Mixing
Surface integrals of second kind
$begingroup$
In the formula for calculating surface integrals of second kind, we have:
But, this integral is denoted by $int int _S vec{F}cdot hat{n}dS $ . So, should we always normalize the expression $ frac{partial vec r }{partial v} times frac{partial vec r}{partial u} $ before substituting it into the formula?
Thanks in advance
!
multivariable-calculus integration
edited Jan 17 at 21:36
Glorfindel
3,41981830
asked Jun 19 '13 at 7:49
czashczash
21227
$endgroup$
add a comment |
$begingroup$
In the formula for calculating surface integrals of second kind, we have:
But, this integral is denoted by $int int _S vec{F}cdot hat{n}dS $ . So, should we always normalize the expression $ frac{partial vec r }{partial v} times frac{partial vec r}{partial u} $ before substituting it into the formula?
Thanks in advance
!
multivariable-calculus integration
edited Jan 17 at 21:36
Glorfindel
3,41981830
asked Jun 19 '13 at 7:49
czashczash
21227
$endgroup$
$begingroup$
A related problem.
$endgroup$
– Mhenni Benghorbal
Jun 19 '13 at 8:24
add a comment |
$begingroup$
In the formula for calculating surface integrals of second kind, we have:
But, this integral is denoted by $int int _S vec{F}cdot hat{n}dS $ . So, should we always normalize the expression $ frac{partial vec r }{partial v} times frac{partial vec r}{partial u} $ before substituting it into the formula?
Thanks in advance
!
multivariable-calculus integration
edited Jan 17 at 21:36
Glorfindel
3,41981830
asked Jun 19 '13 at 7:49
czashczash
21227
$endgroup$
In the formula for calculating surface integrals of second kind, we have:
But, this integral is denoted by $int int _S vec{F}cdot hat{n}dS $ . So, should we always normalize the expression $ frac{partial vec r }{partial v} times frac{partial vec r}{partial u} $ before substituting it into the formula?
Thanks in advance
!
multivariable-calculus integration
multivariable-calculus integration
edited Jan 17 at 21:36
Glorfindel
3,41981830
asked Jun 19 '13 at 7:49
czashczash
21227
edited Jan 17 at 21:36
Glorfindel
3,41981830
asked Jun 19 '13 at 7:49
czashczash
21227
edited Jan 17 at 21:36
Glorfindel
3,41981830
edited Jan 17 at 21:36
Glorfindel
3,41981830
edited Jan 17 at 21:36
Glorfindel
3,41981830
3,41981830
asked Jun 19 '13 at 7:49
czashczash
21227
asked Jun 19 '13 at 7:49
czashczash
21227
asked Jun 19 '13 at 7:49
czashczash
21227
21227
$begingroup$
A related problem.
$endgroup$
– Mhenni Benghorbal
Jun 19 '13 at 8:24
add a comment |
$begingroup$
A related problem.
$endgroup$
– Mhenni Benghorbal
Jun 19 '13 at 8:24
$begingroup$
A related problem.
$endgroup$
– Mhenni Benghorbal
Jun 19 '13 at 8:24
$begingroup$
A related problem.
$endgroup$
– Mhenni Benghorbal
Jun 19 '13 at 8:24
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here's a little derivation of the formula - hopefully it'll help you to answer your own question.
The definition of the surface integral tells us that
$$
iint_S vec F cdot dvec S = iint_S vec F cdot hat n , dS
$$
where $hat n$ is the unit normal vector to the surface. This equivalence is well described in chapter 16 of Stewart's book. Moving on, we know that
$$
vec r_u times vec r_v
$$
is a vector normal to the surface parametrized as $vec r(u,v)$. This vector must to normal to the surface because $vec r_u$ and $vec r_v$ are both curves tangent to the surface and their cross product gives a vector that perpendicular to each of them. To make this a unit vector, we simply divide it by it's length $left| vec r_u times vec r_v right|$ and obtain
$$
hat n = frac{vec r_u times vec r_v}{left| vec r_u times vec r_v right|}
$$
Substituting this into the above integral, we obtain
$$
iint_S vec F cdot left( frac{vec r_u times vec r_v}{left| vec r_u times vec r_v right|} right) , dS
$$
Furthermore, we know that $left| vec r_u times vec r_v right|, dA$ is the infinitesimal surface area element $dS$ (this is also well described in Stewart). Plugging this in, we get
$$
iint_D vec F cdot left( frac{vec r_u times vec r_v}{left| vec r_u times vec r_v right|} right) , left| vec r_u times vec r_v right| , dA
$$
where $D$ is the domain of the parameters $u$ and $v$. This simplifies to
$$
iint_D vec F cdot left( vec r_u times vec r_v right) , dA
$$
Any of these expressions can be used in computing the surface integral. However, using any of them before the final one would just be a recalcitrant implementation of the final formula as they would simplify to the same thing. If you're trying to compute the surface integral over $D$, the parameters of $u$ and $v$, then no, you do not need to scale the normal vector to length one.
Hope this helps!
answered Jun 19 '13 at 8:12
RyanRyan
584211
$endgroup$
$begingroup$
I know I mentioned Stewart's book a couple of times. If you don't use his book, this resource is essentially a reproduction of it and also an invaluable resource: tutorial.math.lamar.edu/Classes/CalcIII/SurfaceIntegrals.aspx.
$endgroup$
– Ryan
Jun 19 '13 at 8:14
$begingroup$
Thanks a lot for your answer, but there is still one more thing I can't understand. In your final formula you showed that we actually do not need to normalize our normal vector. But in your last sentence you wrote that if we are using the formula for double integral over D , we do need to rescale the normal vector... Can you please explain to me why the second sentence doesn't contradict the last formula? Thanks
$endgroup$
– czash
Jun 19 '13 at 9:15
$begingroup$
Typo - sorry about that. You do not need to rescale it.
$endgroup$
– Ryan
Jun 19 '13 at 9:53
add a comment |
Your Answer
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$begingroup$
Here's a little derivation of the formula - hopefully it'll help you to answer your own question.
The definition of the surface integral tells us that
$$
iint_S vec F cdot dvec S = iint_S vec F cdot hat n , dS
$$
where $hat n$ is the unit normal vector to the surface. This equivalence is well described in chapter 16 of Stewart's book. Moving on, we know that
$$
vec r_u times vec r_v
$$
is a vector normal to the surface parametrized as $vec r(u,v)$. This vector must to normal to the surface because $vec r_u$ and $vec r_v$ are both curves tangent to the surface and their cross product gives a vector that perpendicular to each of them. To make this a unit vector, we simply divide it by it's length $left| vec r_u times vec r_v right|$ and obtain
$$
hat n = frac{vec r_u times vec r_v}{left| vec r_u times vec r_v right|}
$$
Substituting this into the above integral, we obtain
$$
iint_S vec F cdot left( frac{vec r_u times vec r_v}{left| vec r_u times vec r_v right|} right) , dS
$$
Furthermore, we know that $left| vec r_u times vec r_v right|, dA$ is the infinitesimal surface area element $dS$ (this is also well described in Stewart). Plugging this in, we get
$$
iint_D vec F cdot left( frac{vec r_u times vec r_v}{left| vec r_u times vec r_v right|} right) , left| vec r_u times vec r_v right| , dA
$$
where $D$ is the domain of the parameters $u$ and $v$. This simplifies to
$$
iint_D vec F cdot left( vec r_u times vec r_v right) , dA
$$
Any of these expressions can be used in computing the surface integral. However, using any of them before the final one would just be a recalcitrant implementation of the final formula as they would simplify to the same thing. If you're trying to compute the surface integral over $D$, the parameters of $u$ and $v$, then no, you do not need to scale the normal vector to length one.
Hope this helps!
answered Jun 19 '13 at 8:12
RyanRyan
584211
$endgroup$
$begingroup$
I know I mentioned Stewart's book a couple of times. If you don't use his book, this resource is essentially a reproduction of it and also an invaluable resource: tutorial.math.lamar.edu/Classes/CalcIII/SurfaceIntegrals.aspx.
$endgroup$
– Ryan
Jun 19 '13 at 8:14
$begingroup$
Thanks a lot for your answer, but there is still one more thing I can't understand. In your final formula you showed that we actually do not need to normalize our normal vector. But in your last sentence you wrote that if we are using the formula for double integral over D , we do need to rescale the normal vector... Can you please explain to me why the second sentence doesn't contradict the last formula? Thanks
$endgroup$
– czash
Jun 19 '13 at 9:15
$begingroup$
Typo - sorry about that. You do not need to rescale it.
$endgroup$
– Ryan
Jun 19 '13 at 9:53
add a comment |
$begingroup$
Here's a little derivation of the formula - hopefully it'll help you to answer your own question.
The definition of the surface integral tells us that
$$
iint_S vec F cdot dvec S = iint_S vec F cdot hat n , dS
$$
where $hat n$ is the unit normal vector to the surface. This equivalence is well described in chapter 16 of Stewart's book. Moving on, we know that
$$
vec r_u times vec r_v
$$
is a vector normal to the surface parametrized as $vec r(u,v)$. This vector must to normal to the surface because $vec r_u$ and $vec r_v$ are both curves tangent to the surface and their cross product gives a vector that perpendicular to each of them. To make this a unit vector, we simply divide it by it's length $left| vec r_u times vec r_v right|$ and obtain
$$
hat n = frac{vec r_u times vec r_v}{left| vec r_u times vec r_v right|}
$$
Substituting this into the above integral, we obtain
$$
iint_S vec F cdot left( frac{vec r_u times vec r_v}{left| vec r_u times vec r_v right|} right) , dS
$$
Furthermore, we know that $left| vec r_u times vec r_v right|, dA$ is the infinitesimal surface area element $dS$ (this is also well described in Stewart). Plugging this in, we get
$$
iint_D vec F cdot left( frac{vec r_u times vec r_v}{left| vec r_u times vec r_v right|} right) , left| vec r_u times vec r_v right| , dA
$$
where $D$ is the domain of the parameters $u$ and $v$. This simplifies to
$$
iint_D vec F cdot left( vec r_u times vec r_v right) , dA
$$
Any of these expressions can be used in computing the surface integral. However, using any of them before the final one would just be a recalcitrant implementation of the final formula as they would simplify to the same thing. If you're trying to compute the surface integral over $D$, the parameters of $u$ and $v$, then no, you do not need to scale the normal vector to length one.
Hope this helps!
answered Jun 19 '13 at 8:12
RyanRyan
584211
$endgroup$
$begingroup$
I know I mentioned Stewart's book a couple of times. If you don't use his book, this resource is essentially a reproduction of it and also an invaluable resource: tutorial.math.lamar.edu/Classes/CalcIII/SurfaceIntegrals.aspx.
$endgroup$
– Ryan
Jun 19 '13 at 8:14
$begingroup$
Thanks a lot for your answer, but there is still one more thing I can't understand. In your final formula you showed that we actually do not need to normalize our normal vector. But in your last sentence you wrote that if we are using the formula for double integral over D , we do need to rescale the normal vector... Can you please explain to me why the second sentence doesn't contradict the last formula? Thanks
$endgroup$
– czash
Jun 19 '13 at 9:15
$begingroup$
Typo - sorry about that. You do not need to rescale it.
$endgroup$
– Ryan
Jun 19 '13 at 9:53
add a comment |
$begingroup$
Here's a little derivation of the formula - hopefully it'll help you to answer your own question.
The definition of the surface integral tells us that
$$
iint_S vec F cdot dvec S = iint_S vec F cdot hat n , dS
$$
where $hat n$ is the unit normal vector to the surface. This equivalence is well described in chapter 16 of Stewart's book. Moving on, we know that
$$
vec r_u times vec r_v
$$
is a vector normal to the surface parametrized as $vec r(u,v)$. This vector must to normal to the surface because $vec r_u$ and $vec r_v$ are both curves tangent to the surface and their cross product gives a vector that perpendicular to each of them. To make this a unit vector, we simply divide it by it's length $left| vec r_u times vec r_v right|$ and obtain
$$
hat n = frac{vec r_u times vec r_v}{left| vec r_u times vec r_v right|}
$$
Substituting this into the above integral, we obtain
$$
iint_S vec F cdot left( frac{vec r_u times vec r_v}{left| vec r_u times vec r_v right|} right) , dS
$$
Furthermore, we know that $left| vec r_u times vec r_v right|, dA$ is the infinitesimal surface area element $dS$ (this is also well described in Stewart). Plugging this in, we get
$$
iint_D vec F cdot left( frac{vec r_u times vec r_v}{left| vec r_u times vec r_v right|} right) , left| vec r_u times vec r_v right| , dA
$$
where $D$ is the domain of the parameters $u$ and $v$. This simplifies to
$$
iint_D vec F cdot left( vec r_u times vec r_v right) , dA
$$
Any of these expressions can be used in computing the surface integral. However, using any of them before the final one would just be a recalcitrant implementation of the final formula as they would simplify to the same thing. If you're trying to compute the surface integral over $D$, the parameters of $u$ and $v$, then no, you do not need to scale the normal vector to length one.
Hope this helps!
answered Jun 19 '13 at 8:12
RyanRyan
584211
$endgroup$
Here's a little derivation of the formula - hopefully it'll help you to answer your own question.
The definition of the surface integral tells us that
$$
iint_S vec F cdot dvec S = iint_S vec F cdot hat n , dS
$$
where $hat n$ is the unit normal vector to the surface. This equivalence is well described in chapter 16 of Stewart's book. Moving on, we know that
$$
vec r_u times vec r_v
$$
is a vector normal to the surface parametrized as $vec r(u,v)$. This vector must to normal to the surface because $vec r_u$ and $vec r_v$ are both curves tangent to the surface and their cross product gives a vector that perpendicular to each of them. To make this a unit vector, we simply divide it by it's length $left| vec r_u times vec r_v right|$ and obtain
$$
hat n = frac{vec r_u times vec r_v}{left| vec r_u times vec r_v right|}
$$
Substituting this into the above integral, we obtain
$$
iint_S vec F cdot left( frac{vec r_u times vec r_v}{left| vec r_u times vec r_v right|} right) , dS
$$
Furthermore, we know that $left| vec r_u times vec r_v right|, dA$ is the infinitesimal surface area element $dS$ (this is also well described in Stewart). Plugging this in, we get
$$
iint_D vec F cdot left( frac{vec r_u times vec r_v}{left| vec r_u times vec r_v right|} right) , left| vec r_u times vec r_v right| , dA
$$
where $D$ is the domain of the parameters $u$ and $v$. This simplifies to
$$
iint_D vec F cdot left( vec r_u times vec r_v right) , dA
$$
Any of these expressions can be used in computing the surface integral. However, using any of them before the final one would just be a recalcitrant implementation of the final formula as they would simplify to the same thing. If you're trying to compute the surface integral over $D$, the parameters of $u$ and $v$, then no, you do not need to scale the normal vector to length one.
Hope this helps!
answered Jun 19 '13 at 8:12
RyanRyan
584211
edited Jun 19 '13 at 9:53
answered Jun 19 '13 at 8:12
RyanRyan
584211
answered Jun 19 '13 at 8:12
RyanRyan
584211
answered Jun 19 '13 at 8:12
RyanRyan
584211
584211
$begingroup$
I know I mentioned Stewart's book a couple of times. If you don't use his book, this resource is essentially a reproduction of it and also an invaluable resource: tutorial.math.lamar.edu/Classes/CalcIII/SurfaceIntegrals.aspx.
$endgroup$
– Ryan
Jun 19 '13 at 8:14
$begingroup$
Thanks a lot for your answer, but there is still one more thing I can't understand. In your final formula you showed that we actually do not need to normalize our normal vector. But in your last sentence you wrote that if we are using the formula for double integral over D , we do need to rescale the normal vector... Can you please explain to me why the second sentence doesn't contradict the last formula? Thanks
$endgroup$
– czash
Jun 19 '13 at 9:15
$begingroup$
Typo - sorry about that. You do not need to rescale it.
$endgroup$
– Ryan
Jun 19 '13 at 9:53
add a comment |
$begingroup$
I know I mentioned Stewart's book a couple of times. If you don't use his book, this resource is essentially a reproduction of it and also an invaluable resource: tutorial.math.lamar.edu/Classes/CalcIII/SurfaceIntegrals.aspx.
$endgroup$
– Ryan
Jun 19 '13 at 8:14
$begingroup$
Thanks a lot for your answer, but there is still one more thing I can't understand. In your final formula you showed that we actually do not need to normalize our normal vector. But in your last sentence you wrote that if we are using the formula for double integral over D , we do need to rescale the normal vector... Can you please explain to me why the second sentence doesn't contradict the last formula? Thanks
$endgroup$
– czash
Jun 19 '13 at 9:15
$begingroup$
Typo - sorry about that. You do not need to rescale it.
$endgroup$
– Ryan
Jun 19 '13 at 9:53
$begingroup$
I know I mentioned Stewart's book a couple of times. If you don't use his book, this resource is essentially a reproduction of it and also an invaluable resource: tutorial.math.lamar.edu/Classes/CalcIII/SurfaceIntegrals.aspx.
$endgroup$
– Ryan
Jun 19 '13 at 8:14
$begingroup$
I know I mentioned Stewart's book a couple of times. If you don't use his book, this resource is essentially a reproduction of it and also an invaluable resource: tutorial.math.lamar.edu/Classes/CalcIII/SurfaceIntegrals.aspx.
$endgroup$
– Ryan
Jun 19 '13 at 8:14
$begingroup$
Thanks a lot for your answer, but there is still one more thing I can't understand. In your final formula you showed that we actually do not need to normalize our normal vector. But in your last sentence you wrote that if we are using the formula for double integral over D , we do need to rescale the normal vector... Can you please explain to me why the second sentence doesn't contradict the last formula? Thanks
$endgroup$
– czash
Jun 19 '13 at 9:15
$begingroup$
Thanks a lot for your answer, but there is still one more thing I can't understand. In your final formula you showed that we actually do not need to normalize our normal vector. But in your last sentence you wrote that if we are using the formula for double integral over D , we do need to rescale the normal vector... Can you please explain to me why the second sentence doesn't contradict the last formula? Thanks
$endgroup$
– czash
Jun 19 '13 at 9:15
$begingroup$
Typo - sorry about that. You do not need to rescale it.
$endgroup$
– Ryan
Jun 19 '13 at 9:53
$begingroup$
Typo - sorry about that. You do not need to rescale it.
$endgroup$
– Ryan
Jun 19 '13 at 9:53
add a comment |
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$begingroup$
A related problem.
$endgroup$
– Mhenni Benghorbal
Jun 19 '13 at 8:24