Parameterize the sphere of radius $5$ in $mathbb{R}^3$ so that in all its points, the normal vector points...











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Parameterize the sphere of radius $5$ in $mathbb{R}^3$ so that in all its points, the normal vector points outwards.



I've thought about doing the following but I do not know if I'm doing things right:



I thought about using spherical coordinates and I get that $x=5sinthetacosphi, y=5sinthetasinphi, z=5costheta$ where $0leq thetaleqpi, 0leqphileq2pi$.



Is this fine? Is this the parameterization that works for me? How can I do to confirm that the normal vector points out? Thank you.










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  • What do you mean by "the normal vector points outwards"?
    – Apocalypse
    Nov 17 at 2:46















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Parameterize the sphere of radius $5$ in $mathbb{R}^3$ so that in all its points, the normal vector points outwards.



I've thought about doing the following but I do not know if I'm doing things right:



I thought about using spherical coordinates and I get that $x=5sinthetacosphi, y=5sinthetasinphi, z=5costheta$ where $0leq thetaleqpi, 0leqphileq2pi$.



Is this fine? Is this the parameterization that works for me? How can I do to confirm that the normal vector points out? Thank you.










share|cite|improve this question






















  • What do you mean by "the normal vector points outwards"?
    – Apocalypse
    Nov 17 at 2:46













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0
down vote

favorite
1









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down vote

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Parameterize the sphere of radius $5$ in $mathbb{R}^3$ so that in all its points, the normal vector points outwards.



I've thought about doing the following but I do not know if I'm doing things right:



I thought about using spherical coordinates and I get that $x=5sinthetacosphi, y=5sinthetasinphi, z=5costheta$ where $0leq thetaleqpi, 0leqphileq2pi$.



Is this fine? Is this the parameterization that works for me? How can I do to confirm that the normal vector points out? Thank you.










share|cite|improve this question













Parameterize the sphere of radius $5$ in $mathbb{R}^3$ so that in all its points, the normal vector points outwards.



I've thought about doing the following but I do not know if I'm doing things right:



I thought about using spherical coordinates and I get that $x=5sinthetacosphi, y=5sinthetasinphi, z=5costheta$ where $0leq thetaleqpi, 0leqphileq2pi$.



Is this fine? Is this the parameterization that works for me? How can I do to confirm that the normal vector points out? Thank you.







calculus integration multivariable-calculus parametrization






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asked Nov 16 at 22:37









user402543

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  • What do you mean by "the normal vector points outwards"?
    – Apocalypse
    Nov 17 at 2:46


















  • What do you mean by "the normal vector points outwards"?
    – Apocalypse
    Nov 17 at 2:46
















What do you mean by "the normal vector points outwards"?
– Apocalypse
Nov 17 at 2:46




What do you mean by "the normal vector points outwards"?
– Apocalypse
Nov 17 at 2:46










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The parametrization is fine. The sphere is centered at the origin. The two partial derivatives would be $$mathbf{u} =left(begin{array}{c}5cos(theta)cos(phi)\5cos(theta)sin(phi)\-5sin(theta)end{array}right)$$ and $$mathbf{v} =left(begin{array}{c}-5sin(theta)sin(phi)\5cos(phi)sin(theta)\0end{array}right)$$ When you cross $mathbf{u}$ and $mathbf{v}$ you get the normal which comes from the origin and thus exits the sphere in the outward direction. To get it at a point on the sphere, you have to select a point, say using $theta$ and $phi$ then use the same values to compute u and v and then just add the cross product vector (i.e. the normal) to the calculated point. It will continue to point the same direction which will be away from the origin and thus outward.






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    The parametrization is fine. The sphere is centered at the origin. The two partial derivatives would be $$mathbf{u} =left(begin{array}{c}5cos(theta)cos(phi)\5cos(theta)sin(phi)\-5sin(theta)end{array}right)$$ and $$mathbf{v} =left(begin{array}{c}-5sin(theta)sin(phi)\5cos(phi)sin(theta)\0end{array}right)$$ When you cross $mathbf{u}$ and $mathbf{v}$ you get the normal which comes from the origin and thus exits the sphere in the outward direction. To get it at a point on the sphere, you have to select a point, say using $theta$ and $phi$ then use the same values to compute u and v and then just add the cross product vector (i.e. the normal) to the calculated point. It will continue to point the same direction which will be away from the origin and thus outward.






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      The parametrization is fine. The sphere is centered at the origin. The two partial derivatives would be $$mathbf{u} =left(begin{array}{c}5cos(theta)cos(phi)\5cos(theta)sin(phi)\-5sin(theta)end{array}right)$$ and $$mathbf{v} =left(begin{array}{c}-5sin(theta)sin(phi)\5cos(phi)sin(theta)\0end{array}right)$$ When you cross $mathbf{u}$ and $mathbf{v}$ you get the normal which comes from the origin and thus exits the sphere in the outward direction. To get it at a point on the sphere, you have to select a point, say using $theta$ and $phi$ then use the same values to compute u and v and then just add the cross product vector (i.e. the normal) to the calculated point. It will continue to point the same direction which will be away from the origin and thus outward.






      share|cite|improve this answer























        up vote
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        up vote
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        down vote









        The parametrization is fine. The sphere is centered at the origin. The two partial derivatives would be $$mathbf{u} =left(begin{array}{c}5cos(theta)cos(phi)\5cos(theta)sin(phi)\-5sin(theta)end{array}right)$$ and $$mathbf{v} =left(begin{array}{c}-5sin(theta)sin(phi)\5cos(phi)sin(theta)\0end{array}right)$$ When you cross $mathbf{u}$ and $mathbf{v}$ you get the normal which comes from the origin and thus exits the sphere in the outward direction. To get it at a point on the sphere, you have to select a point, say using $theta$ and $phi$ then use the same values to compute u and v and then just add the cross product vector (i.e. the normal) to the calculated point. It will continue to point the same direction which will be away from the origin and thus outward.






        share|cite|improve this answer












        The parametrization is fine. The sphere is centered at the origin. The two partial derivatives would be $$mathbf{u} =left(begin{array}{c}5cos(theta)cos(phi)\5cos(theta)sin(phi)\-5sin(theta)end{array}right)$$ and $$mathbf{v} =left(begin{array}{c}-5sin(theta)sin(phi)\5cos(phi)sin(theta)\0end{array}right)$$ When you cross $mathbf{u}$ and $mathbf{v}$ you get the normal which comes from the origin and thus exits the sphere in the outward direction. To get it at a point on the sphere, you have to select a point, say using $theta$ and $phi$ then use the same values to compute u and v and then just add the cross product vector (i.e. the normal) to the calculated point. It will continue to point the same direction which will be away from the origin and thus outward.







        share|cite|improve this answer












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        answered Nov 17 at 3:45









        Narlin

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