parametrize the boundary of a region
$begingroup$
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
asked Jan 17 at 12:03
NPLSNPLS
8112
$endgroup$
add a comment |
$begingroup$
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
asked Jan 17 at 12:03
NPLSNPLS
8112
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Jan 17 at 12:03
NPLSNPLS
8112
$endgroup$
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
asked Jan 17 at 12:03
NPLSNPLS
8112
asked Jan 17 at 12:03
NPLSNPLS
8112
edited Jan 17 at 12:14
NPLS
asked Jan 17 at 12:03
NPLSNPLS
8112
asked Jan 17 at 12:03
NPLSNPLS
8112
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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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
Null real intersection ( disjunct surfaces)
One real closed loop intersection
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!
answered Jan 17 at 12:21
NarasimhamNarasimham
20.8k52158
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
$x=4cos(t),y=4sin(t),z=sqrt{16cos(2t)+18},; 0 leq tleq2pi$.
$x=4cos(t),y=4sin(t),z=-sqrt{16cos(2t)+18},; 0 leq tleq2pi$.
This is surface boundary parametrization.
answered Jan 17 at 12:47
Aleksas DomarkasAleksas Domarkas
1,24916
$endgroup$
$begingroup$
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
add a comment |
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$begingroup$
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
Null real intersection ( disjunct surfaces)
One real closed loop intersection
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!
answered Jan 17 at 12:21
NarasimhamNarasimham
20.8k52158
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
Null real intersection ( disjunct surfaces)
One real closed loop intersection
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!
answered Jan 17 at 12:21
NarasimhamNarasimham
20.8k52158
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
Null real intersection ( disjunct surfaces)
One real closed loop intersection
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!
answered Jan 17 at 12:21
NarasimhamNarasimham
20.8k52158
$endgroup$
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
Null real intersection ( disjunct surfaces)
One real closed loop intersection
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!
answered Jan 17 at 12:21
NarasimhamNarasimham
20.8k52158
edited Jan 18 at 13:28
answered Jan 17 at 12:21
NarasimhamNarasimham
20.8k52158
answered Jan 17 at 12:21
NarasimhamNarasimham
20.8k52158
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
$x=4cos(t),y=4sin(t),z=sqrt{16cos(2t)+18},; 0 leq tleq2pi$.
$x=4cos(t),y=4sin(t),z=-sqrt{16cos(2t)+18},; 0 leq tleq2pi$.
This is surface boundary parametrization.
answered Jan 17 at 12:47
Aleksas DomarkasAleksas Domarkas
1,24916
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
$x=4cos(t),y=4sin(t),z=sqrt{16cos(2t)+18},; 0 leq tleq2pi$.
$x=4cos(t),y=4sin(t),z=-sqrt{16cos(2t)+18},; 0 leq tleq2pi$.
This is surface boundary parametrization.
answered Jan 17 at 12:47
Aleksas DomarkasAleksas Domarkas
1,24916
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
$x=4cos(t),y=4sin(t),z=sqrt{16cos(2t)+18},; 0 leq tleq2pi$.
$x=4cos(t),y=4sin(t),z=-sqrt{16cos(2t)+18},; 0 leq tleq2pi$.
This is surface boundary parametrization.
answered Jan 17 at 12:47
Aleksas DomarkasAleksas Domarkas
1,24916
$endgroup$
$x=4cos(t),y=4sin(t),z=sqrt{16cos(2t)+18},; 0 leq tleq2pi$.
$x=4cos(t),y=4sin(t),z=-sqrt{16cos(2t)+18},; 0 leq tleq2pi$.
This is surface boundary parametrization.
answered Jan 17 at 12:47
Aleksas DomarkasAleksas Domarkas
1,24916
edited Jan 17 at 13:05
answered Jan 17 at 12:47
Aleksas DomarkasAleksas Domarkas
1,24916
answered Jan 17 at 12:47
Aleksas DomarkasAleksas Domarkas
1,24916
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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