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.
calculus integration multivariable-calculus parametrization
asked Nov 16 at 22:37
user402543
375212
add a comment |
up vote
0
down vote
favorite
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
asked Nov 16 at 22:37
user402543
375212
What do you mean by "the normal vector points outwards"?
– Apocalypse
Nov 17 at 2:46
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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
asked Nov 16 at 22:37
user402543
375212
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
calculus integration multivariable-calculus parametrization
asked Nov 16 at 22:37
user402543
375212
asked Nov 16 at 22:37
user402543
375212
asked Nov 16 at 22:37
user402543
375212
asked Nov 16 at 22:37
user402543
375212
asked Nov 16 at 22:37
user402543
375212
375212
What do you mean by "the normal vector points outwards"?
– Apocalypse
Nov 17 at 2:46
add a comment |
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
add a comment |
1 Answer
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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.
answered Nov 17 at 3:45
Narlin
484310
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
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.
answered Nov 17 at 3:45
Narlin
484310
add a comment |
up vote
1
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.
answered Nov 17 at 3:45
Narlin
484310
add a comment |
up vote
1
down vote
up vote
1
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.
answered Nov 17 at 3:45
Narlin
484310
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.
answered Nov 17 at 3:45
Narlin
484310
answered Nov 17 at 3:45
Narlin
484310
answered Nov 17 at 3:45
Narlin
484310
answered Nov 17 at 3:45
Narlin
484310
484310
add a comment |
add a comment |
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What do you mean by "the normal vector points outwards"?
– Apocalypse
Nov 17 at 2:46