Parameterizing the frustum of a cone












0












$begingroup$


Given the frustum of a cone described by $z^2 = x^2 + y^2$ for $2 leq z leq 8$, and the following generalized parameterization of a cone:



enter image description here



I would have thought the following parameterization in two variables would have been correct:



$r(u,v) = langle sqrt{2}/3 cos u, sqrt{2}/3 sin u, v rangle$



for $2 leq v leq 8$ and $0 leq u leq 2pi$



The radius of the top circle of the frustum is $2 sqrt{2} = a$ and the height of the frustum is $8 - 2 = 6 = h$



So, $z = (rh)/a Rightarrow r = (az)/h Rightarrow r = sqrt{2}/3$



And yet, my book tells me the correct parameterization is



$r(u,v) = langle v cos u, v sin u, v rangle$



for $2 leq v leq 8$ and $0 leq u leq 2pi$



Doesn't the book's answer provide the wrong values for r at the top and bottom of the frustum? Am I misunderstanding something?






surface-integrals parametrization







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  • $begingroup$
    What is $langle .,.,.rangle$?
    $endgroup$
    – mvw
    Apr 19 '16 at 18:11










  • $begingroup$
    I'm not really sure what you're asking: its notation for expressing a three dimensional vector, which can also be used as the parametric expression of a surface in 3D space. The one I chose for this surface is written above: r(u,v) = ⟨ (2)^0.5 / 3 cos u,2)^0.5 / 3 sin u, v ⟩
    $endgroup$
    – StudentsTea
    Apr 19 '16 at 18:14










  • $begingroup$
    The notation I see more often is angle brackets for scalar products and round or square brackets for vectors.
    $endgroup$
    – mvw
    Apr 19 '16 at 18:18










  • $begingroup$
    To be sure, r(u,v) is a scalar-valued function.
    $endgroup$
    – StudentsTea
    Apr 19 '16 at 18:25
















0












$begingroup$


Given the frustum of a cone described by $z^2 = x^2 + y^2$ for $2 leq z leq 8$, and the following generalized parameterization of a cone:



enter image description here



I would have thought the following parameterization in two variables would have been correct:



$r(u,v) = langle sqrt{2}/3 cos u, sqrt{2}/3 sin u, v rangle$



for $2 leq v leq 8$ and $0 leq u leq 2pi$



The radius of the top circle of the frustum is $2 sqrt{2} = a$ and the height of the frustum is $8 - 2 = 6 = h$



So, $z = (rh)/a Rightarrow r = (az)/h Rightarrow r = sqrt{2}/3$



And yet, my book tells me the correct parameterization is



$r(u,v) = langle v cos u, v sin u, v rangle$



for $2 leq v leq 8$ and $0 leq u leq 2pi$



Doesn't the book's answer provide the wrong values for r at the top and bottom of the frustum? Am I misunderstanding something?






surface-integrals parametrization







share|cite|improve this question









$endgroup$












  • $begingroup$
    What is $langle .,.,.rangle$?
    $endgroup$
    – mvw
    Apr 19 '16 at 18:11










  • $begingroup$
    I'm not really sure what you're asking: its notation for expressing a three dimensional vector, which can also be used as the parametric expression of a surface in 3D space. The one I chose for this surface is written above: r(u,v) = ⟨ (2)^0.5 / 3 cos u,2)^0.5 / 3 sin u, v ⟩
    $endgroup$
    – StudentsTea
    Apr 19 '16 at 18:14










  • $begingroup$
    The notation I see more often is angle brackets for scalar products and round or square brackets for vectors.
    $endgroup$
    – mvw
    Apr 19 '16 at 18:18










  • $begingroup$
    To be sure, r(u,v) is a scalar-valued function.
    $endgroup$
    – StudentsTea
    Apr 19 '16 at 18:25














0












0








0





$begingroup$


Given the frustum of a cone described by $z^2 = x^2 + y^2$ for $2 leq z leq 8$, and the following generalized parameterization of a cone:



enter image description here



I would have thought the following parameterization in two variables would have been correct:



$r(u,v) = langle sqrt{2}/3 cos u, sqrt{2}/3 sin u, v rangle$



for $2 leq v leq 8$ and $0 leq u leq 2pi$



The radius of the top circle of the frustum is $2 sqrt{2} = a$ and the height of the frustum is $8 - 2 = 6 = h$



So, $z = (rh)/a Rightarrow r = (az)/h Rightarrow r = sqrt{2}/3$



And yet, my book tells me the correct parameterization is



$r(u,v) = langle v cos u, v sin u, v rangle$



for $2 leq v leq 8$ and $0 leq u leq 2pi$



Doesn't the book's answer provide the wrong values for r at the top and bottom of the frustum? Am I misunderstanding something?






surface-integrals parametrization







share|cite|improve this question









$endgroup$




Given the frustum of a cone described by $z^2 = x^2 + y^2$ for $2 leq z leq 8$, and the following generalized parameterization of a cone:



enter image description here



I would have thought the following parameterization in two variables would have been correct:



$r(u,v) = langle sqrt{2}/3 cos u, sqrt{2}/3 sin u, v rangle$



for $2 leq v leq 8$ and $0 leq u leq 2pi$



The radius of the top circle of the frustum is $2 sqrt{2} = a$ and the height of the frustum is $8 - 2 = 6 = h$



So, $z = (rh)/a Rightarrow r = (az)/h Rightarrow r = sqrt{2}/3$



And yet, my book tells me the correct parameterization is



$r(u,v) = langle v cos u, v sin u, v rangle$



for $2 leq v leq 8$ and $0 leq u leq 2pi$



Doesn't the book's answer provide the wrong values for r at the top and bottom of the frustum? Am I misunderstanding something?






surface-integrals parametrization




surface-integrals parametrization






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Apr 19 '16 at 17:55









StudentsTeaStudentsTea

8112824




8112824












  • $begingroup$
    What is $langle .,.,.rangle$?
    $endgroup$
    – mvw
    Apr 19 '16 at 18:11










  • $begingroup$
    I'm not really sure what you're asking: its notation for expressing a three dimensional vector, which can also be used as the parametric expression of a surface in 3D space. The one I chose for this surface is written above: r(u,v) = ⟨ (2)^0.5 / 3 cos u,2)^0.5 / 3 sin u, v ⟩
    $endgroup$
    – StudentsTea
    Apr 19 '16 at 18:14










  • $begingroup$
    The notation I see more often is angle brackets for scalar products and round or square brackets for vectors.
    $endgroup$
    – mvw
    Apr 19 '16 at 18:18










  • $begingroup$
    To be sure, r(u,v) is a scalar-valued function.
    $endgroup$
    – StudentsTea
    Apr 19 '16 at 18:25


















  • $begingroup$
    What is $langle .,.,.rangle$?
    $endgroup$
    – mvw
    Apr 19 '16 at 18:11










  • $begingroup$
    I'm not really sure what you're asking: its notation for expressing a three dimensional vector, which can also be used as the parametric expression of a surface in 3D space. The one I chose for this surface is written above: r(u,v) = ⟨ (2)^0.5 / 3 cos u,2)^0.5 / 3 sin u, v ⟩
    $endgroup$
    – StudentsTea
    Apr 19 '16 at 18:14










  • $begingroup$
    The notation I see more often is angle brackets for scalar products and round or square brackets for vectors.
    $endgroup$
    – mvw
    Apr 19 '16 at 18:18










  • $begingroup$
    To be sure, r(u,v) is a scalar-valued function.
    $endgroup$
    – StudentsTea
    Apr 19 '16 at 18:25
















$begingroup$
What is $langle .,.,.rangle$?
$endgroup$
– mvw
Apr 19 '16 at 18:11




$begingroup$
What is $langle .,.,.rangle$?
$endgroup$
– mvw
Apr 19 '16 at 18:11












$begingroup$
I'm not really sure what you're asking: its notation for expressing a three dimensional vector, which can also be used as the parametric expression of a surface in 3D space. The one I chose for this surface is written above: r(u,v) = ⟨ (2)^0.5 / 3 cos u,2)^0.5 / 3 sin u, v ⟩
$endgroup$
– StudentsTea
Apr 19 '16 at 18:14




$begingroup$
I'm not really sure what you're asking: its notation for expressing a three dimensional vector, which can also be used as the parametric expression of a surface in 3D space. The one I chose for this surface is written above: r(u,v) = ⟨ (2)^0.5 / 3 cos u,2)^0.5 / 3 sin u, v ⟩
$endgroup$
– StudentsTea
Apr 19 '16 at 18:14












$begingroup$
The notation I see more often is angle brackets for scalar products and round or square brackets for vectors.
$endgroup$
– mvw
Apr 19 '16 at 18:18




$begingroup$
The notation I see more often is angle brackets for scalar products and round or square brackets for vectors.
$endgroup$
– mvw
Apr 19 '16 at 18:18












$begingroup$
To be sure, r(u,v) is a scalar-valued function.
$endgroup$
– StudentsTea
Apr 19 '16 at 18:25




$begingroup$
To be sure, r(u,v) is a scalar-valued function.
$endgroup$
– StudentsTea
Apr 19 '16 at 18:25










1 Answer
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$begingroup$

First of all, it is important to mention the parameterization of a surface (or a line) is not unique. There are usually different ways to describe a same surface, so just because you have a different answer as the one in the book does not necessarily mean that you (or the book) is wrong.



This being said, your parameterization is wrong. In fact, yours describes a cylinder (and not a cone). You should always check that when you plug the components of $r(u,v)$, they satisfy the cartesian equation of the surface. In your case,
$$x^2(u,v)+y^2(u,v)=frac{2}{9}neq z^2(u,v).$$



You can check that the answer given by the book does satisfy $x^2(u,v)+y^2(u,v)= z^2(u,v)$, therefore it is correct.






share|cite|improve this answer









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    $begingroup$

    First of all, it is important to mention the parameterization of a surface (or a line) is not unique. There are usually different ways to describe a same surface, so just because you have a different answer as the one in the book does not necessarily mean that you (or the book) is wrong.



    This being said, your parameterization is wrong. In fact, yours describes a cylinder (and not a cone). You should always check that when you plug the components of $r(u,v)$, they satisfy the cartesian equation of the surface. In your case,
    $$x^2(u,v)+y^2(u,v)=frac{2}{9}neq z^2(u,v).$$



    You can check that the answer given by the book does satisfy $x^2(u,v)+y^2(u,v)= z^2(u,v)$, therefore it is correct.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      First of all, it is important to mention the parameterization of a surface (or a line) is not unique. There are usually different ways to describe a same surface, so just because you have a different answer as the one in the book does not necessarily mean that you (or the book) is wrong.



      This being said, your parameterization is wrong. In fact, yours describes a cylinder (and not a cone). You should always check that when you plug the components of $r(u,v)$, they satisfy the cartesian equation of the surface. In your case,
      $$x^2(u,v)+y^2(u,v)=frac{2}{9}neq z^2(u,v).$$



      You can check that the answer given by the book does satisfy $x^2(u,v)+y^2(u,v)= z^2(u,v)$, therefore it is correct.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        First of all, it is important to mention the parameterization of a surface (or a line) is not unique. There are usually different ways to describe a same surface, so just because you have a different answer as the one in the book does not necessarily mean that you (or the book) is wrong.



        This being said, your parameterization is wrong. In fact, yours describes a cylinder (and not a cone). You should always check that when you plug the components of $r(u,v)$, they satisfy the cartesian equation of the surface. In your case,
        $$x^2(u,v)+y^2(u,v)=frac{2}{9}neq z^2(u,v).$$



        You can check that the answer given by the book does satisfy $x^2(u,v)+y^2(u,v)= z^2(u,v)$, therefore it is correct.






        share|cite|improve this answer









        $endgroup$



        First of all, it is important to mention the parameterization of a surface (or a line) is not unique. There are usually different ways to describe a same surface, so just because you have a different answer as the one in the book does not necessarily mean that you (or the book) is wrong.



        This being said, your parameterization is wrong. In fact, yours describes a cylinder (and not a cone). You should always check that when you plug the components of $r(u,v)$, they satisfy the cartesian equation of the surface. In your case,
        $$x^2(u,v)+y^2(u,v)=frac{2}{9}neq z^2(u,v).$$



        You can check that the answer given by the book does satisfy $x^2(u,v)+y^2(u,v)= z^2(u,v)$, therefore it is correct.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Apr 20 '16 at 0:41









        KuifjeKuifje

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