Calculate the flow of $vec{F}(x,y,z) = 3xvec{i} + 3yvec{j} + z^5vec{k}$ over the surface $x^2 + y^2 = 25$ for...
Calculate the flow of $vec{F}(x,y,z) = 3xvec{i} + 3yvec{j} + z^5vec{k}$ over the surface $x^2 + y^2 = 25$ for $0 leq z leq 1$. The normal considered points inside.
The book uses cylindrical coordinates and the final answer is $-150 pi$. I want to see other ways to approach the problem so I could understand better the concept. Must we not use the jacobian in this case? How would I solve the problem with cartesian coordinates?
multivariable-calculus surfaces surface-integrals
asked Nov 29 at 0:00
Carlos Oliveira
847
add a comment |
Calculate the flow of $vec{F}(x,y,z) = 3xvec{i} + 3yvec{j} + z^5vec{k}$ over the surface $x^2 + y^2 = 25$ for $0 leq z leq 1$. The normal considered points inside.
The book uses cylindrical coordinates and the final answer is $-150 pi$. I want to see other ways to approach the problem so I could understand better the concept. Must we not use the jacobian in this case? How would I solve the problem with cartesian coordinates?
multivariable-calculus surfaces surface-integrals
asked Nov 29 at 0:00
Carlos Oliveira
847
add a comment |
Calculate the flow of $vec{F}(x,y,z) = 3xvec{i} + 3yvec{j} + z^5vec{k}$ over the surface $x^2 + y^2 = 25$ for $0 leq z leq 1$. The normal considered points inside.
The book uses cylindrical coordinates and the final answer is $-150 pi$. I want to see other ways to approach the problem so I could understand better the concept. Must we not use the jacobian in this case? How would I solve the problem with cartesian coordinates?
multivariable-calculus surfaces surface-integrals
asked Nov 29 at 0:00
Carlos Oliveira
847
Calculate the flow of $vec{F}(x,y,z) = 3xvec{i} + 3yvec{j} + z^5vec{k}$ over the surface $x^2 + y^2 = 25$ for $0 leq z leq 1$. The normal considered points inside.
The book uses cylindrical coordinates and the final answer is $-150 pi$. I want to see other ways to approach the problem so I could understand better the concept. Must we not use the jacobian in this case? How would I solve the problem with cartesian coordinates?
multivariable-calculus surfaces surface-integrals
multivariable-calculus surfaces surface-integrals
asked Nov 29 at 0:00
Carlos Oliveira
847
asked Nov 29 at 0:00
Carlos Oliveira
847
asked Nov 29 at 0:00
Carlos Oliveira
847
asked Nov 29 at 0:00
Carlos Oliveira
847
asked Nov 29 at 0:00
Carlos Oliveira
847
847
add a comment |
add a comment |
1 Answer
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First parametrize the surface $S$, $vec r=(x(t,s),y(t,s),z(t,s))$. An obvious choice is $y=t$, leading to $x=pmsqrt{25-t^2}$ and $z=s$ free:
$$vec r=(pmsqrt{25-t^2}, t,s);;-5leq t<5;;0leq sleq 1$$
$vec F$ on the surface is $vec F=(pm3sqrt{25-t^2},3t,s^5)$
Now, we need the expression for a vector for the surface element. We get it from the cross product of two vectors tangent to the surface:
$dfrac{partial vec r}{partial t}=left(dfrac{mp t}{sqrt{25-t^2}},1,0right)$
$dfrac{partial vec r}{partial s}=(0,0,1)$
$dfrac{partial vec r}{partial t}timesdfrac{partial vec r}{partial s}=left(1,dfrac{pm t}{sqrt{25-t^2}},0right)$
Then, the flux,
$$Phi=int_Svec F·left(dfrac{partial vec r}{partial t}timesdfrac{partial vec r}{partial s}right);mathbb d t,mathbb ds$$
$$Phi=int_{0}^1int_{-5}^5left(3sqrt{25-t^2}+dfrac{3t^2}{sqrt{25-t^2}}right),mathbb d t,mathbb ds+int_{0}^1int_{5}^{-5}left(-3sqrt{25-t^2}-dfrac{3t^2}{sqrt{25-t^2}}right),mathbb d t,mathbb ds$$
$$Phi=150pi$$
My result is in positive because I've been using the pointing outwards vector for the surface element (and the field $vec F$ points outwards too on the surface)
answered Nov 29 at 17:13
Rafa Budría
5,4701825
There are a lot of details to pay attention to when using cartesian coordinates! Thank you for the alternative solution. I have also noticed you didn't use jacobian and ended up with the same result. I think it is kind strange how different parameterizations give the same result without a factor to correct it. I guess the answer for this is in higher mathematics... although I have seen a deduction for the flux integral formula where the jacobian is canceled out.
– Carlos Oliveira
Nov 30 at 0:59
We don't need the jacobian! we didn't change coordinates :) The, say, part of the jacobian the cylindrical coordinates have in cartesian is given directly from the surface element and the formula 8ninto the parametrization. And you can grasp the reason for your questions thinking geometrically, thinking of the invariant objects involved (like vectors).
– Rafa Budría
Nov 30 at 7:56
add a comment |
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1 Answer
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1 Answer
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First parametrize the surface $S$, $vec r=(x(t,s),y(t,s),z(t,s))$. An obvious choice is $y=t$, leading to $x=pmsqrt{25-t^2}$ and $z=s$ free:
$$vec r=(pmsqrt{25-t^2}, t,s);;-5leq t<5;;0leq sleq 1$$
$vec F$ on the surface is $vec F=(pm3sqrt{25-t^2},3t,s^5)$
Now, we need the expression for a vector for the surface element. We get it from the cross product of two vectors tangent to the surface:
$dfrac{partial vec r}{partial t}=left(dfrac{mp t}{sqrt{25-t^2}},1,0right)$
$dfrac{partial vec r}{partial s}=(0,0,1)$
$dfrac{partial vec r}{partial t}timesdfrac{partial vec r}{partial s}=left(1,dfrac{pm t}{sqrt{25-t^2}},0right)$
Then, the flux,
$$Phi=int_Svec F·left(dfrac{partial vec r}{partial t}timesdfrac{partial vec r}{partial s}right);mathbb d t,mathbb ds$$
$$Phi=int_{0}^1int_{-5}^5left(3sqrt{25-t^2}+dfrac{3t^2}{sqrt{25-t^2}}right),mathbb d t,mathbb ds+int_{0}^1int_{5}^{-5}left(-3sqrt{25-t^2}-dfrac{3t^2}{sqrt{25-t^2}}right),mathbb d t,mathbb ds$$
$$Phi=150pi$$
My result is in positive because I've been using the pointing outwards vector for the surface element (and the field $vec F$ points outwards too on the surface)
answered Nov 29 at 17:13
Rafa Budría
5,4701825
There are a lot of details to pay attention to when using cartesian coordinates! Thank you for the alternative solution. I have also noticed you didn't use jacobian and ended up with the same result. I think it is kind strange how different parameterizations give the same result without a factor to correct it. I guess the answer for this is in higher mathematics... although I have seen a deduction for the flux integral formula where the jacobian is canceled out.
– Carlos Oliveira
Nov 30 at 0:59
We don't need the jacobian! we didn't change coordinates :) The, say, part of the jacobian the cylindrical coordinates have in cartesian is given directly from the surface element and the formula 8ninto the parametrization. And you can grasp the reason for your questions thinking geometrically, thinking of the invariant objects involved (like vectors).
– Rafa Budría
Nov 30 at 7:56
add a comment |
First parametrize the surface $S$, $vec r=(x(t,s),y(t,s),z(t,s))$. An obvious choice is $y=t$, leading to $x=pmsqrt{25-t^2}$ and $z=s$ free:
$$vec r=(pmsqrt{25-t^2}, t,s);;-5leq t<5;;0leq sleq 1$$
$vec F$ on the surface is $vec F=(pm3sqrt{25-t^2},3t,s^5)$
Now, we need the expression for a vector for the surface element. We get it from the cross product of two vectors tangent to the surface:
$dfrac{partial vec r}{partial t}=left(dfrac{mp t}{sqrt{25-t^2}},1,0right)$
$dfrac{partial vec r}{partial s}=(0,0,1)$
$dfrac{partial vec r}{partial t}timesdfrac{partial vec r}{partial s}=left(1,dfrac{pm t}{sqrt{25-t^2}},0right)$
Then, the flux,
$$Phi=int_Svec F·left(dfrac{partial vec r}{partial t}timesdfrac{partial vec r}{partial s}right);mathbb d t,mathbb ds$$
$$Phi=int_{0}^1int_{-5}^5left(3sqrt{25-t^2}+dfrac{3t^2}{sqrt{25-t^2}}right),mathbb d t,mathbb ds+int_{0}^1int_{5}^{-5}left(-3sqrt{25-t^2}-dfrac{3t^2}{sqrt{25-t^2}}right),mathbb d t,mathbb ds$$
$$Phi=150pi$$
My result is in positive because I've been using the pointing outwards vector for the surface element (and the field $vec F$ points outwards too on the surface)
answered Nov 29 at 17:13
Rafa Budría
5,4701825
There are a lot of details to pay attention to when using cartesian coordinates! Thank you for the alternative solution. I have also noticed you didn't use jacobian and ended up with the same result. I think it is kind strange how different parameterizations give the same result without a factor to correct it. I guess the answer for this is in higher mathematics... although I have seen a deduction for the flux integral formula where the jacobian is canceled out.
– Carlos Oliveira
Nov 30 at 0:59
We don't need the jacobian! we didn't change coordinates :) The, say, part of the jacobian the cylindrical coordinates have in cartesian is given directly from the surface element and the formula 8ninto the parametrization. And you can grasp the reason for your questions thinking geometrically, thinking of the invariant objects involved (like vectors).
– Rafa Budría
Nov 30 at 7:56
add a comment |
First parametrize the surface $S$, $vec r=(x(t,s),y(t,s),z(t,s))$. An obvious choice is $y=t$, leading to $x=pmsqrt{25-t^2}$ and $z=s$ free:
$$vec r=(pmsqrt{25-t^2}, t,s);;-5leq t<5;;0leq sleq 1$$
$vec F$ on the surface is $vec F=(pm3sqrt{25-t^2},3t,s^5)$
Now, we need the expression for a vector for the surface element. We get it from the cross product of two vectors tangent to the surface:
$dfrac{partial vec r}{partial t}=left(dfrac{mp t}{sqrt{25-t^2}},1,0right)$
$dfrac{partial vec r}{partial s}=(0,0,1)$
$dfrac{partial vec r}{partial t}timesdfrac{partial vec r}{partial s}=left(1,dfrac{pm t}{sqrt{25-t^2}},0right)$
Then, the flux,
$$Phi=int_Svec F·left(dfrac{partial vec r}{partial t}timesdfrac{partial vec r}{partial s}right);mathbb d t,mathbb ds$$
$$Phi=int_{0}^1int_{-5}^5left(3sqrt{25-t^2}+dfrac{3t^2}{sqrt{25-t^2}}right),mathbb d t,mathbb ds+int_{0}^1int_{5}^{-5}left(-3sqrt{25-t^2}-dfrac{3t^2}{sqrt{25-t^2}}right),mathbb d t,mathbb ds$$
$$Phi=150pi$$
My result is in positive because I've been using the pointing outwards vector for the surface element (and the field $vec F$ points outwards too on the surface)
answered Nov 29 at 17:13
Rafa Budría
5,4701825
First parametrize the surface $S$, $vec r=(x(t,s),y(t,s),z(t,s))$. An obvious choice is $y=t$, leading to $x=pmsqrt{25-t^2}$ and $z=s$ free:
$$vec r=(pmsqrt{25-t^2}, t,s);;-5leq t<5;;0leq sleq 1$$
$vec F$ on the surface is $vec F=(pm3sqrt{25-t^2},3t,s^5)$
Now, we need the expression for a vector for the surface element. We get it from the cross product of two vectors tangent to the surface:
$dfrac{partial vec r}{partial t}=left(dfrac{mp t}{sqrt{25-t^2}},1,0right)$
$dfrac{partial vec r}{partial s}=(0,0,1)$
$dfrac{partial vec r}{partial t}timesdfrac{partial vec r}{partial s}=left(1,dfrac{pm t}{sqrt{25-t^2}},0right)$
Then, the flux,
$$Phi=int_Svec F·left(dfrac{partial vec r}{partial t}timesdfrac{partial vec r}{partial s}right);mathbb d t,mathbb ds$$
$$Phi=int_{0}^1int_{-5}^5left(3sqrt{25-t^2}+dfrac{3t^2}{sqrt{25-t^2}}right),mathbb d t,mathbb ds+int_{0}^1int_{5}^{-5}left(-3sqrt{25-t^2}-dfrac{3t^2}{sqrt{25-t^2}}right),mathbb d t,mathbb ds$$
$$Phi=150pi$$
My result is in positive because I've been using the pointing outwards vector for the surface element (and the field $vec F$ points outwards too on the surface)
answered Nov 29 at 17:13
Rafa Budría
5,4701825
edited Nov 29 at 17:19
answered Nov 29 at 17:13
Rafa Budría
5,4701825
answered Nov 29 at 17:13
Rafa Budría
5,4701825
answered Nov 29 at 17:13
Rafa Budría
5,4701825
5,4701825
There are a lot of details to pay attention to when using cartesian coordinates! Thank you for the alternative solution. I have also noticed you didn't use jacobian and ended up with the same result. I think it is kind strange how different parameterizations give the same result without a factor to correct it. I guess the answer for this is in higher mathematics... although I have seen a deduction for the flux integral formula where the jacobian is canceled out.
– Carlos Oliveira
Nov 30 at 0:59
We don't need the jacobian! we didn't change coordinates :) The, say, part of the jacobian the cylindrical coordinates have in cartesian is given directly from the surface element and the formula 8ninto the parametrization. And you can grasp the reason for your questions thinking geometrically, thinking of the invariant objects involved (like vectors).
– Rafa Budría
Nov 30 at 7:56
add a comment |
There are a lot of details to pay attention to when using cartesian coordinates! Thank you for the alternative solution. I have also noticed you didn't use jacobian and ended up with the same result. I think it is kind strange how different parameterizations give the same result without a factor to correct it. I guess the answer for this is in higher mathematics... although I have seen a deduction for the flux integral formula where the jacobian is canceled out.
– Carlos Oliveira
Nov 30 at 0:59
We don't need the jacobian! we didn't change coordinates :) The, say, part of the jacobian the cylindrical coordinates have in cartesian is given directly from the surface element and the formula 8ninto the parametrization. And you can grasp the reason for your questions thinking geometrically, thinking of the invariant objects involved (like vectors).
– Rafa Budría
Nov 30 at 7:56
There are a lot of details to pay attention to when using cartesian coordinates! Thank you for the alternative solution. I have also noticed you didn't use jacobian and ended up with the same result. I think it is kind strange how different parameterizations give the same result without a factor to correct it. I guess the answer for this is in higher mathematics... although I have seen a deduction for the flux integral formula where the jacobian is canceled out.
– Carlos Oliveira
Nov 30 at 0:59
There are a lot of details to pay attention to when using cartesian coordinates! Thank you for the alternative solution. I have also noticed you didn't use jacobian and ended up with the same result. I think it is kind strange how different parameterizations give the same result without a factor to correct it. I guess the answer for this is in higher mathematics... although I have seen a deduction for the flux integral formula where the jacobian is canceled out.
– Carlos Oliveira
Nov 30 at 0:59
We don't need the jacobian! we didn't change coordinates :) The, say, part of the jacobian the cylindrical coordinates have in cartesian is given directly from the surface element and the formula 8ninto the parametrization. And you can grasp the reason for your questions thinking geometrically, thinking of the invariant objects involved (like vectors).
– Rafa Budría
Nov 30 at 7:56
We don't need the jacobian! we didn't change coordinates :) The, say, part of the jacobian the cylindrical coordinates have in cartesian is given directly from the surface element and the formula 8ninto the parametrization. And you can grasp the reason for your questions thinking geometrically, thinking of the invariant objects involved (like vectors).
– Rafa Budría
Nov 30 at 7:56
add a comment |
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