Does this 3d geometric shape has a name? Three mutually-perpendicular disks
Clash Royale CLAN TAG#URR8PPP
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Someone knows if this shape has a name?
It is the resulting figure of making a circle in each plane of the three axes, each of the same radius.
Sorry for the image, from any angle one can see three semi circles cutting each other.
Edit: I'm asking for a name in english, of course not for de equation.
Thanks
geometry terminology
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up vote
-1
down vote
favorite
Someone knows if this shape has a name?
It is the resulting figure of making a circle in each plane of the three axes, each of the same radius.
Sorry for the image, from any angle one can see three semi circles cutting each other.
Edit: I'm asking for a name in english, of course not for de equation.
Thanks
geometry terminology
So basically this is $x^2+y^2=r^2$, $x^2+z^2=r^2$ and $y^2+z^2=r^2$?
â Mohammad Zuhair Khan
Aug 30 at 8:07
2
en.m.wikipedia.org/wiki/Armillary_sphere.
â Oscar Lanzi
Aug 30 at 8:09
@MohammadZuhairKhan $x^2+y^2=r^2$ is a cylinder.
â lisyarus
Aug 30 at 8:21
Oh right. I should probably have added $z=0$ to it. Thanks for the heads up @lisyarus
â Mohammad Zuhair Khan
Aug 30 at 8:22
@MohammadZuhairKhan Then it would be a circle, not a disk :) It should be $x^2+y^2<r^2 wedge z=0$, and similar so two other disks.
â lisyarus
Aug 30 at 8:23
 |Â
show 3 more comments
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Someone knows if this shape has a name?
It is the resulting figure of making a circle in each plane of the three axes, each of the same radius.
Sorry for the image, from any angle one can see three semi circles cutting each other.
Edit: I'm asking for a name in english, of course not for de equation.
Thanks
geometry terminology
Someone knows if this shape has a name?
It is the resulting figure of making a circle in each plane of the three axes, each of the same radius.
Sorry for the image, from any angle one can see three semi circles cutting each other.
Edit: I'm asking for a name in english, of course not for de equation.
Thanks
geometry terminology
geometry terminology
edited Aug 30 at 9:26
joriki
167k10180333
167k10180333
asked Aug 30 at 8:05
Perico Cruel
245
245
So basically this is $x^2+y^2=r^2$, $x^2+z^2=r^2$ and $y^2+z^2=r^2$?
â Mohammad Zuhair Khan
Aug 30 at 8:07
2
en.m.wikipedia.org/wiki/Armillary_sphere.
â Oscar Lanzi
Aug 30 at 8:09
@MohammadZuhairKhan $x^2+y^2=r^2$ is a cylinder.
â lisyarus
Aug 30 at 8:21
Oh right. I should probably have added $z=0$ to it. Thanks for the heads up @lisyarus
â Mohammad Zuhair Khan
Aug 30 at 8:22
@MohammadZuhairKhan Then it would be a circle, not a disk :) It should be $x^2+y^2<r^2 wedge z=0$, and similar so two other disks.
â lisyarus
Aug 30 at 8:23
 |Â
show 3 more comments
So basically this is $x^2+y^2=r^2$, $x^2+z^2=r^2$ and $y^2+z^2=r^2$?
â Mohammad Zuhair Khan
Aug 30 at 8:07
2
en.m.wikipedia.org/wiki/Armillary_sphere.
â Oscar Lanzi
Aug 30 at 8:09
@MohammadZuhairKhan $x^2+y^2=r^2$ is a cylinder.
â lisyarus
Aug 30 at 8:21
Oh right. I should probably have added $z=0$ to it. Thanks for the heads up @lisyarus
â Mohammad Zuhair Khan
Aug 30 at 8:22
@MohammadZuhairKhan Then it would be a circle, not a disk :) It should be $x^2+y^2<r^2 wedge z=0$, and similar so two other disks.
â lisyarus
Aug 30 at 8:23
So basically this is $x^2+y^2=r^2$, $x^2+z^2=r^2$ and $y^2+z^2=r^2$?
â Mohammad Zuhair Khan
Aug 30 at 8:07
So basically this is $x^2+y^2=r^2$, $x^2+z^2=r^2$ and $y^2+z^2=r^2$?
â Mohammad Zuhair Khan
Aug 30 at 8:07
2
2
en.m.wikipedia.org/wiki/Armillary_sphere.
â Oscar Lanzi
Aug 30 at 8:09
en.m.wikipedia.org/wiki/Armillary_sphere.
â Oscar Lanzi
Aug 30 at 8:09
@MohammadZuhairKhan $x^2+y^2=r^2$ is a cylinder.
â lisyarus
Aug 30 at 8:21
@MohammadZuhairKhan $x^2+y^2=r^2$ is a cylinder.
â lisyarus
Aug 30 at 8:21
Oh right. I should probably have added $z=0$ to it. Thanks for the heads up @lisyarus
â Mohammad Zuhair Khan
Aug 30 at 8:22
Oh right. I should probably have added $z=0$ to it. Thanks for the heads up @lisyarus
â Mohammad Zuhair Khan
Aug 30 at 8:22
@MohammadZuhairKhan Then it would be a circle, not a disk :) It should be $x^2+y^2<r^2 wedge z=0$, and similar so two other disks.
â lisyarus
Aug 30 at 8:23
@MohammadZuhairKhan Then it would be a circle, not a disk :) It should be $x^2+y^2<r^2 wedge z=0$, and similar so two other disks.
â lisyarus
Aug 30 at 8:23
 |Â
show 3 more comments
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So basically this is $x^2+y^2=r^2$, $x^2+z^2=r^2$ and $y^2+z^2=r^2$?
â Mohammad Zuhair Khan
Aug 30 at 8:07
2
en.m.wikipedia.org/wiki/Armillary_sphere.
â Oscar Lanzi
Aug 30 at 8:09
@MohammadZuhairKhan $x^2+y^2=r^2$ is a cylinder.
â lisyarus
Aug 30 at 8:21
Oh right. I should probably have added $z=0$ to it. Thanks for the heads up @lisyarus
â Mohammad Zuhair Khan
Aug 30 at 8:22
@MohammadZuhairKhan Then it would be a circle, not a disk :) It should be $x^2+y^2<r^2 wedge z=0$, and similar so two other disks.
â lisyarus
Aug 30 at 8:23