Derivation of formula for surface integral
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Let $S$ be a surface given by the parametrization $vec{r}(u,v)= x(u,v)vec{i} + y(u,v)vec{j} + z(u,v)vec{k}$ for $(u,v)in D$. If $f(x,y,z)$ is a function defined on $S$, then the surface integral can be found using the formula
$$
iint_S f(x,y,z) dS = iint_D f(vec{r}(u,v))lvert vec{r}_utimesvec{r}_urvert dA
$$
I have seen how to derive this by drawing some pictures, but I am wondering if there is a proof that makes use of the change of variable formula. I have tried a few things, but I can't find anything that gives a Jacobian of $lvert vec{r}_utimesvec{r}_urvert$.
Is this possible?
calculus integration surfaces alternative-proof
asked Nov 19 at 12:31
John Doe
23621346
add a comment |
up vote
0
down vote
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Let $S$ be a surface given by the parametrization $vec{r}(u,v)= x(u,v)vec{i} + y(u,v)vec{j} + z(u,v)vec{k}$ for $(u,v)in D$. If $f(x,y,z)$ is a function defined on $S$, then the surface integral can be found using the formula
$$
iint_S f(x,y,z) dS = iint_D f(vec{r}(u,v))lvert vec{r}_utimesvec{r}_urvert dA
$$
I have seen how to derive this by drawing some pictures, but I am wondering if there is a proof that makes use of the change of variable formula. I have tried a few things, but I can't find anything that gives a Jacobian of $lvert vec{r}_utimesvec{r}_urvert$.
Is this possible?
calculus integration surfaces alternative-proof
asked Nov 19 at 12:31
John Doe
23621346
Every derivation I have seen of this is geometric. The $|vec r_utimesvec r_v|dA$ term is the calculation of $dS$ so strictly speaking, we are not changing variables. We are given a parametrization of a surface and calculating the surface integral based on that.
– John Douma
Nov 19 at 13:15
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $S$ be a surface given by the parametrization $vec{r}(u,v)= x(u,v)vec{i} + y(u,v)vec{j} + z(u,v)vec{k}$ for $(u,v)in D$. If $f(x,y,z)$ is a function defined on $S$, then the surface integral can be found using the formula
$$
iint_S f(x,y,z) dS = iint_D f(vec{r}(u,v))lvert vec{r}_utimesvec{r}_urvert dA
$$
I have seen how to derive this by drawing some pictures, but I am wondering if there is a proof that makes use of the change of variable formula. I have tried a few things, but I can't find anything that gives a Jacobian of $lvert vec{r}_utimesvec{r}_urvert$.
Is this possible?
calculus integration surfaces alternative-proof
asked Nov 19 at 12:31
John Doe
23621346
Let $S$ be a surface given by the parametrization $vec{r}(u,v)= x(u,v)vec{i} + y(u,v)vec{j} + z(u,v)vec{k}$ for $(u,v)in D$. If $f(x,y,z)$ is a function defined on $S$, then the surface integral can be found using the formula
$$
iint_S f(x,y,z) dS = iint_D f(vec{r}(u,v))lvert vec{r}_utimesvec{r}_urvert dA
$$
I have seen how to derive this by drawing some pictures, but I am wondering if there is a proof that makes use of the change of variable formula. I have tried a few things, but I can't find anything that gives a Jacobian of $lvert vec{r}_utimesvec{r}_urvert$.
Is this possible?
calculus integration surfaces alternative-proof
calculus integration surfaces alternative-proof
asked Nov 19 at 12:31
John Doe
23621346
asked Nov 19 at 12:31
John Doe
23621346
asked Nov 19 at 12:31
John Doe
23621346
asked Nov 19 at 12:31
John Doe
23621346
asked Nov 19 at 12:31
John Doe
23621346
23621346
Every derivation I have seen of this is geometric. The $|vec r_utimesvec r_v|dA$ term is the calculation of $dS$ so strictly speaking, we are not changing variables. We are given a parametrization of a surface and calculating the surface integral based on that.
– John Douma
Nov 19 at 13:15
add a comment |
Every derivation I have seen of this is geometric. The $|vec r_utimesvec r_v|dA$ term is the calculation of $dS$ so strictly speaking, we are not changing variables. We are given a parametrization of a surface and calculating the surface integral based on that.
– John Douma
Nov 19 at 13:15
Every derivation I have seen of this is geometric. The $|vec r_utimesvec r_v|dA$ term is the calculation of $dS$ so strictly speaking, we are not changing variables. We are given a parametrization of a surface and calculating the surface integral based on that.
– John Douma
Nov 19 at 13:15
Every derivation I have seen of this is geometric. The $|vec r_utimesvec r_v|dA$ term is the calculation of $dS$ so strictly speaking, we are not changing variables. We are given a parametrization of a surface and calculating the surface integral based on that.
– John Douma
Nov 19 at 13:15
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
The formula you have written down is merely the definition of the surface integral(and as such doesn't need any proof). Ok, to be slightly more precise, if you define the integral as the sum of surface "area elements" times the value of the function at any point inside that surface, then you see that this gives you the above formula after noting that the vector product is the same as the area of the surface element.
The change of variable formula just shows that you may parametrize your surface S in any way you like and still get the same answer. The Jacobian comes up when we have an integral of the type $int_{a}^{b} int_{c}^{d}f(x,y) ,dx,dy$ and some other params $u,v$ and $u=u(x,y)$ and $v=v(x,y)$ and then $int_{a}^{b} int_{c}^{d}f(x,y) ,dx,dy = int_{r}^{s} int_{m}^{n}J(u,v)f(u,v) ,dx,dy$.
In our case, assuming you have some other set of parameters $(p,q)$ such that $vec{r}(p,q) = x(p,q)vec{i}+y(p,q)vec{j}+z(p,q)vec{k}$ , then
$|vec{r}_u wedge vec{r}_v|dudv= J(u,v)|vec{r}_p wedge vec{r}_q|dpdq$(write down this computation in full if you don't believe me!) and by the change of variable formula, the two integrals(one in which we parametrized using $p,q$ and the other with $u,v$) will be equal.
answered Nov 19 at 12:37
Sorin Tirc
77210
1
The above formula was presented to me as a theorem. The definition was the things with "area elements". So there is no way to choose some change of variables so that the $lvert vec{r}_utimesvec{r}_urvert$ becomes the jacobian?
– John Doe
Nov 19 at 12:43
2
@John Doe: I edited my answer to include a small discussion about the Jacobian, but I have to insist that the formula above is a consequence of the definition of the integral and has, strictly speaking, nothing to do with a Jacobian. However, given that when you define an integral you don't define it in terms of a certain parametrization, you may actually deduce the formula of change of variables(which does indeed involve the Jacobian) from this integral formula(but not the other way round).
– Sorin Tirc
Nov 19 at 13:25
Ok, thanks for the answer. It just really looked to me like the $lvert vec{r}_utimesvec{r}_vrvert$ factor is some kind of Jacobian.
– John Doe
Nov 19 at 17:01
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The formula you have written down is merely the definition of the surface integral(and as such doesn't need any proof). Ok, to be slightly more precise, if you define the integral as the sum of surface "area elements" times the value of the function at any point inside that surface, then you see that this gives you the above formula after noting that the vector product is the same as the area of the surface element.
The change of variable formula just shows that you may parametrize your surface S in any way you like and still get the same answer. The Jacobian comes up when we have an integral of the type $int_{a}^{b} int_{c}^{d}f(x,y) ,dx,dy$ and some other params $u,v$ and $u=u(x,y)$ and $v=v(x,y)$ and then $int_{a}^{b} int_{c}^{d}f(x,y) ,dx,dy = int_{r}^{s} int_{m}^{n}J(u,v)f(u,v) ,dx,dy$.
In our case, assuming you have some other set of parameters $(p,q)$ such that $vec{r}(p,q) = x(p,q)vec{i}+y(p,q)vec{j}+z(p,q)vec{k}$ , then
$|vec{r}_u wedge vec{r}_v|dudv= J(u,v)|vec{r}_p wedge vec{r}_q|dpdq$(write down this computation in full if you don't believe me!) and by the change of variable formula, the two integrals(one in which we parametrized using $p,q$ and the other with $u,v$) will be equal.
answered Nov 19 at 12:37
Sorin Tirc
77210
1
The above formula was presented to me as a theorem. The definition was the things with "area elements". So there is no way to choose some change of variables so that the $lvert vec{r}_utimesvec{r}_urvert$ becomes the jacobian?
– John Doe
Nov 19 at 12:43
2
@John Doe: I edited my answer to include a small discussion about the Jacobian, but I have to insist that the formula above is a consequence of the definition of the integral and has, strictly speaking, nothing to do with a Jacobian. However, given that when you define an integral you don't define it in terms of a certain parametrization, you may actually deduce the formula of change of variables(which does indeed involve the Jacobian) from this integral formula(but not the other way round).
– Sorin Tirc
Nov 19 at 13:25
Ok, thanks for the answer. It just really looked to me like the $lvert vec{r}_utimesvec{r}_vrvert$ factor is some kind of Jacobian.
– John Doe
Nov 19 at 17:01
add a comment |
up vote
2
down vote
accepted
The formula you have written down is merely the definition of the surface integral(and as such doesn't need any proof). Ok, to be slightly more precise, if you define the integral as the sum of surface "area elements" times the value of the function at any point inside that surface, then you see that this gives you the above formula after noting that the vector product is the same as the area of the surface element.
The change of variable formula just shows that you may parametrize your surface S in any way you like and still get the same answer. The Jacobian comes up when we have an integral of the type $int_{a}^{b} int_{c}^{d}f(x,y) ,dx,dy$ and some other params $u,v$ and $u=u(x,y)$ and $v=v(x,y)$ and then $int_{a}^{b} int_{c}^{d}f(x,y) ,dx,dy = int_{r}^{s} int_{m}^{n}J(u,v)f(u,v) ,dx,dy$.
In our case, assuming you have some other set of parameters $(p,q)$ such that $vec{r}(p,q) = x(p,q)vec{i}+y(p,q)vec{j}+z(p,q)vec{k}$ , then
$|vec{r}_u wedge vec{r}_v|dudv= J(u,v)|vec{r}_p wedge vec{r}_q|dpdq$(write down this computation in full if you don't believe me!) and by the change of variable formula, the two integrals(one in which we parametrized using $p,q$ and the other with $u,v$) will be equal.
answered Nov 19 at 12:37
Sorin Tirc
77210
1
The above formula was presented to me as a theorem. The definition was the things with "area elements". So there is no way to choose some change of variables so that the $lvert vec{r}_utimesvec{r}_urvert$ becomes the jacobian?
– John Doe
Nov 19 at 12:43
2
@John Doe: I edited my answer to include a small discussion about the Jacobian, but I have to insist that the formula above is a consequence of the definition of the integral and has, strictly speaking, nothing to do with a Jacobian. However, given that when you define an integral you don't define it in terms of a certain parametrization, you may actually deduce the formula of change of variables(which does indeed involve the Jacobian) from this integral formula(but not the other way round).
– Sorin Tirc
Nov 19 at 13:25
Ok, thanks for the answer. It just really looked to me like the $lvert vec{r}_utimesvec{r}_vrvert$ factor is some kind of Jacobian.
– John Doe
Nov 19 at 17:01
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The formula you have written down is merely the definition of the surface integral(and as such doesn't need any proof). Ok, to be slightly more precise, if you define the integral as the sum of surface "area elements" times the value of the function at any point inside that surface, then you see that this gives you the above formula after noting that the vector product is the same as the area of the surface element.
The change of variable formula just shows that you may parametrize your surface S in any way you like and still get the same answer. The Jacobian comes up when we have an integral of the type $int_{a}^{b} int_{c}^{d}f(x,y) ,dx,dy$ and some other params $u,v$ and $u=u(x,y)$ and $v=v(x,y)$ and then $int_{a}^{b} int_{c}^{d}f(x,y) ,dx,dy = int_{r}^{s} int_{m}^{n}J(u,v)f(u,v) ,dx,dy$.
In our case, assuming you have some other set of parameters $(p,q)$ such that $vec{r}(p,q) = x(p,q)vec{i}+y(p,q)vec{j}+z(p,q)vec{k}$ , then
$|vec{r}_u wedge vec{r}_v|dudv= J(u,v)|vec{r}_p wedge vec{r}_q|dpdq$(write down this computation in full if you don't believe me!) and by the change of variable formula, the two integrals(one in which we parametrized using $p,q$ and the other with $u,v$) will be equal.
answered Nov 19 at 12:37
Sorin Tirc
77210
The formula you have written down is merely the definition of the surface integral(and as such doesn't need any proof). Ok, to be slightly more precise, if you define the integral as the sum of surface "area elements" times the value of the function at any point inside that surface, then you see that this gives you the above formula after noting that the vector product is the same as the area of the surface element.
The change of variable formula just shows that you may parametrize your surface S in any way you like and still get the same answer. The Jacobian comes up when we have an integral of the type $int_{a}^{b} int_{c}^{d}f(x,y) ,dx,dy$ and some other params $u,v$ and $u=u(x,y)$ and $v=v(x,y)$ and then $int_{a}^{b} int_{c}^{d}f(x,y) ,dx,dy = int_{r}^{s} int_{m}^{n}J(u,v)f(u,v) ,dx,dy$.
In our case, assuming you have some other set of parameters $(p,q)$ such that $vec{r}(p,q) = x(p,q)vec{i}+y(p,q)vec{j}+z(p,q)vec{k}$ , then
$|vec{r}_u wedge vec{r}_v|dudv= J(u,v)|vec{r}_p wedge vec{r}_q|dpdq$(write down this computation in full if you don't believe me!) and by the change of variable formula, the two integrals(one in which we parametrized using $p,q$ and the other with $u,v$) will be equal.
answered Nov 19 at 12:37
Sorin Tirc
77210
edited Nov 19 at 13:38
answered Nov 19 at 12:37
Sorin Tirc
77210
answered Nov 19 at 12:37
Sorin Tirc
77210
answered Nov 19 at 12:37
Sorin Tirc
77210
77210
1
The above formula was presented to me as a theorem. The definition was the things with "area elements". So there is no way to choose some change of variables so that the $lvert vec{r}_utimesvec{r}_urvert$ becomes the jacobian?
– John Doe
Nov 19 at 12:43
2
@John Doe: I edited my answer to include a small discussion about the Jacobian, but I have to insist that the formula above is a consequence of the definition of the integral and has, strictly speaking, nothing to do with a Jacobian. However, given that when you define an integral you don't define it in terms of a certain parametrization, you may actually deduce the formula of change of variables(which does indeed involve the Jacobian) from this integral formula(but not the other way round).
– Sorin Tirc
Nov 19 at 13:25
Ok, thanks for the answer. It just really looked to me like the $lvert vec{r}_utimesvec{r}_vrvert$ factor is some kind of Jacobian.
– John Doe
Nov 19 at 17:01
add a comment |
1
The above formula was presented to me as a theorem. The definition was the things with "area elements". So there is no way to choose some change of variables so that the $lvert vec{r}_utimesvec{r}_urvert$ becomes the jacobian?
– John Doe
Nov 19 at 12:43
2
@John Doe: I edited my answer to include a small discussion about the Jacobian, but I have to insist that the formula above is a consequence of the definition of the integral and has, strictly speaking, nothing to do with a Jacobian. However, given that when you define an integral you don't define it in terms of a certain parametrization, you may actually deduce the formula of change of variables(which does indeed involve the Jacobian) from this integral formula(but not the other way round).
– Sorin Tirc
Nov 19 at 13:25
Ok, thanks for the answer. It just really looked to me like the $lvert vec{r}_utimesvec{r}_vrvert$ factor is some kind of Jacobian.
– John Doe
Nov 19 at 17:01
1
1
The above formula was presented to me as a theorem. The definition was the things with "area elements". So there is no way to choose some change of variables so that the $lvert vec{r}_utimesvec{r}_urvert$ becomes the jacobian?
– John Doe
Nov 19 at 12:43
The above formula was presented to me as a theorem. The definition was the things with "area elements". So there is no way to choose some change of variables so that the $lvert vec{r}_utimesvec{r}_urvert$ becomes the jacobian?
– John Doe
Nov 19 at 12:43
2
2
@John Doe: I edited my answer to include a small discussion about the Jacobian, but I have to insist that the formula above is a consequence of the definition of the integral and has, strictly speaking, nothing to do with a Jacobian. However, given that when you define an integral you don't define it in terms of a certain parametrization, you may actually deduce the formula of change of variables(which does indeed involve the Jacobian) from this integral formula(but not the other way round).
– Sorin Tirc
Nov 19 at 13:25
@John Doe: I edited my answer to include a small discussion about the Jacobian, but I have to insist that the formula above is a consequence of the definition of the integral and has, strictly speaking, nothing to do with a Jacobian. However, given that when you define an integral you don't define it in terms of a certain parametrization, you may actually deduce the formula of change of variables(which does indeed involve the Jacobian) from this integral formula(but not the other way round).
– Sorin Tirc
Nov 19 at 13:25
Ok, thanks for the answer. It just really looked to me like the $lvert vec{r}_utimesvec{r}_vrvert$ factor is some kind of Jacobian.
– John Doe
Nov 19 at 17:01
Ok, thanks for the answer. It just really looked to me like the $lvert vec{r}_utimesvec{r}_vrvert$ factor is some kind of Jacobian.
– John Doe
Nov 19 at 17:01
add a comment |
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Every derivation I have seen of this is geometric. The $|vec r_utimesvec r_v|dA$ term is the calculation of $dS$ so strictly speaking, we are not changing variables. We are given a parametrization of a surface and calculating the surface integral based on that.
– John Douma
Nov 19 at 13:15