parametrize the boundary of a region












1












$begingroup$


I need to parametrize the boundary of this region :



$D={y^2+z^2le x^2+18,x^2+y^2le 16}$



So It's a one-sheet hyperboloid (radius=$sqrt{18}$)+ cylinder with radius 4



I know how to parametrize them separately:



hyperboloid : $r(theta,phi)=(sinhphi,sqrt{18}sinthetacosh{phi},sqrt{18}costhetacosh{phi})$



cylinder: $r(theta,z)=(4costheta,4sintheta,z)$



But how do I connect them? Ho do I parametrize the surface boundary?



Or maybe "parametrize the boundary" implies giving two parametrizations?If it is so, are my parametrizations right?






multivariable-calculus parametric surface-integrals parametrization







share|cite|improve this question











$endgroup$












  • $begingroup$
    Apologies about delayed post. Shall post parameterizations and image tomorrow. Basically you are drilling/ intersecting a hyperboloid of 1 sheet with a cylinder.. There are two distorted "circular" disks produced. You can consider 1) the remaining continuous shell after drilling out the cylindrical holes or 2) the two discs themselves as the region depending on what you want to evaluate..
    $endgroup$
    – Narasimham
    Jan 17 at 23:13


















1












$begingroup$


I need to parametrize the boundary of this region :



$D={y^2+z^2le x^2+18,x^2+y^2le 16}$



So It's a one-sheet hyperboloid (radius=$sqrt{18}$)+ cylinder with radius 4



I know how to parametrize them separately:



hyperboloid : $r(theta,phi)=(sinhphi,sqrt{18}sinthetacosh{phi},sqrt{18}costhetacosh{phi})$



cylinder: $r(theta,z)=(4costheta,4sintheta,z)$



But how do I connect them? Ho do I parametrize the surface boundary?



Or maybe "parametrize the boundary" implies giving two parametrizations?If it is so, are my parametrizations right?






multivariable-calculus parametric surface-integrals parametrization







share|cite|improve this question











$endgroup$












  • $begingroup$
    Apologies about delayed post. Shall post parameterizations and image tomorrow. Basically you are drilling/ intersecting a hyperboloid of 1 sheet with a cylinder.. There are two distorted "circular" disks produced. You can consider 1) the remaining continuous shell after drilling out the cylindrical holes or 2) the two discs themselves as the region depending on what you want to evaluate..
    $endgroup$
    – Narasimham
    Jan 17 at 23:13
















1












1








1





$begingroup$


I need to parametrize the boundary of this region :



$D={y^2+z^2le x^2+18,x^2+y^2le 16}$



So It's a one-sheet hyperboloid (radius=$sqrt{18}$)+ cylinder with radius 4



I know how to parametrize them separately:



hyperboloid : $r(theta,phi)=(sinhphi,sqrt{18}sinthetacosh{phi},sqrt{18}costhetacosh{phi})$



cylinder: $r(theta,z)=(4costheta,4sintheta,z)$



But how do I connect them? Ho do I parametrize the surface boundary?



Or maybe "parametrize the boundary" implies giving two parametrizations?If it is so, are my parametrizations right?






multivariable-calculus parametric surface-integrals parametrization







share|cite|improve this question











$endgroup$




I need to parametrize the boundary of this region :



$D={y^2+z^2le x^2+18,x^2+y^2le 16}$



So It's a one-sheet hyperboloid (radius=$sqrt{18}$)+ cylinder with radius 4



I know how to parametrize them separately:



hyperboloid : $r(theta,phi)=(sinhphi,sqrt{18}sinthetacosh{phi},sqrt{18}costhetacosh{phi})$



cylinder: $r(theta,z)=(4costheta,4sintheta,z)$



But how do I connect them? Ho do I parametrize the surface boundary?



Or maybe "parametrize the boundary" implies giving two parametrizations?If it is so, are my parametrizations right?






multivariable-calculus parametric surface-integrals parametrization




multivariable-calculus parametric surface-integrals parametrization






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 17 at 12:14







NPLS

















asked Jan 17 at 12:03









NPLSNPLS

8112




8112












  • $begingroup$
    Apologies about delayed post. Shall post parameterizations and image tomorrow. Basically you are drilling/ intersecting a hyperboloid of 1 sheet with a cylinder.. There are two distorted "circular" disks produced. You can consider 1) the remaining continuous shell after drilling out the cylindrical holes or 2) the two discs themselves as the region depending on what you want to evaluate..
    $endgroup$
    – Narasimham
    Jan 17 at 23:13




















  • $begingroup$
    Apologies about delayed post. Shall post parameterizations and image tomorrow. Basically you are drilling/ intersecting a hyperboloid of 1 sheet with a cylinder.. There are two distorted "circular" disks produced. You can consider 1) the remaining continuous shell after drilling out the cylindrical holes or 2) the two discs themselves as the region depending on what you want to evaluate..
    $endgroup$
    – Narasimham
    Jan 17 at 23:13


















$begingroup$
Apologies about delayed post. Shall post parameterizations and image tomorrow. Basically you are drilling/ intersecting a hyperboloid of 1 sheet with a cylinder.. There are two distorted "circular" disks produced. You can consider 1) the remaining continuous shell after drilling out the cylindrical holes or 2) the two discs themselves as the region depending on what you want to evaluate..
$endgroup$
– Narasimham
Jan 17 at 23:13






$begingroup$
Apologies about delayed post. Shall post parameterizations and image tomorrow. Basically you are drilling/ intersecting a hyperboloid of 1 sheet with a cylinder.. There are two distorted "circular" disks produced. You can consider 1) the remaining continuous shell after drilling out the cylindrical holes or 2) the two discs themselves as the region depending on what you want to evaluate..
$endgroup$
– Narasimham
Jan 17 at 23:13












2 Answers
2






active

oldest

votes


















1












$begingroup$

We consider at first equalities. There are two surfaces of revolution $(a=sqrt{18},b=4 )$



$$ y^2+z^2-x^2=a^2 $$



$$ y^2+z^2=b^2 $$



which are hyperboloid of one sheet and a circular cylinder respectively.



Intersection of surfaces is the boundary of region for further purposes of integration etc.



Eliminate $y$ we get for $z$



$$ z = sqrt{ 2x^2+a^2-b^2 } tag1 $$



Circle of radius $b$ in projection is parameterized as $ (x,y)= b(cos theta, sin theta) $ and for $z$ we obtain by plugging in from above (1)



$$ z=pm sqrt{ 2 b^2 cos^2 theta +(a^2-b^2)}$$



The connected regions and the surfaces that created them along with (white) lines of intersection are plotted in Mathematica below .



Note that when two conicoids intersect we can generally have




  1. Null real intersection ( disjunct surfaces)


  2. One real closed loop intersection


  3. Two real intersections



depending on distance of centers of central conicoids. In each case we can refer to the patches so created arbitrarily connecting them as internal or external to intersection contour made or defined by the equality. Each sign defines its own boundary of demarcated intersection, which enables determining regions where inequalities apply.



To find demarcation and connection a sketch is often helpful in guiding the region continuity of the patch in reference.



So the parameterizations are same but for $pm$ sign where each sign represents a region.



The situation is no different if you consider intersection of a big cylinder and a smaller cylinder regions so created!



Boundary of Intrxn






share|cite|improve this answer











$endgroup$













  • $begingroup$
    could you give me the actual parametrization, please?
    $endgroup$
    – NPLS
    Jan 17 at 12:30










  • $begingroup$
    They are given after (1) above.
    $endgroup$
    – Narasimham
    Jan 18 at 13:06










  • $begingroup$
    Thank you! I got it!
    $endgroup$
    – NPLS
    Jan 18 at 14:49



















0












$begingroup$

$x=4cos(t),y=4sin(t),z=sqrt{16cos(2t)+18},; 0 leq tleq2pi$.
enter image description here



$x=4cos(t),y=4sin(t),z=-sqrt{16cos(2t)+18},; 0 leq tleq2pi$.
enter image description here



This is surface boundary parametrization.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    this is a curve parametrization? or a surface parametrization?
    $endgroup$
    – NPLS
    Jan 17 at 12:55










  • $begingroup$
    how did you find z?
    $endgroup$
    – NPLS
    Jan 17 at 13:10












  • $begingroup$
    @NPLS $z^2=x^2-y^2+18=16cos^2t-16sin^2t+18=16cos2t+18$.
    $endgroup$
    – Aleksas Domarkas
    Jan 17 at 13:32











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2 Answers
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2 Answers
2






active

oldest

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active

oldest

votes






active

oldest

votes









1












$begingroup$

We consider at first equalities. There are two surfaces of revolution $(a=sqrt{18},b=4 )$



$$ y^2+z^2-x^2=a^2 $$



$$ y^2+z^2=b^2 $$



which are hyperboloid of one sheet and a circular cylinder respectively.



Intersection of surfaces is the boundary of region for further purposes of integration etc.



Eliminate $y$ we get for $z$



$$ z = sqrt{ 2x^2+a^2-b^2 } tag1 $$



Circle of radius $b$ in projection is parameterized as $ (x,y)= b(cos theta, sin theta) $ and for $z$ we obtain by plugging in from above (1)



$$ z=pm sqrt{ 2 b^2 cos^2 theta +(a^2-b^2)}$$



The connected regions and the surfaces that created them along with (white) lines of intersection are plotted in Mathematica below .



Note that when two conicoids intersect we can generally have




  1. Null real intersection ( disjunct surfaces)


  2. One real closed loop intersection


  3. Two real intersections



depending on distance of centers of central conicoids. In each case we can refer to the patches so created arbitrarily connecting them as internal or external to intersection contour made or defined by the equality. Each sign defines its own boundary of demarcated intersection, which enables determining regions where inequalities apply.



To find demarcation and connection a sketch is often helpful in guiding the region continuity of the patch in reference.



So the parameterizations are same but for $pm$ sign where each sign represents a region.



The situation is no different if you consider intersection of a big cylinder and a smaller cylinder regions so created!



Boundary of Intrxn






share|cite|improve this answer











$endgroup$













  • $begingroup$
    could you give me the actual parametrization, please?
    $endgroup$
    – NPLS
    Jan 17 at 12:30










  • $begingroup$
    They are given after (1) above.
    $endgroup$
    – Narasimham
    Jan 18 at 13:06










  • $begingroup$
    Thank you! I got it!
    $endgroup$
    – NPLS
    Jan 18 at 14:49
















1












$begingroup$

We consider at first equalities. There are two surfaces of revolution $(a=sqrt{18},b=4 )$



$$ y^2+z^2-x^2=a^2 $$



$$ y^2+z^2=b^2 $$



which are hyperboloid of one sheet and a circular cylinder respectively.



Intersection of surfaces is the boundary of region for further purposes of integration etc.



Eliminate $y$ we get for $z$



$$ z = sqrt{ 2x^2+a^2-b^2 } tag1 $$



Circle of radius $b$ in projection is parameterized as $ (x,y)= b(cos theta, sin theta) $ and for $z$ we obtain by plugging in from above (1)



$$ z=pm sqrt{ 2 b^2 cos^2 theta +(a^2-b^2)}$$



The connected regions and the surfaces that created them along with (white) lines of intersection are plotted in Mathematica below .



Note that when two conicoids intersect we can generally have




  1. Null real intersection ( disjunct surfaces)


  2. One real closed loop intersection


  3. Two real intersections



depending on distance of centers of central conicoids. In each case we can refer to the patches so created arbitrarily connecting them as internal or external to intersection contour made or defined by the equality. Each sign defines its own boundary of demarcated intersection, which enables determining regions where inequalities apply.



To find demarcation and connection a sketch is often helpful in guiding the region continuity of the patch in reference.



So the parameterizations are same but for $pm$ sign where each sign represents a region.



The situation is no different if you consider intersection of a big cylinder and a smaller cylinder regions so created!



Boundary of Intrxn






share|cite|improve this answer











$endgroup$













  • $begingroup$
    could you give me the actual parametrization, please?
    $endgroup$
    – NPLS
    Jan 17 at 12:30










  • $begingroup$
    They are given after (1) above.
    $endgroup$
    – Narasimham
    Jan 18 at 13:06










  • $begingroup$
    Thank you! I got it!
    $endgroup$
    – NPLS
    Jan 18 at 14:49














1












1








1





$begingroup$

We consider at first equalities. There are two surfaces of revolution $(a=sqrt{18},b=4 )$



$$ y^2+z^2-x^2=a^2 $$



$$ y^2+z^2=b^2 $$



which are hyperboloid of one sheet and a circular cylinder respectively.



Intersection of surfaces is the boundary of region for further purposes of integration etc.



Eliminate $y$ we get for $z$



$$ z = sqrt{ 2x^2+a^2-b^2 } tag1 $$



Circle of radius $b$ in projection is parameterized as $ (x,y)= b(cos theta, sin theta) $ and for $z$ we obtain by plugging in from above (1)



$$ z=pm sqrt{ 2 b^2 cos^2 theta +(a^2-b^2)}$$



The connected regions and the surfaces that created them along with (white) lines of intersection are plotted in Mathematica below .



Note that when two conicoids intersect we can generally have




  1. Null real intersection ( disjunct surfaces)


  2. One real closed loop intersection


  3. Two real intersections



depending on distance of centers of central conicoids. In each case we can refer to the patches so created arbitrarily connecting them as internal or external to intersection contour made or defined by the equality. Each sign defines its own boundary of demarcated intersection, which enables determining regions where inequalities apply.



To find demarcation and connection a sketch is often helpful in guiding the region continuity of the patch in reference.



So the parameterizations are same but for $pm$ sign where each sign represents a region.



The situation is no different if you consider intersection of a big cylinder and a smaller cylinder regions so created!



Boundary of Intrxn






share|cite|improve this answer











$endgroup$



We consider at first equalities. There are two surfaces of revolution $(a=sqrt{18},b=4 )$



$$ y^2+z^2-x^2=a^2 $$



$$ y^2+z^2=b^2 $$



which are hyperboloid of one sheet and a circular cylinder respectively.



Intersection of surfaces is the boundary of region for further purposes of integration etc.



Eliminate $y$ we get for $z$



$$ z = sqrt{ 2x^2+a^2-b^2 } tag1 $$



Circle of radius $b$ in projection is parameterized as $ (x,y)= b(cos theta, sin theta) $ and for $z$ we obtain by plugging in from above (1)



$$ z=pm sqrt{ 2 b^2 cos^2 theta +(a^2-b^2)}$$



The connected regions and the surfaces that created them along with (white) lines of intersection are plotted in Mathematica below .



Note that when two conicoids intersect we can generally have




  1. Null real intersection ( disjunct surfaces)


  2. One real closed loop intersection


  3. Two real intersections



depending on distance of centers of central conicoids. In each case we can refer to the patches so created arbitrarily connecting them as internal or external to intersection contour made or defined by the equality. Each sign defines its own boundary of demarcated intersection, which enables determining regions where inequalities apply.



To find demarcation and connection a sketch is often helpful in guiding the region continuity of the patch in reference.



So the parameterizations are same but for $pm$ sign where each sign represents a region.



The situation is no different if you consider intersection of a big cylinder and a smaller cylinder regions so created!



Boundary of Intrxn







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 18 at 13:28

























answered Jan 17 at 12:21









NarasimhamNarasimham

20.8k52158




20.8k52158












  • $begingroup$
    could you give me the actual parametrization, please?
    $endgroup$
    – NPLS
    Jan 17 at 12:30










  • $begingroup$
    They are given after (1) above.
    $endgroup$
    – Narasimham
    Jan 18 at 13:06










  • $begingroup$
    Thank you! I got it!
    $endgroup$
    – NPLS
    Jan 18 at 14:49


















  • $begingroup$
    could you give me the actual parametrization, please?
    $endgroup$
    – NPLS
    Jan 17 at 12:30










  • $begingroup$
    They are given after (1) above.
    $endgroup$
    – Narasimham
    Jan 18 at 13:06










  • $begingroup$
    Thank you! I got it!
    $endgroup$
    – NPLS
    Jan 18 at 14:49
















$begingroup$
could you give me the actual parametrization, please?
$endgroup$
– NPLS
Jan 17 at 12:30




$begingroup$
could you give me the actual parametrization, please?
$endgroup$
– NPLS
Jan 17 at 12:30












$begingroup$
They are given after (1) above.
$endgroup$
– Narasimham
Jan 18 at 13:06




$begingroup$
They are given after (1) above.
$endgroup$
– Narasimham
Jan 18 at 13:06












$begingroup$
Thank you! I got it!
$endgroup$
– NPLS
Jan 18 at 14:49




$begingroup$
Thank you! I got it!
$endgroup$
– NPLS
Jan 18 at 14:49











0












$begingroup$

$x=4cos(t),y=4sin(t),z=sqrt{16cos(2t)+18},; 0 leq tleq2pi$.
enter image description here



$x=4cos(t),y=4sin(t),z=-sqrt{16cos(2t)+18},; 0 leq tleq2pi$.
enter image description here



This is surface boundary parametrization.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    this is a curve parametrization? or a surface parametrization?
    $endgroup$
    – NPLS
    Jan 17 at 12:55










  • $begingroup$
    how did you find z?
    $endgroup$
    – NPLS
    Jan 17 at 13:10












  • $begingroup$
    @NPLS $z^2=x^2-y^2+18=16cos^2t-16sin^2t+18=16cos2t+18$.
    $endgroup$
    – Aleksas Domarkas
    Jan 17 at 13:32
















0












$begingroup$

$x=4cos(t),y=4sin(t),z=sqrt{16cos(2t)+18},; 0 leq tleq2pi$.
enter image description here



$x=4cos(t),y=4sin(t),z=-sqrt{16cos(2t)+18},; 0 leq tleq2pi$.
enter image description here



This is surface boundary parametrization.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    this is a curve parametrization? or a surface parametrization?
    $endgroup$
    – NPLS
    Jan 17 at 12:55










  • $begingroup$
    how did you find z?
    $endgroup$
    – NPLS
    Jan 17 at 13:10












  • $begingroup$
    @NPLS $z^2=x^2-y^2+18=16cos^2t-16sin^2t+18=16cos2t+18$.
    $endgroup$
    – Aleksas Domarkas
    Jan 17 at 13:32














0












0








0





$begingroup$

$x=4cos(t),y=4sin(t),z=sqrt{16cos(2t)+18},; 0 leq tleq2pi$.
enter image description here



$x=4cos(t),y=4sin(t),z=-sqrt{16cos(2t)+18},; 0 leq tleq2pi$.
enter image description here



This is surface boundary parametrization.






share|cite|improve this answer











$endgroup$



$x=4cos(t),y=4sin(t),z=sqrt{16cos(2t)+18},; 0 leq tleq2pi$.
enter image description here



$x=4cos(t),y=4sin(t),z=-sqrt{16cos(2t)+18},; 0 leq tleq2pi$.
enter image description here



This is surface boundary parametrization.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 17 at 13:05

























answered Jan 17 at 12:47









Aleksas DomarkasAleksas Domarkas

1,24916




1,24916












  • $begingroup$
    this is a curve parametrization? or a surface parametrization?
    $endgroup$
    – NPLS
    Jan 17 at 12:55










  • $begingroup$
    how did you find z?
    $endgroup$
    – NPLS
    Jan 17 at 13:10












  • $begingroup$
    @NPLS $z^2=x^2-y^2+18=16cos^2t-16sin^2t+18=16cos2t+18$.
    $endgroup$
    – Aleksas Domarkas
    Jan 17 at 13:32


















  • $begingroup$
    this is a curve parametrization? or a surface parametrization?
    $endgroup$
    – NPLS
    Jan 17 at 12:55










  • $begingroup$
    how did you find z?
    $endgroup$
    – NPLS
    Jan 17 at 13:10












  • $begingroup$
    @NPLS $z^2=x^2-y^2+18=16cos^2t-16sin^2t+18=16cos2t+18$.
    $endgroup$
    – Aleksas Domarkas
    Jan 17 at 13:32
















$begingroup$
this is a curve parametrization? or a surface parametrization?
$endgroup$
– NPLS
Jan 17 at 12:55




$begingroup$
this is a curve parametrization? or a surface parametrization?
$endgroup$
– NPLS
Jan 17 at 12:55












$begingroup$
how did you find z?
$endgroup$
– NPLS
Jan 17 at 13:10






$begingroup$
how did you find z?
$endgroup$
– NPLS
Jan 17 at 13:10














$begingroup$
@NPLS $z^2=x^2-y^2+18=16cos^2t-16sin^2t+18=16cos2t+18$.
$endgroup$
– Aleksas Domarkas
Jan 17 at 13:32




$begingroup$
@NPLS $z^2=x^2-y^2+18=16cos^2t-16sin^2t+18=16cos2t+18$.
$endgroup$
– Aleksas Domarkas
Jan 17 at 13:32


















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The Binding of Isaac: Rebirth/Afterbirth

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Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
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What does “Dominus providebit” mean?

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5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
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