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Evaluating area of $frac34$ of a disc
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Evaluate $$int_gamma_1cup gamma_2xdx+x^2ydy$$
Where
$gamma_1(t)=(2cos t,2sin t),tin [-fracpi2,pi]$
$gamma_2(t)=(cos t,sin t),tin [-fracpi2,pi]$
Using Green's theorem
Visualising the curve it is a there quarters disc
So the curves will be as followed
$gamma_a(t)=(0,1+t),tin[0,1]$
$gamma_1(t)=(2cos t,2sin t),tin [-fracpi2,pi]$
$gamma_b(t)=(2-t,0),tin[0,1]$
$gamma_2(t)=(cos t,sin t),tin [-fracpi2,pi]$
Using Green's theorem
In the case of $gamma_a(t)$ and $gamma_b(t)$ both vanishes as $Q_y=2xy$ and once $x=0$ and once $y=0$ so the integrand vanishes
$$I_1=int_gamma_1xdx+x^2ydy=intint_D 2xy dx dy=int_0^2int_-fracpi2^pi(2rho cos theta *rho sin theta *rho )d theta drho=\=int_0^2int_-fracpi2^pi(2rho^3 cos thetasin theta)d theta drho=int_0^2int_-fracpi2^pi(rho^3 2cos thetasin theta)d theta drho=\=int_0^2int_-fracpi2^pi(rho^3 sin 2theta)d theta drho=int_0^2rho^3*(-fraccos2theta2|_-fracpi2^pi)drho=-int_0^2rho^3 drho=-4$$
The same happen to
$$I_2=-int_gamma_2xdx+x^2ydy=-int_0^1int_-fracpi2^pi(2rho^3 cos thetasin theta)d theta drho=int_0^1rho^3 drho=1$$
So overall we have $-4+1=-3$ which is obviously wrong, what have I missed?
Which is obviously wrong, what have I missed?
multivariable-calculus greens-theorem
asked Aug 2 at 22:46
gbox
5,30851841
 |Â
show 10 more comments
up vote
0
down vote
favorite
Evaluate $$int_gamma_1cup gamma_2xdx+x^2ydy$$
Where
$gamma_1(t)=(2cos t,2sin t),tin [-fracpi2,pi]$
$gamma_2(t)=(cos t,sin t),tin [-fracpi2,pi]$
Using Green's theorem
Visualising the curve it is a there quarters disc
So the curves will be as followed
$gamma_a(t)=(0,1+t),tin[0,1]$
$gamma_1(t)=(2cos t,2sin t),tin [-fracpi2,pi]$
$gamma_b(t)=(2-t,0),tin[0,1]$
$gamma_2(t)=(cos t,sin t),tin [-fracpi2,pi]$
Using Green's theorem
In the case of $gamma_a(t)$ and $gamma_b(t)$ both vanishes as $Q_y=2xy$ and once $x=0$ and once $y=0$ so the integrand vanishes
$$I_1=int_gamma_1xdx+x^2ydy=intint_D 2xy dx dy=int_0^2int_-fracpi2^pi(2rho cos theta *rho sin theta *rho )d theta drho=\=int_0^2int_-fracpi2^pi(2rho^3 cos thetasin theta)d theta drho=int_0^2int_-fracpi2^pi(rho^3 2cos thetasin theta)d theta drho=\=int_0^2int_-fracpi2^pi(rho^3 sin 2theta)d theta drho=int_0^2rho^3*(-fraccos2theta2|_-fracpi2^pi)drho=-int_0^2rho^3 drho=-4$$
The same happen to
$$I_2=-int_gamma_2xdx+x^2ydy=-int_0^1int_-fracpi2^pi(2rho^3 cos thetasin theta)d theta drho=int_0^1rho^3 drho=1$$
So overall we have $-4+1=-3$ which is obviously wrong, what have I missed?
Which is obviously wrong, what have I missed?
multivariable-calculus greens-theorem
asked Aug 2 at 22:46
gbox
5,30851841
Thanks! but I will still get $-frac154+1$
– gbox
Aug 3 at 7:25
What are $gamma_a, gamma_b$ doing in this question?
– xbh
Aug 3 at 7:27
I wanted to create a closed curve so I can use Green's theorem
– gbox
Aug 3 at 7:29
Using Green's theorem need to the area be closed but $gamma_1cupgamma_2$ isn't close in the main question.
– user 108128
Aug 3 at 7:29
Also what's the orientation?
– xbh
Aug 3 at 7:30
 |Â
show 10 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Evaluate $$int_gamma_1cup gamma_2xdx+x^2ydy$$
Where
$gamma_1(t)=(2cos t,2sin t),tin [-fracpi2,pi]$
$gamma_2(t)=(cos t,sin t),tin [-fracpi2,pi]$
Using Green's theorem
Visualising the curve it is a there quarters disc
So the curves will be as followed
$gamma_a(t)=(0,1+t),tin[0,1]$
$gamma_1(t)=(2cos t,2sin t),tin [-fracpi2,pi]$
$gamma_b(t)=(2-t,0),tin[0,1]$
$gamma_2(t)=(cos t,sin t),tin [-fracpi2,pi]$
Using Green's theorem
In the case of $gamma_a(t)$ and $gamma_b(t)$ both vanishes as $Q_y=2xy$ and once $x=0$ and once $y=0$ so the integrand vanishes
$$I_1=int_gamma_1xdx+x^2ydy=intint_D 2xy dx dy=int_0^2int_-fracpi2^pi(2rho cos theta *rho sin theta *rho )d theta drho=\=int_0^2int_-fracpi2^pi(2rho^3 cos thetasin theta)d theta drho=int_0^2int_-fracpi2^pi(rho^3 2cos thetasin theta)d theta drho=\=int_0^2int_-fracpi2^pi(rho^3 sin 2theta)d theta drho=int_0^2rho^3*(-fraccos2theta2|_-fracpi2^pi)drho=-int_0^2rho^3 drho=-4$$
The same happen to
$$I_2=-int_gamma_2xdx+x^2ydy=-int_0^1int_-fracpi2^pi(2rho^3 cos thetasin theta)d theta drho=int_0^1rho^3 drho=1$$
So overall we have $-4+1=-3$ which is obviously wrong, what have I missed?
Which is obviously wrong, what have I missed?
multivariable-calculus greens-theorem
asked Aug 2 at 22:46
gbox
5,30851841
Evaluate $$int_gamma_1cup gamma_2xdx+x^2ydy$$
Where
$gamma_1(t)=(2cos t,2sin t),tin [-fracpi2,pi]$
$gamma_2(t)=(cos t,sin t),tin [-fracpi2,pi]$
Using Green's theorem
Visualising the curve it is a there quarters disc
So the curves will be as followed
$gamma_a(t)=(0,1+t),tin[0,1]$
$gamma_1(t)=(2cos t,2sin t),tin [-fracpi2,pi]$
$gamma_b(t)=(2-t,0),tin[0,1]$
$gamma_2(t)=(cos t,sin t),tin [-fracpi2,pi]$
Using Green's theorem
In the case of $gamma_a(t)$ and $gamma_b(t)$ both vanishes as $Q_y=2xy$ and once $x=0$ and once $y=0$ so the integrand vanishes
$$I_1=int_gamma_1xdx+x^2ydy=intint_D 2xy dx dy=int_0^2int_-fracpi2^pi(2rho cos theta *rho sin theta *rho )d theta drho=\=int_0^2int_-fracpi2^pi(2rho^3 cos thetasin theta)d theta drho=int_0^2int_-fracpi2^pi(rho^3 2cos thetasin theta)d theta drho=\=int_0^2int_-fracpi2^pi(rho^3 sin 2theta)d theta drho=int_0^2rho^3*(-fraccos2theta2|_-fracpi2^pi)drho=-int_0^2rho^3 drho=-4$$
The same happen to
$$I_2=-int_gamma_2xdx+x^2ydy=-int_0^1int_-fracpi2^pi(2rho^3 cos thetasin theta)d theta drho=int_0^1rho^3 drho=1$$
So overall we have $-4+1=-3$ which is obviously wrong, what have I missed?
Which is obviously wrong, what have I missed?
multivariable-calculus greens-theorem
asked Aug 2 at 22:46
gbox
5,30851841
edited Aug 3 at 7:33
asked Aug 2 at 22:46
gbox
5,30851841
asked Aug 2 at 22:46
gbox
5,30851841
asked Aug 2 at 22:46
gbox
5,30851841
5,30851841
Thanks! but I will still get $-frac154+1$
– gbox
Aug 3 at 7:25
What are $gamma_a, gamma_b$ doing in this question?
– xbh
Aug 3 at 7:27
I wanted to create a closed curve so I can use Green's theorem
– gbox
Aug 3 at 7:29
Using Green's theorem need to the area be closed but $gamma_1cupgamma_2$ isn't close in the main question.
– user 108128
Aug 3 at 7:29
Also what's the orientation?
– xbh
Aug 3 at 7:30
 |Â
show 10 more comments
Thanks! but I will still get $-frac154+1$
– gbox
Aug 3 at 7:25
What are $gamma_a, gamma_b$ doing in this question?
– xbh
Aug 3 at 7:27
I wanted to create a closed curve so I can use Green's theorem
– gbox
Aug 3 at 7:29
Using Green's theorem need to the area be closed but $gamma_1cupgamma_2$ isn't close in the main question.
– user 108128
Aug 3 at 7:29
Also what's the orientation?
– xbh
Aug 3 at 7:30
Thanks! but I will still get $-frac154+1$
– gbox
Aug 3 at 7:25
Thanks! but I will still get $-frac154+1$
– gbox
Aug 3 at 7:25
What are $gamma_a, gamma_b$ doing in this question?
– xbh
Aug 3 at 7:27
What are $gamma_a, gamma_b$ doing in this question?
– xbh
Aug 3 at 7:27
I wanted to create a closed curve so I can use Green's theorem
– gbox
Aug 3 at 7:29
I wanted to create a closed curve so I can use Green's theorem
– gbox
Aug 3 at 7:29
Using Green's theorem need to the area be closed but $gamma_1cupgamma_2$ isn't close in the main question.
– user 108128
Aug 3 at 7:29
Using Green's theorem need to the area be closed but $gamma_1cupgamma_2$ isn't close in the main question.
– user 108128
Aug 3 at 7:29
Also what's the orientation?
– xbh
Aug 3 at 7:30
Also what's the orientation?
– xbh
Aug 3 at 7:30
 |Â
show 10 more comments
3 Answers
3
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up vote
1
down vote
Hint: Consider follwoing paths to close the area with positive directions
$gamma_a(t)=(0,-1-t),tin[0,1]$,
$gamma_1(t)=(2cos t,2sin t),tin [-dfracpi2,pi]$,
$gamma_b(t)=(-2+t,0),tin[0,1]$,
$gamma_2(t)=(cos t,sin t),tin [-dfracpi2,pi]$.
Now use direct way with path $gamma_a+gamma_1+gamma_b-gamma_2$. With Green's theorem
$$int_1^2int_-fracpi2^pi2 r^3 sin tcos t dt dr=colorblue-dfrac154$$
Edit: Details in direct way are
$$displaystyleint_gamma_a=int_0^10d0+0dt=0$$
$$displaystyleint_gamma_1=int_-fracpi2^pi-4sin tcos t+16sin tcos^3 t dt=-2$$
$$displaystyleint_gamma_b=int_0^1(-2+t) dt=-dfrac32$$
$$displaystyleint_gamma_2=int_-fracpi2^pi-sin tcos t+sin tcos^3 t dt=dfrac14$$
these conclude $0+(-2)+(-dfrac32)-(dfrac14)=colorblue-dfrac154$
answered Aug 3 at 7:46
user 108128
18.7k41544
$gamma_a,gamma_b$ vanishes and $gamma_1cup gamma_2$ is $int_1^2int_frac-pi2^pi sr^3 sin tcos t dt dr$?
– gbox
Aug 3 at 8:43
Nah. Only $displaystyleint_gamma_a=0$ .
– user 108128
Aug 3 at 8:45
But, we have $Pdx=x, Qdy=x^2y$ so $Q_x-P_y=2xy-0$ and on $gamma_b$, $y=0$ And how can the area be negative?
– gbox
Aug 3 at 8:54
1
with $gamma_b$, $y=0$ so it's $int_0^1 xdx=int_0^1 (-2+t)dt=-dfrac32$
– user 108128
Aug 3 at 9:15
1
@user108128 The question states $int_gamma_1cup gamma_2xdx+x^2ydy$ using Green theorem and $gamma_1cup gamma_2$ is not coled. Thus, in mine interpretation, we need to close the area, apply Green theorem and subtract the line integral added.
– gimusi
Aug 3 at 12:09
 |Â
show 13 more comments
up vote
1
down vote
Let define
- $gamma_3(t)=(-2+2t,0),tin[0,1]$
- $gamma_4(t)=(0,-2t),tin[0,1]$
- $gamma_5(t)=(-1+t,0),tin[0,1]$
- $gamma_6(t)=(0,-t),tin[0,1]$
then we have
$$int_gamma_1cup gamma_2xdx+x^2ydy=int_gamma_1cup gamma_3cup gamma_4xdx+x^2ydy+int_gamma_2cup gamma_5cup gamma_6xdx+x^2ydy-int_gamma_3xdx+x^2ydy-int_gamma_4xdx+x^2ydy-int_gamma_5xdx+x^2ydy-int_gamma_6xdx+x^2ydy$$
and by Green theorem
$$int_gamma_1cup gamma_3cup gamma_4xdx+x^2ydy+int_gamma_2cup gamma_5cup gamma_6xdx+x^2ydy=iint_D_1 2xy,dx,dy+iint_D_22xy,dx,dy$$
and by polar coordinates
$$iint_D_12xy,dx,dy=int_-pi/2^pi dtheta int_0^2 2rho^3cos theta sin theta, drho=-frac14$$
$$iint_D_22xy,dx,dy=int_-pi/2^pi dtheta int_0^1 2rho^3cos theta sin theta, drho=-4$$
and
$$int_gamma_3xdx+x^2ydy=int_0^1 2(-2+2t)dt=left[-4t+2t^2right]_0^1=-2$$
$$int_gamma_4xdx+x^2ydy=0$$
$$int_gamma_5xdx+x^2ydy=int_0^1 (-1+t)dt=left[-t+frac12 t^2right]_0^1=-frac12$$
$$int_gamma_6xdx+x^2ydy=0$$
therefore
$$int_gamma_1cup gamma_2xdx+x^2ydy=-frac14-4+2+frac12=frac-1-16+8+24=-frac74$$
which is consistent with direct evaluation for
integral 1 $int_gamma_1xdx+x^2ydy=-2$
integral 2 $int_gamma_1xdx+x^2ydy=frac14$
answered Aug 3 at 7:47
gimusi
63.8k73480
Can not we use just $gamma_a,gamma_b$ as I did? you use two curves to describe describe a line?
– gbox
Aug 3 at 10:21
@gbox I've made in that way to use Green theorem as requested, but of course we can also proceed in other equivalent ways.
– gimusi
Aug 3 at 11:59
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You are wrong in the limits of the outer integral. It should be from 1 to 2 therefore:$$int_1^2int_-fracpi2^pir^3sin2theta dtheta dr=int_1^2-r^3dr=-dfrac154$$
answered Aug 3 at 8:55
Mostafa Ayaz
8,5183630
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Hint: Consider follwoing paths to close the area with positive directions
$gamma_a(t)=(0,-1-t),tin[0,1]$,
$gamma_1(t)=(2cos t,2sin t),tin [-dfracpi2,pi]$,
$gamma_b(t)=(-2+t,0),tin[0,1]$,
$gamma_2(t)=(cos t,sin t),tin [-dfracpi2,pi]$.
Now use direct way with path $gamma_a+gamma_1+gamma_b-gamma_2$. With Green's theorem
$$int_1^2int_-fracpi2^pi2 r^3 sin tcos t dt dr=colorblue-dfrac154$$
Edit: Details in direct way are
$$displaystyleint_gamma_a=int_0^10d0+0dt=0$$
$$displaystyleint_gamma_1=int_-fracpi2^pi-4sin tcos t+16sin tcos^3 t dt=-2$$
$$displaystyleint_gamma_b=int_0^1(-2+t) dt=-dfrac32$$
$$displaystyleint_gamma_2=int_-fracpi2^pi-sin tcos t+sin tcos^3 t dt=dfrac14$$
these conclude $0+(-2)+(-dfrac32)-(dfrac14)=colorblue-dfrac154$
answered Aug 3 at 7:46
user 108128
18.7k41544
$gamma_a,gamma_b$ vanishes and $gamma_1cup gamma_2$ is $int_1^2int_frac-pi2^pi sr^3 sin tcos t dt dr$?
– gbox
Aug 3 at 8:43
Nah. Only $displaystyleint_gamma_a=0$ .
– user 108128
Aug 3 at 8:45
But, we have $Pdx=x, Qdy=x^2y$ so $Q_x-P_y=2xy-0$ and on $gamma_b$, $y=0$ And how can the area be negative?
– gbox
Aug 3 at 8:54
1
with $gamma_b$, $y=0$ so it's $int_0^1 xdx=int_0^1 (-2+t)dt=-dfrac32$
– user 108128
Aug 3 at 9:15
1
@user108128 The question states $int_gamma_1cup gamma_2xdx+x^2ydy$ using Green theorem and $gamma_1cup gamma_2$ is not coled. Thus, in mine interpretation, we need to close the area, apply Green theorem and subtract the line integral added.
– gimusi
Aug 3 at 12:09
 |Â
show 13 more comments
up vote
1
down vote
Hint: Consider follwoing paths to close the area with positive directions
$gamma_a(t)=(0,-1-t),tin[0,1]$,
$gamma_1(t)=(2cos t,2sin t),tin [-dfracpi2,pi]$,
$gamma_b(t)=(-2+t,0),tin[0,1]$,
$gamma_2(t)=(cos t,sin t),tin [-dfracpi2,pi]$.
Now use direct way with path $gamma_a+gamma_1+gamma_b-gamma_2$. With Green's theorem
$$int_1^2int_-fracpi2^pi2 r^3 sin tcos t dt dr=colorblue-dfrac154$$
Edit: Details in direct way are
$$displaystyleint_gamma_a=int_0^10d0+0dt=0$$
$$displaystyleint_gamma_1=int_-fracpi2^pi-4sin tcos t+16sin tcos^3 t dt=-2$$
$$displaystyleint_gamma_b=int_0^1(-2+t) dt=-dfrac32$$
$$displaystyleint_gamma_2=int_-fracpi2^pi-sin tcos t+sin tcos^3 t dt=dfrac14$$
these conclude $0+(-2)+(-dfrac32)-(dfrac14)=colorblue-dfrac154$
answered Aug 3 at 7:46
user 108128
18.7k41544
$gamma_a,gamma_b$ vanishes and $gamma_1cup gamma_2$ is $int_1^2int_frac-pi2^pi sr^3 sin tcos t dt dr$?
– gbox
Aug 3 at 8:43
Nah. Only $displaystyleint_gamma_a=0$ .
– user 108128
Aug 3 at 8:45
But, we have $Pdx=x, Qdy=x^2y$ so $Q_x-P_y=2xy-0$ and on $gamma_b$, $y=0$ And how can the area be negative?
– gbox
Aug 3 at 8:54
1
with $gamma_b$, $y=0$ so it's $int_0^1 xdx=int_0^1 (-2+t)dt=-dfrac32$
– user 108128
Aug 3 at 9:15
1
@user108128 The question states $int_gamma_1cup gamma_2xdx+x^2ydy$ using Green theorem and $gamma_1cup gamma_2$ is not coled. Thus, in mine interpretation, we need to close the area, apply Green theorem and subtract the line integral added.
– gimusi
Aug 3 at 12:09
 |Â
show 13 more comments
up vote
1
down vote
up vote
1
down vote
Hint: Consider follwoing paths to close the area with positive directions
$gamma_a(t)=(0,-1-t),tin[0,1]$,
$gamma_1(t)=(2cos t,2sin t),tin [-dfracpi2,pi]$,
$gamma_b(t)=(-2+t,0),tin[0,1]$,
$gamma_2(t)=(cos t,sin t),tin [-dfracpi2,pi]$.
Now use direct way with path $gamma_a+gamma_1+gamma_b-gamma_2$. With Green's theorem
$$int_1^2int_-fracpi2^pi2 r^3 sin tcos t dt dr=colorblue-dfrac154$$
Edit: Details in direct way are
$$displaystyleint_gamma_a=int_0^10d0+0dt=0$$
$$displaystyleint_gamma_1=int_-fracpi2^pi-4sin tcos t+16sin tcos^3 t dt=-2$$
$$displaystyleint_gamma_b=int_0^1(-2+t) dt=-dfrac32$$
$$displaystyleint_gamma_2=int_-fracpi2^pi-sin tcos t+sin tcos^3 t dt=dfrac14$$
these conclude $0+(-2)+(-dfrac32)-(dfrac14)=colorblue-dfrac154$
answered Aug 3 at 7:46
user 108128
18.7k41544
Hint: Consider follwoing paths to close the area with positive directions
$gamma_a(t)=(0,-1-t),tin[0,1]$,
$gamma_1(t)=(2cos t,2sin t),tin [-dfracpi2,pi]$,
$gamma_b(t)=(-2+t,0),tin[0,1]$,
$gamma_2(t)=(cos t,sin t),tin [-dfracpi2,pi]$.
Now use direct way with path $gamma_a+gamma_1+gamma_b-gamma_2$. With Green's theorem
$$int_1^2int_-fracpi2^pi2 r^3 sin tcos t dt dr=colorblue-dfrac154$$
Edit: Details in direct way are
$$displaystyleint_gamma_a=int_0^10d0+0dt=0$$
$$displaystyleint_gamma_1=int_-fracpi2^pi-4sin tcos t+16sin tcos^3 t dt=-2$$
$$displaystyleint_gamma_b=int_0^1(-2+t) dt=-dfrac32$$
$$displaystyleint_gamma_2=int_-fracpi2^pi-sin tcos t+sin tcos^3 t dt=dfrac14$$
these conclude $0+(-2)+(-dfrac32)-(dfrac14)=colorblue-dfrac154$
answered Aug 3 at 7:46
user 108128
18.7k41544
edited Aug 3 at 11:49
answered Aug 3 at 7:46
user 108128
18.7k41544
answered Aug 3 at 7:46
user 108128
18.7k41544
answered Aug 3 at 7:46
user 108128
18.7k41544
18.7k41544
$gamma_a,gamma_b$ vanishes and $gamma_1cup gamma_2$ is $int_1^2int_frac-pi2^pi sr^3 sin tcos t dt dr$?
– gbox
Aug 3 at 8:43
Nah. Only $displaystyleint_gamma_a=0$ .
– user 108128
Aug 3 at 8:45
But, we have $Pdx=x, Qdy=x^2y$ so $Q_x-P_y=2xy-0$ and on $gamma_b$, $y=0$ And how can the area be negative?
– gbox
Aug 3 at 8:54
1
with $gamma_b$, $y=0$ so it's $int_0^1 xdx=int_0^1 (-2+t)dt=-dfrac32$
– user 108128
Aug 3 at 9:15
1
@user108128 The question states $int_gamma_1cup gamma_2xdx+x^2ydy$ using Green theorem and $gamma_1cup gamma_2$ is not coled. Thus, in mine interpretation, we need to close the area, apply Green theorem and subtract the line integral added.
– gimusi
Aug 3 at 12:09
 |Â
show 13 more comments
$gamma_a,gamma_b$ vanishes and $gamma_1cup gamma_2$ is $int_1^2int_frac-pi2^pi sr^3 sin tcos t dt dr$?
– gbox
Aug 3 at 8:43
Nah. Only $displaystyleint_gamma_a=0$ .
– user 108128
Aug 3 at 8:45
But, we have $Pdx=x, Qdy=x^2y$ so $Q_x-P_y=2xy-0$ and on $gamma_b$, $y=0$ And how can the area be negative?
– gbox
Aug 3 at 8:54
1
with $gamma_b$, $y=0$ so it's $int_0^1 xdx=int_0^1 (-2+t)dt=-dfrac32$
– user 108128
Aug 3 at 9:15
1
@user108128 The question states $int_gamma_1cup gamma_2xdx+x^2ydy$ using Green theorem and $gamma_1cup gamma_2$ is not coled. Thus, in mine interpretation, we need to close the area, apply Green theorem and subtract the line integral added.
– gimusi
Aug 3 at 12:09
$gamma_a,gamma_b$ vanishes and $gamma_1cup gamma_2$ is $int_1^2int_frac-pi2^pi sr^3 sin tcos t dt dr$?
– gbox
Aug 3 at 8:43
$gamma_a,gamma_b$ vanishes and $gamma_1cup gamma_2$ is $int_1^2int_frac-pi2^pi sr^3 sin tcos t dt dr$?
– gbox
Aug 3 at 8:43
Nah. Only $displaystyleint_gamma_a=0$ .
– user 108128
Aug 3 at 8:45
Nah. Only $displaystyleint_gamma_a=0$ .
– user 108128
Aug 3 at 8:45
But, we have $Pdx=x, Qdy=x^2y$ so $Q_x-P_y=2xy-0$ and on $gamma_b$, $y=0$ And how can the area be negative?
– gbox
Aug 3 at 8:54
But, we have $Pdx=x, Qdy=x^2y$ so $Q_x-P_y=2xy-0$ and on $gamma_b$, $y=0$ And how can the area be negative?
– gbox
Aug 3 at 8:54
1
1
with $gamma_b$, $y=0$ so it's $int_0^1 xdx=int_0^1 (-2+t)dt=-dfrac32$
– user 108128
Aug 3 at 9:15
with $gamma_b$, $y=0$ so it's $int_0^1 xdx=int_0^1 (-2+t)dt=-dfrac32$
– user 108128
Aug 3 at 9:15
1
1
@user108128 The question states $int_gamma_1cup gamma_2xdx+x^2ydy$ using Green theorem and $gamma_1cup gamma_2$ is not coled. Thus, in mine interpretation, we need to close the area, apply Green theorem and subtract the line integral added.
– gimusi
Aug 3 at 12:09
@user108128 The question states $int_gamma_1cup gamma_2xdx+x^2ydy$ using Green theorem and $gamma_1cup gamma_2$ is not coled. Thus, in mine interpretation, we need to close the area, apply Green theorem and subtract the line integral added.
– gimusi
Aug 3 at 12:09
 |Â
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up vote
1
down vote
Let define
- $gamma_3(t)=(-2+2t,0),tin[0,1]$
- $gamma_4(t)=(0,-2t),tin[0,1]$
- $gamma_5(t)=(-1+t,0),tin[0,1]$
- $gamma_6(t)=(0,-t),tin[0,1]$
then we have
$$int_gamma_1cup gamma_2xdx+x^2ydy=int_gamma_1cup gamma_3cup gamma_4xdx+x^2ydy+int_gamma_2cup gamma_5cup gamma_6xdx+x^2ydy-int_gamma_3xdx+x^2ydy-int_gamma_4xdx+x^2ydy-int_gamma_5xdx+x^2ydy-int_gamma_6xdx+x^2ydy$$
and by Green theorem
$$int_gamma_1cup gamma_3cup gamma_4xdx+x^2ydy+int_gamma_2cup gamma_5cup gamma_6xdx+x^2ydy=iint_D_1 2xy,dx,dy+iint_D_22xy,dx,dy$$
and by polar coordinates
$$iint_D_12xy,dx,dy=int_-pi/2^pi dtheta int_0^2 2rho^3cos theta sin theta, drho=-frac14$$
$$iint_D_22xy,dx,dy=int_-pi/2^pi dtheta int_0^1 2rho^3cos theta sin theta, drho=-4$$
and
$$int_gamma_3xdx+x^2ydy=int_0^1 2(-2+2t)dt=left[-4t+2t^2right]_0^1=-2$$
$$int_gamma_4xdx+x^2ydy=0$$
$$int_gamma_5xdx+x^2ydy=int_0^1 (-1+t)dt=left[-t+frac12 t^2right]_0^1=-frac12$$
$$int_gamma_6xdx+x^2ydy=0$$
therefore
$$int_gamma_1cup gamma_2xdx+x^2ydy=-frac14-4+2+frac12=frac-1-16+8+24=-frac74$$
which is consistent with direct evaluation for
integral 1 $int_gamma_1xdx+x^2ydy=-2$
integral 2 $int_gamma_1xdx+x^2ydy=frac14$
answered Aug 3 at 7:47
gimusi
63.8k73480
Can not we use just $gamma_a,gamma_b$ as I did? you use two curves to describe describe a line?
– gbox
Aug 3 at 10:21
@gbox I've made in that way to use Green theorem as requested, but of course we can also proceed in other equivalent ways.
– gimusi
Aug 3 at 11:59
add a comment |Â
up vote
1
down vote
Let define
- $gamma_3(t)=(-2+2t,0),tin[0,1]$
- $gamma_4(t)=(0,-2t),tin[0,1]$
- $gamma_5(t)=(-1+t,0),tin[0,1]$
- $gamma_6(t)=(0,-t),tin[0,1]$
then we have
$$int_gamma_1cup gamma_2xdx+x^2ydy=int_gamma_1cup gamma_3cup gamma_4xdx+x^2ydy+int_gamma_2cup gamma_5cup gamma_6xdx+x^2ydy-int_gamma_3xdx+x^2ydy-int_gamma_4xdx+x^2ydy-int_gamma_5xdx+x^2ydy-int_gamma_6xdx+x^2ydy$$
and by Green theorem
$$int_gamma_1cup gamma_3cup gamma_4xdx+x^2ydy+int_gamma_2cup gamma_5cup gamma_6xdx+x^2ydy=iint_D_1 2xy,dx,dy+iint_D_22xy,dx,dy$$
and by polar coordinates
$$iint_D_12xy,dx,dy=int_-pi/2^pi dtheta int_0^2 2rho^3cos theta sin theta, drho=-frac14$$
$$iint_D_22xy,dx,dy=int_-pi/2^pi dtheta int_0^1 2rho^3cos theta sin theta, drho=-4$$
and
$$int_gamma_3xdx+x^2ydy=int_0^1 2(-2+2t)dt=left[-4t+2t^2right]_0^1=-2$$
$$int_gamma_4xdx+x^2ydy=0$$
$$int_gamma_5xdx+x^2ydy=int_0^1 (-1+t)dt=left[-t+frac12 t^2right]_0^1=-frac12$$
$$int_gamma_6xdx+x^2ydy=0$$
therefore
$$int_gamma_1cup gamma_2xdx+x^2ydy=-frac14-4+2+frac12=frac-1-16+8+24=-frac74$$
which is consistent with direct evaluation for
integral 1 $int_gamma_1xdx+x^2ydy=-2$
integral 2 $int_gamma_1xdx+x^2ydy=frac14$
answered Aug 3 at 7:47
gimusi
63.8k73480
Can not we use just $gamma_a,gamma_b$ as I did? you use two curves to describe describe a line?
– gbox
Aug 3 at 10:21
@gbox I've made in that way to use Green theorem as requested, but of course we can also proceed in other equivalent ways.
– gimusi
Aug 3 at 11:59
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let define
- $gamma_3(t)=(-2+2t,0),tin[0,1]$
- $gamma_4(t)=(0,-2t),tin[0,1]$
- $gamma_5(t)=(-1+t,0),tin[0,1]$
- $gamma_6(t)=(0,-t),tin[0,1]$
then we have
$$int_gamma_1cup gamma_2xdx+x^2ydy=int_gamma_1cup gamma_3cup gamma_4xdx+x^2ydy+int_gamma_2cup gamma_5cup gamma_6xdx+x^2ydy-int_gamma_3xdx+x^2ydy-int_gamma_4xdx+x^2ydy-int_gamma_5xdx+x^2ydy-int_gamma_6xdx+x^2ydy$$
and by Green theorem
$$int_gamma_1cup gamma_3cup gamma_4xdx+x^2ydy+int_gamma_2cup gamma_5cup gamma_6xdx+x^2ydy=iint_D_1 2xy,dx,dy+iint_D_22xy,dx,dy$$
and by polar coordinates
$$iint_D_12xy,dx,dy=int_-pi/2^pi dtheta int_0^2 2rho^3cos theta sin theta, drho=-frac14$$
$$iint_D_22xy,dx,dy=int_-pi/2^pi dtheta int_0^1 2rho^3cos theta sin theta, drho=-4$$
and
$$int_gamma_3xdx+x^2ydy=int_0^1 2(-2+2t)dt=left[-4t+2t^2right]_0^1=-2$$
$$int_gamma_4xdx+x^2ydy=0$$
$$int_gamma_5xdx+x^2ydy=int_0^1 (-1+t)dt=left[-t+frac12 t^2right]_0^1=-frac12$$
$$int_gamma_6xdx+x^2ydy=0$$
therefore
$$int_gamma_1cup gamma_2xdx+x^2ydy=-frac14-4+2+frac12=frac-1-16+8+24=-frac74$$
which is consistent with direct evaluation for
integral 1 $int_gamma_1xdx+x^2ydy=-2$
integral 2 $int_gamma_1xdx+x^2ydy=frac14$
answered Aug 3 at 7:47
gimusi
63.8k73480
Let define
- $gamma_3(t)=(-2+2t,0),tin[0,1]$
- $gamma_4(t)=(0,-2t),tin[0,1]$
- $gamma_5(t)=(-1+t,0),tin[0,1]$
- $gamma_6(t)=(0,-t),tin[0,1]$
then we have
$$int_gamma_1cup gamma_2xdx+x^2ydy=int_gamma_1cup gamma_3cup gamma_4xdx+x^2ydy+int_gamma_2cup gamma_5cup gamma_6xdx+x^2ydy-int_gamma_3xdx+x^2ydy-int_gamma_4xdx+x^2ydy-int_gamma_5xdx+x^2ydy-int_gamma_6xdx+x^2ydy$$
and by Green theorem
$$int_gamma_1cup gamma_3cup gamma_4xdx+x^2ydy+int_gamma_2cup gamma_5cup gamma_6xdx+x^2ydy=iint_D_1 2xy,dx,dy+iint_D_22xy,dx,dy$$
and by polar coordinates
$$iint_D_12xy,dx,dy=int_-pi/2^pi dtheta int_0^2 2rho^3cos theta sin theta, drho=-frac14$$
$$iint_D_22xy,dx,dy=int_-pi/2^pi dtheta int_0^1 2rho^3cos theta sin theta, drho=-4$$
and
$$int_gamma_3xdx+x^2ydy=int_0^1 2(-2+2t)dt=left[-4t+2t^2right]_0^1=-2$$
$$int_gamma_4xdx+x^2ydy=0$$
$$int_gamma_5xdx+x^2ydy=int_0^1 (-1+t)dt=left[-t+frac12 t^2right]_0^1=-frac12$$
$$int_gamma_6xdx+x^2ydy=0$$
therefore
$$int_gamma_1cup gamma_2xdx+x^2ydy=-frac14-4+2+frac12=frac-1-16+8+24=-frac74$$
which is consistent with direct evaluation for
integral 1 $int_gamma_1xdx+x^2ydy=-2$
integral 2 $int_gamma_1xdx+x^2ydy=frac14$
answered Aug 3 at 7:47
gimusi
63.8k73480
edited Aug 3 at 12:06
answered Aug 3 at 7:47
gimusi
63.8k73480
answered Aug 3 at 7:47
gimusi
63.8k73480
answered Aug 3 at 7:47
gimusi
63.8k73480
63.8k73480
Can not we use just $gamma_a,gamma_b$ as I did? you use two curves to describe describe a line?
– gbox
Aug 3 at 10:21
@gbox I've made in that way to use Green theorem as requested, but of course we can also proceed in other equivalent ways.
– gimusi
Aug 3 at 11:59
add a comment |Â
Can not we use just $gamma_a,gamma_b$ as I did? you use two curves to describe describe a line?
– gbox
Aug 3 at 10:21
@gbox I've made in that way to use Green theorem as requested, but of course we can also proceed in other equivalent ways.
– gimusi
Aug 3 at 11:59
Can not we use just $gamma_a,gamma_b$ as I did? you use two curves to describe describe a line?
– gbox
Aug 3 at 10:21
Can not we use just $gamma_a,gamma_b$ as I did? you use two curves to describe describe a line?
– gbox
Aug 3 at 10:21
@gbox I've made in that way to use Green theorem as requested, but of course we can also proceed in other equivalent ways.
– gimusi
Aug 3 at 11:59
@gbox I've made in that way to use Green theorem as requested, but of course we can also proceed in other equivalent ways.
– gimusi
Aug 3 at 11:59
add a comment |Â
up vote
0
down vote
You are wrong in the limits of the outer integral. It should be from 1 to 2 therefore:$$int_1^2int_-fracpi2^pir^3sin2theta dtheta dr=int_1^2-r^3dr=-dfrac154$$
answered Aug 3 at 8:55
Mostafa Ayaz
8,5183630
add a comment |Â
up vote
0
down vote
You are wrong in the limits of the outer integral. It should be from 1 to 2 therefore:$$int_1^2int_-fracpi2^pir^3sin2theta dtheta dr=int_1^2-r^3dr=-dfrac154$$
answered Aug 3 at 8:55
Mostafa Ayaz
8,5183630
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You are wrong in the limits of the outer integral. It should be from 1 to 2 therefore:$$int_1^2int_-fracpi2^pir^3sin2theta dtheta dr=int_1^2-r^3dr=-dfrac154$$
answered Aug 3 at 8:55
Mostafa Ayaz
8,5183630
You are wrong in the limits of the outer integral. It should be from 1 to 2 therefore:$$int_1^2int_-fracpi2^pir^3sin2theta dtheta dr=int_1^2-r^3dr=-dfrac154$$
answered Aug 3 at 8:55
Mostafa Ayaz
8,5183630
edited Aug 3 at 9:00
answered Aug 3 at 8:55
Mostafa Ayaz
8,5183630
answered Aug 3 at 8:55
Mostafa Ayaz
8,5183630
answered Aug 3 at 8:55
Mostafa Ayaz
8,5183630
8,5183630
add a comment |Â
add a comment |Â
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Thanks! but I will still get $-frac154+1$
– gbox
Aug 3 at 7:25
What are $gamma_a, gamma_b$ doing in this question?
– xbh
Aug 3 at 7:27
I wanted to create a closed curve so I can use Green's theorem
– gbox
Aug 3 at 7:29
Using Green's theorem need to the area be closed but $gamma_1cupgamma_2$ isn't close in the main question.
– user 108128
Aug 3 at 7:29
Also what's the orientation?
– xbh
Aug 3 at 7:30