2, and 2) Volume of sphere = (4/3) * pi * radius3
Note: Solved by the mathematician Archimedes (287 B.C. - 212 B.C.), waaay before calculus came around!!
Please tell me how to solve this puzzle! Is there a way to do this without setting up a Riemann sum and finding a limit, essentially evaluating an integral?
geometry euclidean-geometry puzzle volume faq
edited Jan 22 at 10:47
Lee David Chung Lin
4,37031241
asked Nov 15 '12 at 12:11
littleOlittleO
29.9k647110
$endgroup$
add a comment |
$begingroup$
I found this puzzle here. (It's labeled "crossed cylinders".) Here's the description:
Two cylinders of equal radius are intersected at right angles as shown at left. Find the volume of the intersection between the two cylinders, without using calculus! A 3D picture of the intersection is shown at right.
Hint (medium hint - exactly which high school formulae you need): 1) Area of circle = pi * radius2, and 2) Volume of sphere = (4/3) * pi * radius3
Note: Solved by the mathematician Archimedes (287 B.C. - 212 B.C.), waaay before calculus came around!!
Please tell me how to solve this puzzle! Is there a way to do this without setting up a Riemann sum and finding a limit, essentially evaluating an integral?
geometry euclidean-geometry puzzle volume faq
edited Jan 22 at 10:47
Lee David Chung Lin
4,37031241
asked Nov 15 '12 at 12:11
littleOlittleO
29.9k647110
$endgroup$
$begingroup$
This post is chosen to be the target for duplicates for finding the volume of Steinmetz solid without using calculus. For the the calculus approach, please see this.
$endgroup$
– Lee David Chung Lin
Jan 22 at 10:47
$begingroup$
Also see this answer with arguably better content.
$endgroup$
– Lee David Chung Lin
Jan 22 at 14:25
add a comment |
$begingroup$
I found this puzzle here. (It's labeled "crossed cylinders".) Here's the description:
Two cylinders of equal radius are intersected at right angles as shown at left. Find the volume of the intersection between the two cylinders, without using calculus! A 3D picture of the intersection is shown at right.
Hint (medium hint - exactly which high school formulae you need): 1) Area of circle = pi * radius2, and 2) Volume of sphere = (4/3) * pi * radius3
Note: Solved by the mathematician Archimedes (287 B.C. - 212 B.C.), waaay before calculus came around!!
Please tell me how to solve this puzzle! Is there a way to do this without setting up a Riemann sum and finding a limit, essentially evaluating an integral?
geometry euclidean-geometry puzzle volume faq
edited Jan 22 at 10:47
Lee David Chung Lin
4,37031241
asked Nov 15 '12 at 12:11
littleOlittleO
29.9k647110
$endgroup$
I found this puzzle here. (It's labeled "crossed cylinders".) Here's the description:
Two cylinders of equal radius are intersected at right angles as shown at left. Find the volume of the intersection between the two cylinders, without using calculus! A 3D picture of the intersection is shown at right.
Hint (medium hint - exactly which high school formulae you need): 1) Area of circle = pi * radius2, and 2) Volume of sphere = (4/3) * pi * radius3
Note: Solved by the mathematician Archimedes (287 B.C. - 212 B.C.), waaay before calculus came around!!
Please tell me how to solve this puzzle! Is there a way to do this without setting up a Riemann sum and finding a limit, essentially evaluating an integral?
geometry euclidean-geometry puzzle volume faq
geometry euclidean-geometry puzzle volume faq
edited Jan 22 at 10:47
Lee David Chung Lin
4,37031241
asked Nov 15 '12 at 12:11
littleOlittleO
29.9k647110
edited Jan 22 at 10:47
Lee David Chung Lin
4,37031241
asked Nov 15 '12 at 12:11
littleOlittleO
29.9k647110
edited Jan 22 at 10:47
Lee David Chung Lin
4,37031241
edited Jan 22 at 10:47
Lee David Chung Lin
4,37031241
edited Jan 22 at 10:47
Lee David Chung Lin
4,37031241
4,37031241
asked Nov 15 '12 at 12:11
littleOlittleO
29.9k647110
asked Nov 15 '12 at 12:11
littleOlittleO
29.9k647110
asked Nov 15 '12 at 12:11
littleOlittleO
29.9k647110
29.9k647110
$begingroup$
This post is chosen to be the target for duplicates for finding the volume of Steinmetz solid without using calculus. For the the calculus approach, please see this.
$endgroup$
– Lee David Chung Lin
Jan 22 at 10:47
$begingroup$
Also see this answer with arguably better content.
$endgroup$
– Lee David Chung Lin
Jan 22 at 14:25
add a comment |
$begingroup$
This post is chosen to be the target for duplicates for finding the volume of Steinmetz solid without using calculus. For the the calculus approach, please see this.
$endgroup$
– Lee David Chung Lin
Jan 22 at 10:47
$begingroup$
Also see this answer with arguably better content.
$endgroup$
– Lee David Chung Lin
Jan 22 at 14:25
$begingroup$
This post is chosen to be the target for duplicates for finding the volume of Steinmetz solid without using calculus. For the the calculus approach, please see this.
$endgroup$
– Lee David Chung Lin
Jan 22 at 10:47
$begingroup$
This post is chosen to be the target for duplicates for finding the volume of Steinmetz solid without using calculus. For the the calculus approach, please see this.
$endgroup$
– Lee David Chung Lin
Jan 22 at 10:47
$begingroup$
Also see this answer with arguably better content.
$endgroup$
– Lee David Chung Lin
Jan 22 at 14:25
$begingroup$
Also see this answer with arguably better content.
$endgroup$
– Lee David Chung Lin
Jan 22 at 14:25
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Compare the solid to a sphere of the same size. The slices of the solid are squares to the sphere's circles, so the ratio of areas is always $4/pi$. The volumes must therefore be in the same ratio, giving $(4/pi) cdot (4/3)pi r^3 = (16/3) r^3$.
answered Nov 15 '12 at 12:26
RahulRahul
33.2k568173
$endgroup$
$begingroup$
Wow, that's brilliant, thank you!
$endgroup$
– littleO
Nov 15 '12 at 12:28
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Compare the solid to a sphere of the same size. The slices of the solid are squares to the sphere's circles, so the ratio of areas is always $4/pi$. The volumes must therefore be in the same ratio, giving $(4/pi) cdot (4/3)pi r^3 = (16/3) r^3$.
answered Nov 15 '12 at 12:26
RahulRahul
33.2k568173
$endgroup$
$begingroup$
Wow, that's brilliant, thank you!
$endgroup$
– littleO
Nov 15 '12 at 12:28
add a comment |
$begingroup$
Compare the solid to a sphere of the same size. The slices of the solid are squares to the sphere's circles, so the ratio of areas is always $4/pi$. The volumes must therefore be in the same ratio, giving $(4/pi) cdot (4/3)pi r^3 = (16/3) r^3$.
answered Nov 15 '12 at 12:26
RahulRahul
33.2k568173
$endgroup$
$begingroup$
Wow, that's brilliant, thank you!
$endgroup$
– littleO
Nov 15 '12 at 12:28
add a comment |
$begingroup$
Compare the solid to a sphere of the same size. The slices of the solid are squares to the sphere's circles, so the ratio of areas is always $4/pi$. The volumes must therefore be in the same ratio, giving $(4/pi) cdot (4/3)pi r^3 = (16/3) r^3$.
answered Nov 15 '12 at 12:26
RahulRahul
33.2k568173
$endgroup$
Compare the solid to a sphere of the same size. The slices of the solid are squares to the sphere's circles, so the ratio of areas is always $4/pi$. The volumes must therefore be in the same ratio, giving $(4/pi) cdot (4/3)pi r^3 = (16/3) r^3$.
answered Nov 15 '12 at 12:26
RahulRahul
33.2k568173
answered Nov 15 '12 at 12:26
RahulRahul
33.2k568173
answered Nov 15 '12 at 12:26
RahulRahul
33.2k568173
answered Nov 15 '12 at 12:26
RahulRahul
33.2k568173
33.2k568173
$begingroup$
Wow, that's brilliant, thank you!
$endgroup$
– littleO
Nov 15 '12 at 12:28
add a comment |
$begingroup$
Wow, that's brilliant, thank you!
$endgroup$
– littleO
Nov 15 '12 at 12:28
$begingroup$
Wow, that's brilliant, thank you!
$endgroup$
– littleO
Nov 15 '12 at 12:28
$begingroup$
Wow, that's brilliant, thank you!
$endgroup$
– littleO
Nov 15 '12 at 12:28
add a comment |
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$begingroup$
This post is chosen to be the target for duplicates for finding the volume of Steinmetz solid without using calculus. For the the calculus approach, please see this.
$endgroup$
– Lee David Chung Lin
Jan 22 at 10:47
$begingroup$
Also see this answer with arguably better content.
$endgroup$
– Lee David Chung Lin
Jan 22 at 14:25