Does this 3d geometric shape has a name? Three mutually-perpendicular disks

The name of the pictureThe name of the pictureThe name of the pictureClash 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.



The shape is the same from every angle



Thanks










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  • So basically this is $x^2+y^2=r^2$, $x^2+z^2=r^2$ and $y^2+z^2=r^2$?
    – Mohammad Zuhair Khan
    Aug 30 at 8:07






  • 2




    en.m.wikipedia.org/wiki/Armillary_sphere.
    – Oscar Lanzi
    Aug 30 at 8:09










  • @MohammadZuhairKhan $x^2+y^2=r^2$ is a cylinder.
    – lisyarus
    Aug 30 at 8:21










  • Oh right. I should probably have added $z=0$ to it. Thanks for the heads up @lisyarus
    – 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.
    – lisyarus
    Aug 30 at 8:23














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.



The shape is the same from every angle



Thanks










share|cite|improve this question























  • So basically this is $x^2+y^2=r^2$, $x^2+z^2=r^2$ and $y^2+z^2=r^2$?
    – Mohammad Zuhair Khan
    Aug 30 at 8:07






  • 2




    en.m.wikipedia.org/wiki/Armillary_sphere.
    – Oscar Lanzi
    Aug 30 at 8:09










  • @MohammadZuhairKhan $x^2+y^2=r^2$ is a cylinder.
    – lisyarus
    Aug 30 at 8:21










  • Oh right. I should probably have added $z=0$ to it. Thanks for the heads up @lisyarus
    – 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.
    – lisyarus
    Aug 30 at 8:23












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.



The shape is the same from every angle



Thanks










share|cite|improve this question















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.



The shape is the same from every angle



Thanks







geometry terminology






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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 $y^2+z^2=r^2$?
    – Mohammad Zuhair Khan
    Aug 30 at 8:07






  • 2




    en.m.wikipedia.org/wiki/Armillary_sphere.
    – Oscar Lanzi
    Aug 30 at 8:09










  • @MohammadZuhairKhan $x^2+y^2=r^2$ is a cylinder.
    – lisyarus
    Aug 30 at 8:21










  • Oh right. I should probably have added $z=0$ to it. Thanks for the heads up @lisyarus
    – 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.
    – lisyarus
    Aug 30 at 8:23
















  • So basically this is $x^2+y^2=r^2$, $x^2+z^2=r^2$ and $y^2+z^2=r^2$?
    – Mohammad Zuhair Khan
    Aug 30 at 8:07






  • 2




    en.m.wikipedia.org/wiki/Armillary_sphere.
    – Oscar Lanzi
    Aug 30 at 8:09










  • @MohammadZuhairKhan $x^2+y^2=r^2$ is a cylinder.
    – lisyarus
    Aug 30 at 8:21










  • Oh right. I should probably have added $z=0$ to it. Thanks for the heads up @lisyarus
    – 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.
    – lisyarus
    Aug 30 at 8:23















So basically this is $x^2+y^2=r^2$, $x^2+z^2=r^2$ and $y^2+z^2=r^2$?
– 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 $y^2+z^2=r^2$?
– Mohammad Zuhair Khan
Aug 30 at 8:07




2




2




en.m.wikipedia.org/wiki/Armillary_sphere.
– Oscar Lanzi
Aug 30 at 8:09




en.m.wikipedia.org/wiki/Armillary_sphere.
– Oscar Lanzi
Aug 30 at 8:09












@MohammadZuhairKhan $x^2+y^2=r^2$ is a cylinder.
– lisyarus
Aug 30 at 8:21




@MohammadZuhairKhan $x^2+y^2=r^2$ is a cylinder.
– lisyarus
Aug 30 at 8:21












Oh right. I should probably have added $z=0$ to it. Thanks for the heads up @lisyarus
– 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
– 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.
– 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.
– lisyarus
Aug 30 at 8:23















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