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Some questions regarding a formula that's supposed to be the integral formula of Gauss and Green.
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According to my professor, this formula is called the integral formula of Gauss and Green. I've tried searching for it online but I could find nothing similar to it.
B here is a region that's bounded by the simple closed curve γ(t).
I don't really understand what is it that we get after applying this formula. I mean, when we calculate the definite integral of f over the Intervall [a,b], we get the area under the curve of f from x=a to x=b. And double integrals are used either to calculate the area of a region or the volume above it that's bounded by the surface of a multi-variable function. But what is it that we calculate using this formula?
This formula is supposed to be something slimier to the fundamental theorem of calculus but in multi-variable calculus. I cannot see how.
Thanks for all your help. And forgive me for the grammatical mistakes, if there is any. English isn't my first language.
calculus analysis multivariable-calculus greens-theorem
asked Jul 24 at 22:39
Ahmad Al-Mekhlafy
1
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up vote
0
down vote
favorite
According to my professor, this formula is called the integral formula of Gauss and Green. I've tried searching for it online but I could find nothing similar to it.
B here is a region that's bounded by the simple closed curve γ(t).
I don't really understand what is it that we get after applying this formula. I mean, when we calculate the definite integral of f over the Intervall [a,b], we get the area under the curve of f from x=a to x=b. And double integrals are used either to calculate the area of a region or the volume above it that's bounded by the surface of a multi-variable function. But what is it that we calculate using this formula?
This formula is supposed to be something slimier to the fundamental theorem of calculus but in multi-variable calculus. I cannot see how.
Thanks for all your help. And forgive me for the grammatical mistakes, if there is any. English isn't my first language.
calculus analysis multivariable-calculus greens-theorem
asked Jul 24 at 22:39
Ahmad Al-Mekhlafy
1
It is the green's theorem. Maybe you can notice that for a two by two matrix $A=(a|b)$ with columns $a,b$ that $det(a|b) = acdot b^perp$, where $b^perp$ Is $b$ clockwise rotated by 90 degrees
– Calvin Khor
Jul 24 at 22:54
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
According to my professor, this formula is called the integral formula of Gauss and Green. I've tried searching for it online but I could find nothing similar to it.
B here is a region that's bounded by the simple closed curve γ(t).
I don't really understand what is it that we get after applying this formula. I mean, when we calculate the definite integral of f over the Intervall [a,b], we get the area under the curve of f from x=a to x=b. And double integrals are used either to calculate the area of a region or the volume above it that's bounded by the surface of a multi-variable function. But what is it that we calculate using this formula?
This formula is supposed to be something slimier to the fundamental theorem of calculus but in multi-variable calculus. I cannot see how.
Thanks for all your help. And forgive me for the grammatical mistakes, if there is any. English isn't my first language.
calculus analysis multivariable-calculus greens-theorem
asked Jul 24 at 22:39
Ahmad Al-Mekhlafy
1
According to my professor, this formula is called the integral formula of Gauss and Green. I've tried searching for it online but I could find nothing similar to it.
B here is a region that's bounded by the simple closed curve γ(t).
I don't really understand what is it that we get after applying this formula. I mean, when we calculate the definite integral of f over the Intervall [a,b], we get the area under the curve of f from x=a to x=b. And double integrals are used either to calculate the area of a region or the volume above it that's bounded by the surface of a multi-variable function. But what is it that we calculate using this formula?
This formula is supposed to be something slimier to the fundamental theorem of calculus but in multi-variable calculus. I cannot see how.
Thanks for all your help. And forgive me for the grammatical mistakes, if there is any. English isn't my first language.
calculus analysis multivariable-calculus greens-theorem
asked Jul 24 at 22:39
Ahmad Al-Mekhlafy
1
asked Jul 24 at 22:39
Ahmad Al-Mekhlafy
1
asked Jul 24 at 22:39
Ahmad Al-Mekhlafy
1
asked Jul 24 at 22:39
Ahmad Al-Mekhlafy
1
1
It is the green's theorem. Maybe you can notice that for a two by two matrix $A=(a|b)$ with columns $a,b$ that $det(a|b) = acdot b^perp$, where $b^perp$ Is $b$ clockwise rotated by 90 degrees
– Calvin Khor
Jul 24 at 22:54
add a comment |Â
It is the green's theorem. Maybe you can notice that for a two by two matrix $A=(a|b)$ with columns $a,b$ that $det(a|b) = acdot b^perp$, where $b^perp$ Is $b$ clockwise rotated by 90 degrees
– Calvin Khor
Jul 24 at 22:54
It is the green's theorem. Maybe you can notice that for a two by two matrix $A=(a|b)$ with columns $a,b$ that $det(a|b) = acdot b^perp$, where $b^perp$ Is $b$ clockwise rotated by 90 degrees
– Calvin Khor
Jul 24 at 22:54
It is the green's theorem. Maybe you can notice that for a two by two matrix $A=(a|b)$ with columns $a,b$ that $det(a|b) = acdot b^perp$, where $b^perp$ Is $b$ clockwise rotated by 90 degrees
– Calvin Khor
Jul 24 at 22:54
add a comment |Â
1 Answer
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Rewrite the right-hand side as a line integral $$
int_a^bdetpmatrixf_1&gamma_1'\f_2&gamma_2'mathrmdt=
int_A^b f_1gamma_2'-f_2gamma_1'mathrmdt=int_partial Bf_1mathrmdy-f_2mathrmdx
$$
where $partial B$ is the positively-oriented boundary of $B,$ parameterized as $(gamma_1(t),gamma_2(t)),a leq tleq b$
Now it's somewhat reminiscent of the fundamental theorem of calculus isn't it? On the right hand side, we're integrating some function over the boundary. On the left-hand side we're integrating some kind of derivative of that function over the whole area.
This idea can be made precise and generalized to any number of dimensions and in other ways, but that takes a lot of work.
Integrals have many more interpretations and uses than simply areas or volumes. If they didn't, they wouldn't be nearly as important as they are.
answered Jul 24 at 23:37
saulspatz
10.4k21323
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Rewrite the right-hand side as a line integral $$
int_a^bdetpmatrixf_1&gamma_1'\f_2&gamma_2'mathrmdt=
int_A^b f_1gamma_2'-f_2gamma_1'mathrmdt=int_partial Bf_1mathrmdy-f_2mathrmdx
$$
where $partial B$ is the positively-oriented boundary of $B,$ parameterized as $(gamma_1(t),gamma_2(t)),a leq tleq b$
Now it's somewhat reminiscent of the fundamental theorem of calculus isn't it? On the right hand side, we're integrating some function over the boundary. On the left-hand side we're integrating some kind of derivative of that function over the whole area.
This idea can be made precise and generalized to any number of dimensions and in other ways, but that takes a lot of work.
Integrals have many more interpretations and uses than simply areas or volumes. If they didn't, they wouldn't be nearly as important as they are.
answered Jul 24 at 23:37
saulspatz
10.4k21323
add a comment |Â
up vote
0
down vote
Rewrite the right-hand side as a line integral $$
int_a^bdetpmatrixf_1&gamma_1'\f_2&gamma_2'mathrmdt=
int_A^b f_1gamma_2'-f_2gamma_1'mathrmdt=int_partial Bf_1mathrmdy-f_2mathrmdx
$$
where $partial B$ is the positively-oriented boundary of $B,$ parameterized as $(gamma_1(t),gamma_2(t)),a leq tleq b$
Now it's somewhat reminiscent of the fundamental theorem of calculus isn't it? On the right hand side, we're integrating some function over the boundary. On the left-hand side we're integrating some kind of derivative of that function over the whole area.
This idea can be made precise and generalized to any number of dimensions and in other ways, but that takes a lot of work.
Integrals have many more interpretations and uses than simply areas or volumes. If they didn't, they wouldn't be nearly as important as they are.
answered Jul 24 at 23:37
saulspatz
10.4k21323
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Rewrite the right-hand side as a line integral $$
int_a^bdetpmatrixf_1&gamma_1'\f_2&gamma_2'mathrmdt=
int_A^b f_1gamma_2'-f_2gamma_1'mathrmdt=int_partial Bf_1mathrmdy-f_2mathrmdx
$$
where $partial B$ is the positively-oriented boundary of $B,$ parameterized as $(gamma_1(t),gamma_2(t)),a leq tleq b$
Now it's somewhat reminiscent of the fundamental theorem of calculus isn't it? On the right hand side, we're integrating some function over the boundary. On the left-hand side we're integrating some kind of derivative of that function over the whole area.
This idea can be made precise and generalized to any number of dimensions and in other ways, but that takes a lot of work.
Integrals have many more interpretations and uses than simply areas or volumes. If they didn't, they wouldn't be nearly as important as they are.
answered Jul 24 at 23:37
saulspatz
10.4k21323
Rewrite the right-hand side as a line integral $$
int_a^bdetpmatrixf_1&gamma_1'\f_2&gamma_2'mathrmdt=
int_A^b f_1gamma_2'-f_2gamma_1'mathrmdt=int_partial Bf_1mathrmdy-f_2mathrmdx
$$
where $partial B$ is the positively-oriented boundary of $B,$ parameterized as $(gamma_1(t),gamma_2(t)),a leq tleq b$
Now it's somewhat reminiscent of the fundamental theorem of calculus isn't it? On the right hand side, we're integrating some function over the boundary. On the left-hand side we're integrating some kind of derivative of that function over the whole area.
This idea can be made precise and generalized to any number of dimensions and in other ways, but that takes a lot of work.
Integrals have many more interpretations and uses than simply areas or volumes. If they didn't, they wouldn't be nearly as important as they are.
answered Jul 24 at 23:37
saulspatz
10.4k21323
answered Jul 24 at 23:37
saulspatz
10.4k21323
answered Jul 24 at 23:37
saulspatz
10.4k21323
answered Jul 24 at 23:37
saulspatz
10.4k21323
10.4k21323
add a comment |Â
add a comment |Â
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It is the green's theorem. Maybe you can notice that for a two by two matrix $A=(a|b)$ with columns $a,b$ that $det(a|b) = acdot b^perp$, where $b^perp$ Is $b$ clockwise rotated by 90 degrees
– Calvin Khor
Jul 24 at 22:54