Mixing
![determinant rule proof [closed]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhZCTB4-bb-s-32SAm6ytjzFOU06cH9o8q-a_wq_UnH6lcd8hvws23CmBWHM6g256LzTX9yK1EiCCzTC1NSgtxlNk_J9hdr9bA3tD0Nqntbl521j4bLAm_uXPANuWLBhcCKTKzns2gaUodx/s72-c/1.jpg)
Integral of differential forms $int (-y+sin x^2)dx + xdy$.
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
I should calculate
$$int (-y+sin x^2)dx + xdy$$
on the curve $c=c_1+c_2-c_3-c_4$ where it doesn't give me any parametrisation mappings; only the normal
$$
begincases
c_1:& [0,1] rightarrow R^2, c_1(t)= (t,0) ,\
c_2:& [0,1] rightarrow R^2, c_2(t)= (1,t) ,\
c_3:& [0,1] rightarrow R^2, c_3(t)= (t,1) ,\
c_4:& [0,1] rightarrow R^2, c_4(t)= (0,t) .
endcases
$$
should I just use them, for example, for $c_1$: $int^1_0 sin(t^2)dt $ or am I missing something here?
integration multivariable-calculus differential-forms greens-theorem
edited Aug 1 at 20:32
user 108128
18.9k41544
asked Aug 1 at 10:38
Mohamed Mossad
102
add a comment |Â
up vote
0
down vote
favorite
I should calculate
$$int (-y+sin x^2)dx + xdy$$
on the curve $c=c_1+c_2-c_3-c_4$ where it doesn't give me any parametrisation mappings; only the normal
$$
begincases
c_1:& [0,1] rightarrow R^2, c_1(t)= (t,0) ,\
c_2:& [0,1] rightarrow R^2, c_2(t)= (1,t) ,\
c_3:& [0,1] rightarrow R^2, c_3(t)= (t,1) ,\
c_4:& [0,1] rightarrow R^2, c_4(t)= (0,t) .
endcases
$$
should I just use them, for example, for $c_1$: $int^1_0 sin(t^2)dt $ or am I missing something here?
integration multivariable-calculus differential-forms greens-theorem
edited Aug 1 at 20:32
user 108128
18.9k41544
asked Aug 1 at 10:38
Mohamed Mossad
102
1
Why "the normal" ? you want $int_c$ so do as you said. What about green's theorem!
– user 108128
Aug 1 at 11:22
I meant we normally get that curve c for a singular 2-cube but in addition, we get a parametrizing mapping, for example, $(t_1,t_2)$ $rightarrow$ $(t_1^2, t_1t_2, t_2^2)$ which we use to do the integral. So should I just calculate the integral above? because we're not supposed to know how to solve such integrand. Also, what about Green's theorem?
– Mohamed Mossad
Aug 1 at 12:44
1
This smells like Green
– Kuifje
Aug 1 at 13:22
Could you elaborate more, please?
– Mohamed Mossad
Aug 1 at 15:58
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I should calculate
$$int (-y+sin x^2)dx + xdy$$
on the curve $c=c_1+c_2-c_3-c_4$ where it doesn't give me any parametrisation mappings; only the normal
$$
begincases
c_1:& [0,1] rightarrow R^2, c_1(t)= (t,0) ,\
c_2:& [0,1] rightarrow R^2, c_2(t)= (1,t) ,\
c_3:& [0,1] rightarrow R^2, c_3(t)= (t,1) ,\
c_4:& [0,1] rightarrow R^2, c_4(t)= (0,t) .
endcases
$$
should I just use them, for example, for $c_1$: $int^1_0 sin(t^2)dt $ or am I missing something here?
integration multivariable-calculus differential-forms greens-theorem
edited Aug 1 at 20:32
user 108128
18.9k41544
asked Aug 1 at 10:38
Mohamed Mossad
102
I should calculate
$$int (-y+sin x^2)dx + xdy$$
on the curve $c=c_1+c_2-c_3-c_4$ where it doesn't give me any parametrisation mappings; only the normal
$$
begincases
c_1:& [0,1] rightarrow R^2, c_1(t)= (t,0) ,\
c_2:& [0,1] rightarrow R^2, c_2(t)= (1,t) ,\
c_3:& [0,1] rightarrow R^2, c_3(t)= (t,1) ,\
c_4:& [0,1] rightarrow R^2, c_4(t)= (0,t) .
endcases
$$
should I just use them, for example, for $c_1$: $int^1_0 sin(t^2)dt $ or am I missing something here?
integration multivariable-calculus differential-forms greens-theorem
edited Aug 1 at 20:32
user 108128
18.9k41544
asked Aug 1 at 10:38
Mohamed Mossad
102
edited Aug 1 at 20:32
user 108128
18.9k41544
edited Aug 1 at 20:32
user 108128
18.9k41544
edited Aug 1 at 20:32
user 108128
18.9k41544
18.9k41544
asked Aug 1 at 10:38
Mohamed Mossad
102
asked Aug 1 at 10:38
Mohamed Mossad
102
asked Aug 1 at 10:38
Mohamed Mossad
102
102
1
Why "the normal" ? you want $int_c$ so do as you said. What about green's theorem!
– user 108128
Aug 1 at 11:22
I meant we normally get that curve c for a singular 2-cube but in addition, we get a parametrizing mapping, for example, $(t_1,t_2)$ $rightarrow$ $(t_1^2, t_1t_2, t_2^2)$ which we use to do the integral. So should I just calculate the integral above? because we're not supposed to know how to solve such integrand. Also, what about Green's theorem?
– Mohamed Mossad
Aug 1 at 12:44
1
This smells like Green
– Kuifje
Aug 1 at 13:22
Could you elaborate more, please?
– Mohamed Mossad
Aug 1 at 15:58
add a comment |Â
1
Why "the normal" ? you want $int_c$ so do as you said. What about green's theorem!
– user 108128
Aug 1 at 11:22
I meant we normally get that curve c for a singular 2-cube but in addition, we get a parametrizing mapping, for example, $(t_1,t_2)$ $rightarrow$ $(t_1^2, t_1t_2, t_2^2)$ which we use to do the integral. So should I just calculate the integral above? because we're not supposed to know how to solve such integrand. Also, what about Green's theorem?
– Mohamed Mossad
Aug 1 at 12:44
1
This smells like Green
– Kuifje
Aug 1 at 13:22
Could you elaborate more, please?
– Mohamed Mossad
Aug 1 at 15:58
1
1
Why "the normal" ? you want $int_c$ so do as you said. What about green's theorem!
– user 108128
Aug 1 at 11:22
Why "the normal" ? you want $int_c$ so do as you said. What about green's theorem!
– user 108128
Aug 1 at 11:22
I meant we normally get that curve c for a singular 2-cube but in addition, we get a parametrizing mapping, for example, $(t_1,t_2)$ $rightarrow$ $(t_1^2, t_1t_2, t_2^2)$ which we use to do the integral. So should I just calculate the integral above? because we're not supposed to know how to solve such integrand. Also, what about Green's theorem?
– Mohamed Mossad
Aug 1 at 12:44
I meant we normally get that curve c for a singular 2-cube but in addition, we get a parametrizing mapping, for example, $(t_1,t_2)$ $rightarrow$ $(t_1^2, t_1t_2, t_2^2)$ which we use to do the integral. So should I just calculate the integral above? because we're not supposed to know how to solve such integrand. Also, what about Green's theorem?
– Mohamed Mossad
Aug 1 at 12:44
1
1
This smells like Green
– Kuifje
Aug 1 at 13:22
This smells like Green
– Kuifje
Aug 1 at 13:22
Could you elaborate more, please?
– Mohamed Mossad
Aug 1 at 15:58
Could you elaborate more, please?
– Mohamed Mossad
Aug 1 at 15:58
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
With direct way
$$int_c(-y+sin x^2)dx + xdy=int_c_1sin t^2dt + int_c_2dt - int_c_3(-1+sin t^2)dt - int_c_40dt=2$$
where all integrals are on $[0,1]$.
With Green's theorem
$$oint_partial D(P dx+Q dy)=iint_D left ( fracpartial Qpartial x - fracpartial Ppartial yright) dx dy$$
where $D$ is the area of the region, that is $=1$, then
$$int_c(-y+sin x^2)dx + xdy=int_D 2 dx dy=2$$
answered Aug 1 at 20:27
user 108128
18.9k41544
Thank you, it appears to be quite easy to calculate using Green's theorem.
– Mohamed Mossad
Aug 2 at 13:28
Of course green is better here and you are welcome.
– user 108128
Aug 2 at 13:32
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
With direct way
$$int_c(-y+sin x^2)dx + xdy=int_c_1sin t^2dt + int_c_2dt - int_c_3(-1+sin t^2)dt - int_c_40dt=2$$
where all integrals are on $[0,1]$.
With Green's theorem
$$oint_partial D(P dx+Q dy)=iint_D left ( fracpartial Qpartial x - fracpartial Ppartial yright) dx dy$$
where $D$ is the area of the region, that is $=1$, then
$$int_c(-y+sin x^2)dx + xdy=int_D 2 dx dy=2$$
answered Aug 1 at 20:27
user 108128
18.9k41544
Thank you, it appears to be quite easy to calculate using Green's theorem.
– Mohamed Mossad
Aug 2 at 13:28
Of course green is better here and you are welcome.
– user 108128
Aug 2 at 13:32
add a comment |Â
up vote
1
down vote
accepted
With direct way
$$int_c(-y+sin x^2)dx + xdy=int_c_1sin t^2dt + int_c_2dt - int_c_3(-1+sin t^2)dt - int_c_40dt=2$$
where all integrals are on $[0,1]$.
With Green's theorem
$$oint_partial D(P dx+Q dy)=iint_D left ( fracpartial Qpartial x - fracpartial Ppartial yright) dx dy$$
where $D$ is the area of the region, that is $=1$, then
$$int_c(-y+sin x^2)dx + xdy=int_D 2 dx dy=2$$
answered Aug 1 at 20:27
user 108128
18.9k41544
Thank you, it appears to be quite easy to calculate using Green's theorem.
– Mohamed Mossad
Aug 2 at 13:28
Of course green is better here and you are welcome.
– user 108128
Aug 2 at 13:32
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
With direct way
$$int_c(-y+sin x^2)dx + xdy=int_c_1sin t^2dt + int_c_2dt - int_c_3(-1+sin t^2)dt - int_c_40dt=2$$
where all integrals are on $[0,1]$.
With Green's theorem
$$oint_partial D(P dx+Q dy)=iint_D left ( fracpartial Qpartial x - fracpartial Ppartial yright) dx dy$$
where $D$ is the area of the region, that is $=1$, then
$$int_c(-y+sin x^2)dx + xdy=int_D 2 dx dy=2$$
answered Aug 1 at 20:27
user 108128
18.9k41544
With direct way
$$int_c(-y+sin x^2)dx + xdy=int_c_1sin t^2dt + int_c_2dt - int_c_3(-1+sin t^2)dt - int_c_40dt=2$$
where all integrals are on $[0,1]$.
With Green's theorem
$$oint_partial D(P dx+Q dy)=iint_D left ( fracpartial Qpartial x - fracpartial Ppartial yright) dx dy$$
where $D$ is the area of the region, that is $=1$, then
$$int_c(-y+sin x^2)dx + xdy=int_D 2 dx dy=2$$
answered Aug 1 at 20:27
user 108128
18.9k41544
answered Aug 1 at 20:27
user 108128
18.9k41544
answered Aug 1 at 20:27
user 108128
18.9k41544
answered Aug 1 at 20:27
user 108128
18.9k41544
18.9k41544
Thank you, it appears to be quite easy to calculate using Green's theorem.
– Mohamed Mossad
Aug 2 at 13:28
Of course green is better here and you are welcome.
– user 108128
Aug 2 at 13:32
add a comment |Â
Thank you, it appears to be quite easy to calculate using Green's theorem.
– Mohamed Mossad
Aug 2 at 13:28
Of course green is better here and you are welcome.
– user 108128
Aug 2 at 13:32
Thank you, it appears to be quite easy to calculate using Green's theorem.
– Mohamed Mossad
Aug 2 at 13:28
Thank you, it appears to be quite easy to calculate using Green's theorem.
– Mohamed Mossad
Aug 2 at 13:28
Of course green is better here and you are welcome.
– user 108128
Aug 2 at 13:32
Of course green is better here and you are welcome.
– user 108128
Aug 2 at 13:32
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2868937%2fintegral-of-differential-forms-int-y-sin-x2dx-xdy%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
Why "the normal" ? you want $int_c$ so do as you said. What about green's theorem!
– user 108128
Aug 1 at 11:22
I meant we normally get that curve c for a singular 2-cube but in addition, we get a parametrizing mapping, for example, $(t_1,t_2)$ $rightarrow$ $(t_1^2, t_1t_2, t_2^2)$ which we use to do the integral. So should I just calculate the integral above? because we're not supposed to know how to solve such integrand. Also, what about Green's theorem?
– Mohamed Mossad
Aug 1 at 12:44
1
This smells like Green
– Kuifje
Aug 1 at 13:22
Could you elaborate more, please?
– Mohamed Mossad
Aug 1 at 15:58