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what are the ranges of $ u ,v $?
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Let T be the upper half of the torus with parametrization $ r(u,v)=left((2+cos v) cos u, (2+cos v) sin u, sin v right) $
what are the ranges of $ u ,v $ ?
Let $ f(x,y,z)=y^2z $ , then calculate the surface integral over $ T $.
Answer:
The formula is $ iint_S f(x,y,z)dS=iint_D f(r(u,v)) ||r_u times r_v|| dA $
So to evaluate the surface integral I just need to find the ranges of $ u , v $
Given , T is the upper half of the torus.
So I think $ 0 leq u leq pi , 0 leq v leq 2 pi $
Is it true range of $ u,v $ ?
help me with the ranges of $ u,v $.
calculus
asked Jul 27 at 1:31
MONJUR ALAM
647
add a comment |Â
up vote
0
down vote
favorite
Let T be the upper half of the torus with parametrization $ r(u,v)=left((2+cos v) cos u, (2+cos v) sin u, sin v right) $
what are the ranges of $ u ,v $ ?
Let $ f(x,y,z)=y^2z $ , then calculate the surface integral over $ T $.
Answer:
The formula is $ iint_S f(x,y,z)dS=iint_D f(r(u,v)) ||r_u times r_v|| dA $
So to evaluate the surface integral I just need to find the ranges of $ u , v $
Given , T is the upper half of the torus.
So I think $ 0 leq u leq pi , 0 leq v leq 2 pi $
Is it true range of $ u,v $ ?
help me with the ranges of $ u,v $.
calculus
asked Jul 27 at 1:31
MONJUR ALAM
647
You need to explore the parameterization further to understand it instead of blindly asking for the answer.
– MPW
Jul 27 at 1:34
sir I need the ranges of $ u , v $ in order to integrate over the torus . But I am not sure. can you help me ?
– MONJUR ALAM
Jul 27 at 1:41
1
Imagine if the $v$ terms didn’t exist. Then the parametrization would trace out a full circle in the $xy$ plane which hints that $v$ controls the longitude, so $0 leq v leq 2 pi$. The upper half of the torus will be given by $0 leq u leq pi$ so it seems you have the right intuition!
– Osama Ghani
Jul 27 at 6:08
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let T be the upper half of the torus with parametrization $ r(u,v)=left((2+cos v) cos u, (2+cos v) sin u, sin v right) $
what are the ranges of $ u ,v $ ?
Let $ f(x,y,z)=y^2z $ , then calculate the surface integral over $ T $.
Answer:
The formula is $ iint_S f(x,y,z)dS=iint_D f(r(u,v)) ||r_u times r_v|| dA $
So to evaluate the surface integral I just need to find the ranges of $ u , v $
Given , T is the upper half of the torus.
So I think $ 0 leq u leq pi , 0 leq v leq 2 pi $
Is it true range of $ u,v $ ?
help me with the ranges of $ u,v $.
calculus
asked Jul 27 at 1:31
MONJUR ALAM
647
Let T be the upper half of the torus with parametrization $ r(u,v)=left((2+cos v) cos u, (2+cos v) sin u, sin v right) $
what are the ranges of $ u ,v $ ?
Let $ f(x,y,z)=y^2z $ , then calculate the surface integral over $ T $.
Answer:
The formula is $ iint_S f(x,y,z)dS=iint_D f(r(u,v)) ||r_u times r_v|| dA $
So to evaluate the surface integral I just need to find the ranges of $ u , v $
Given , T is the upper half of the torus.
So I think $ 0 leq u leq pi , 0 leq v leq 2 pi $
Is it true range of $ u,v $ ?
help me with the ranges of $ u,v $.
calculus
asked Jul 27 at 1:31
MONJUR ALAM
647
asked Jul 27 at 1:31
MONJUR ALAM
647
asked Jul 27 at 1:31
MONJUR ALAM
647
asked Jul 27 at 1:31
MONJUR ALAM
647
647
You need to explore the parameterization further to understand it instead of blindly asking for the answer.
– MPW
Jul 27 at 1:34
sir I need the ranges of $ u , v $ in order to integrate over the torus . But I am not sure. can you help me ?
– MONJUR ALAM
Jul 27 at 1:41
1
Imagine if the $v$ terms didn’t exist. Then the parametrization would trace out a full circle in the $xy$ plane which hints that $v$ controls the longitude, so $0 leq v leq 2 pi$. The upper half of the torus will be given by $0 leq u leq pi$ so it seems you have the right intuition!
– Osama Ghani
Jul 27 at 6:08
add a comment |Â
You need to explore the parameterization further to understand it instead of blindly asking for the answer.
– MPW
Jul 27 at 1:34
sir I need the ranges of $ u , v $ in order to integrate over the torus . But I am not sure. can you help me ?
– MONJUR ALAM
Jul 27 at 1:41
1
Imagine if the $v$ terms didn’t exist. Then the parametrization would trace out a full circle in the $xy$ plane which hints that $v$ controls the longitude, so $0 leq v leq 2 pi$. The upper half of the torus will be given by $0 leq u leq pi$ so it seems you have the right intuition!
– Osama Ghani
Jul 27 at 6:08
You need to explore the parameterization further to understand it instead of blindly asking for the answer.
– MPW
Jul 27 at 1:34
You need to explore the parameterization further to understand it instead of blindly asking for the answer.
– MPW
Jul 27 at 1:34
sir I need the ranges of $ u , v $ in order to integrate over the torus . But I am not sure. can you help me ?
– MONJUR ALAM
Jul 27 at 1:41
sir I need the ranges of $ u , v $ in order to integrate over the torus . But I am not sure. can you help me ?
– MONJUR ALAM
Jul 27 at 1:41
1
1
Imagine if the $v$ terms didn’t exist. Then the parametrization would trace out a full circle in the $xy$ plane which hints that $v$ controls the longitude, so $0 leq v leq 2 pi$. The upper half of the torus will be given by $0 leq u leq pi$ so it seems you have the right intuition!
– Osama Ghani
Jul 27 at 6:08
Imagine if the $v$ terms didn’t exist. Then the parametrization would trace out a full circle in the $xy$ plane which hints that $v$ controls the longitude, so $0 leq v leq 2 pi$. The upper half of the torus will be given by $0 leq u leq pi$ so it seems you have the right intuition!
– Osama Ghani
Jul 27 at 6:08
add a comment |Â
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You need to explore the parameterization further to understand it instead of blindly asking for the answer.
– MPW
Jul 27 at 1:34
sir I need the ranges of $ u , v $ in order to integrate over the torus . But I am not sure. can you help me ?
– MONJUR ALAM
Jul 27 at 1:41
1
Imagine if the $v$ terms didn’t exist. Then the parametrization would trace out a full circle in the $xy$ plane which hints that $v$ controls the longitude, so $0 leq v leq 2 pi$. The upper half of the torus will be given by $0 leq u leq pi$ so it seems you have the right intuition!
– Osama Ghani
Jul 27 at 6:08