calculating an integral with green's theorem
up vote
0
down vote
favorite
Let
$ M:{ (x,y) in mathbb{R}_+^{2} : x^2+ 4y^2 leq 4, 1 leq x^2-y^2 } $
I want to use Green's Theorem:
$ I(M)= frac{1}{2} int_{partial M} x_1 dx_2 -x_2 dx_1 $
M is the region where the ellipse and hyperbole are overlapping above the x axis,
so $ int_{partial M }$ must be $ int_1^2 $ right?
now, I am not sure what to use as function I need to integrate.I could not find any similar examples.
Your help is very apreciated :-)
real-analysis integration
asked Nov 27 at 15:17
wondering1123
10011
add a comment |
up vote
0
down vote
favorite
Let
$ M:{ (x,y) in mathbb{R}_+^{2} : x^2+ 4y^2 leq 4, 1 leq x^2-y^2 } $
I want to use Green's Theorem:
$ I(M)= frac{1}{2} int_{partial M} x_1 dx_2 -x_2 dx_1 $
M is the region where the ellipse and hyperbole are overlapping above the x axis,
so $ int_{partial M }$ must be $ int_1^2 $ right?
now, I am not sure what to use as function I need to integrate.I could not find any similar examples.
Your help is very apreciated :-)
real-analysis integration
asked Nov 27 at 15:17
wondering1123
10011
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let
$ M:{ (x,y) in mathbb{R}_+^{2} : x^2+ 4y^2 leq 4, 1 leq x^2-y^2 } $
I want to use Green's Theorem:
$ I(M)= frac{1}{2} int_{partial M} x_1 dx_2 -x_2 dx_1 $
M is the region where the ellipse and hyperbole are overlapping above the x axis,
so $ int_{partial M }$ must be $ int_1^2 $ right?
now, I am not sure what to use as function I need to integrate.I could not find any similar examples.
Your help is very apreciated :-)
real-analysis integration
asked Nov 27 at 15:17
wondering1123
10011
Let
$ M:{ (x,y) in mathbb{R}_+^{2} : x^2+ 4y^2 leq 4, 1 leq x^2-y^2 } $
I want to use Green's Theorem:
$ I(M)= frac{1}{2} int_{partial M} x_1 dx_2 -x_2 dx_1 $
M is the region where the ellipse and hyperbole are overlapping above the x axis,
so $ int_{partial M }$ must be $ int_1^2 $ right?
now, I am not sure what to use as function I need to integrate.I could not find any similar examples.
Your help is very apreciated :-)
real-analysis integration
real-analysis integration
asked Nov 27 at 15:17
wondering1123
10011
asked Nov 27 at 15:17
wondering1123
10011
edited Nov 28 at 19:28
asked Nov 27 at 15:17
wondering1123
10011
asked Nov 27 at 15:17
wondering1123
10011
asked Nov 27 at 15:17
wondering1123
10011
10011
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
I think you mean $displaystyle I( M) =frac{1}{2}int _{partial M} xdy-ydx$
Taking $displaystyle I( M)$ as written above, by green's theorem this is the area of region M
Blue shaded region is the Region M.
begin{gather*}
therefore Area of M=I( M) =2int ^{frac{3}{5}}_{-frac{3}{5}}int ^{2sqrt{1-y^{2}}}_{^{sqrt{y^{2} +1}}} dxdy\
or I( M) =left[ ysqrt{1-y^{2}} -frac{1}{2}sqrt{y^{2} +1} +arcsin( y) -frac{1}{2} arcsinh( y)right]^{y=frac{3}{5}}_{y=-frac{3}{5}}\
or I( M) =frac{24}{25} -frac{3sqrt{34}}{25} +2arcsinleft(frac{3}{5}right) - arcsinhleft(frac{3}{5}right)\
or I( M) approx 0.9785
end{gather*}
answered Nov 27 at 16:14
Dikshit Gautam
795
thank you for your answere! Since $ (x,y) in mathbb{R^+}^2 $ are the bounds not quite right, right? Where does the Green's Theorem flows in ?
– wondering1123
Nov 27 at 20:54
@wondering1123 The formula for area,i.e.,I(M) has been derived from Green's theorem
– Dikshit Gautam
Nov 28 at 4:53
I am sorry, but thats the way i would normally integrate it..I thought about it and I still can not explain it with the formular $I(M)$ above. would you mind explaining it?
– wondering1123
Nov 28 at 19:27
You haven't used Green's theorem to evaluate the area. If you use Green's theorem, you will evaluate a line integral, not a double integral.
– saulspatz
Nov 28 at 19:38
add a comment |
up vote
1
down vote
Green's theorem says that $$int_{partial S}{Pdx+Qdy}=intint_S(Q_y-P_x)dxdy$$ with suitable hypotheses on $P,Q,S$. So if we set $P(x,y)=0, Q(x,y)=y,$ we see that the area of $S$ is $$int_{partial S}{y dy}$$ You simply have to parametrize the curves bounding $S$ and integrate the expression for $ydy$.
$S$ breaks into two congruent regions, so it's enough to compute the area of one of them.
The region on the right is bounded below by a segment the x-axis. If we parameterize $y$ on this axis, we will have $y=0$ of course, so the line integral will be $0$ and we can ignore it. The blue boundary curve runs from $(2,0)$ to the red dot at $(sqrt{1.6},sqrt{.4})$ and we can parameterize along this curve as $y$ as $$
y=sqrt{{4-x^2over4}}implies dy= {-xdxover 2sqrt{4-x^2}},$$ so you have to integrate ${-xover4}dx$ from $x=2$ to x=$sqrt{1.6}.$ The direction of the integral is determined because in Green's theorem, the contour of the line integral is oriented counter-clockwise.
We have $ydy={-xover8}dx,$ so we compute $$int_2^{sqrt{.6}}{-xover8}dx={-x^2over16}biggrrvert_2^{sqrt{.6}}={3.4over16}$$
Do the same thing for the arc of the hyperbola that bounds the region on the right.
answered Nov 28 at 20:10
saulspatz
13.7k21328
thanks for the detailled explanation! It is still quite hard for me to get it..and I need to integrate only for positive $x,y$
– wondering1123
Nov 28 at 21:27
I'll add a few more details. Wait a bit.
– saulspatz
Nov 28 at 21:48
I added some details. Note that the contour satisfies $xge0, yge0$ so we are considering only positive $x$ and $y$. What makes you think we aren't?
– saulspatz
Nov 28 at 22:01
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
I think you mean $displaystyle I( M) =frac{1}{2}int _{partial M} xdy-ydx$
Taking $displaystyle I( M)$ as written above, by green's theorem this is the area of region M
Blue shaded region is the Region M.
begin{gather*}
therefore Area of M=I( M) =2int ^{frac{3}{5}}_{-frac{3}{5}}int ^{2sqrt{1-y^{2}}}_{^{sqrt{y^{2} +1}}} dxdy\
or I( M) =left[ ysqrt{1-y^{2}} -frac{1}{2}sqrt{y^{2} +1} +arcsin( y) -frac{1}{2} arcsinh( y)right]^{y=frac{3}{5}}_{y=-frac{3}{5}}\
or I( M) =frac{24}{25} -frac{3sqrt{34}}{25} +2arcsinleft(frac{3}{5}right) - arcsinhleft(frac{3}{5}right)\
or I( M) approx 0.9785
end{gather*}
answered Nov 27 at 16:14
Dikshit Gautam
795
thank you for your answere! Since $ (x,y) in mathbb{R^+}^2 $ are the bounds not quite right, right? Where does the Green's Theorem flows in ?
– wondering1123
Nov 27 at 20:54
@wondering1123 The formula for area,i.e.,I(M) has been derived from Green's theorem
– Dikshit Gautam
Nov 28 at 4:53
I am sorry, but thats the way i would normally integrate it..I thought about it and I still can not explain it with the formular $I(M)$ above. would you mind explaining it?
– wondering1123
Nov 28 at 19:27
You haven't used Green's theorem to evaluate the area. If you use Green's theorem, you will evaluate a line integral, not a double integral.
– saulspatz
Nov 28 at 19:38
add a comment |
up vote
2
down vote
accepted
I think you mean $displaystyle I( M) =frac{1}{2}int _{partial M} xdy-ydx$
Taking $displaystyle I( M)$ as written above, by green's theorem this is the area of region M
Blue shaded region is the Region M.
begin{gather*}
therefore Area of M=I( M) =2int ^{frac{3}{5}}_{-frac{3}{5}}int ^{2sqrt{1-y^{2}}}_{^{sqrt{y^{2} +1}}} dxdy\
or I( M) =left[ ysqrt{1-y^{2}} -frac{1}{2}sqrt{y^{2} +1} +arcsin( y) -frac{1}{2} arcsinh( y)right]^{y=frac{3}{5}}_{y=-frac{3}{5}}\
or I( M) =frac{24}{25} -frac{3sqrt{34}}{25} +2arcsinleft(frac{3}{5}right) - arcsinhleft(frac{3}{5}right)\
or I( M) approx 0.9785
end{gather*}
answered Nov 27 at 16:14
Dikshit Gautam
795
thank you for your answere! Since $ (x,y) in mathbb{R^+}^2 $ are the bounds not quite right, right? Where does the Green's Theorem flows in ?
– wondering1123
Nov 27 at 20:54
@wondering1123 The formula for area,i.e.,I(M) has been derived from Green's theorem
– Dikshit Gautam
Nov 28 at 4:53
I am sorry, but thats the way i would normally integrate it..I thought about it and I still can not explain it with the formular $I(M)$ above. would you mind explaining it?
– wondering1123
Nov 28 at 19:27
You haven't used Green's theorem to evaluate the area. If you use Green's theorem, you will evaluate a line integral, not a double integral.
– saulspatz
Nov 28 at 19:38
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
I think you mean $displaystyle I( M) =frac{1}{2}int _{partial M} xdy-ydx$
Taking $displaystyle I( M)$ as written above, by green's theorem this is the area of region M
Blue shaded region is the Region M.
begin{gather*}
therefore Area of M=I( M) =2int ^{frac{3}{5}}_{-frac{3}{5}}int ^{2sqrt{1-y^{2}}}_{^{sqrt{y^{2} +1}}} dxdy\
or I( M) =left[ ysqrt{1-y^{2}} -frac{1}{2}sqrt{y^{2} +1} +arcsin( y) -frac{1}{2} arcsinh( y)right]^{y=frac{3}{5}}_{y=-frac{3}{5}}\
or I( M) =frac{24}{25} -frac{3sqrt{34}}{25} +2arcsinleft(frac{3}{5}right) - arcsinhleft(frac{3}{5}right)\
or I( M) approx 0.9785
end{gather*}
answered Nov 27 at 16:14
Dikshit Gautam
795
I think you mean $displaystyle I( M) =frac{1}{2}int _{partial M} xdy-ydx$
Taking $displaystyle I( M)$ as written above, by green's theorem this is the area of region M
Blue shaded region is the Region M.
begin{gather*}
therefore Area of M=I( M) =2int ^{frac{3}{5}}_{-frac{3}{5}}int ^{2sqrt{1-y^{2}}}_{^{sqrt{y^{2} +1}}} dxdy\
or I( M) =left[ ysqrt{1-y^{2}} -frac{1}{2}sqrt{y^{2} +1} +arcsin( y) -frac{1}{2} arcsinh( y)right]^{y=frac{3}{5}}_{y=-frac{3}{5}}\
or I( M) =frac{24}{25} -frac{3sqrt{34}}{25} +2arcsinleft(frac{3}{5}right) - arcsinhleft(frac{3}{5}right)\
or I( M) approx 0.9785
end{gather*}
answered Nov 27 at 16:14
Dikshit Gautam
795
answered Nov 27 at 16:14
Dikshit Gautam
795
answered Nov 27 at 16:14
Dikshit Gautam
795
answered Nov 27 at 16:14
Dikshit Gautam
795
795
thank you for your answere! Since $ (x,y) in mathbb{R^+}^2 $ are the bounds not quite right, right? Where does the Green's Theorem flows in ?
– wondering1123
Nov 27 at 20:54
@wondering1123 The formula for area,i.e.,I(M) has been derived from Green's theorem
– Dikshit Gautam
Nov 28 at 4:53
I am sorry, but thats the way i would normally integrate it..I thought about it and I still can not explain it with the formular $I(M)$ above. would you mind explaining it?
– wondering1123
Nov 28 at 19:27
You haven't used Green's theorem to evaluate the area. If you use Green's theorem, you will evaluate a line integral, not a double integral.
– saulspatz
Nov 28 at 19:38
add a comment |
thank you for your answere! Since $ (x,y) in mathbb{R^+}^2 $ are the bounds not quite right, right? Where does the Green's Theorem flows in ?
– wondering1123
Nov 27 at 20:54
@wondering1123 The formula for area,i.e.,I(M) has been derived from Green's theorem
– Dikshit Gautam
Nov 28 at 4:53
I am sorry, but thats the way i would normally integrate it..I thought about it and I still can not explain it with the formular $I(M)$ above. would you mind explaining it?
– wondering1123
Nov 28 at 19:27
You haven't used Green's theorem to evaluate the area. If you use Green's theorem, you will evaluate a line integral, not a double integral.
– saulspatz
Nov 28 at 19:38
thank you for your answere! Since $ (x,y) in mathbb{R^+}^2 $ are the bounds not quite right, right? Where does the Green's Theorem flows in ?
– wondering1123
Nov 27 at 20:54
thank you for your answere! Since $ (x,y) in mathbb{R^+}^2 $ are the bounds not quite right, right? Where does the Green's Theorem flows in ?
– wondering1123
Nov 27 at 20:54
@wondering1123 The formula for area,i.e.,I(M) has been derived from Green's theorem
– Dikshit Gautam
Nov 28 at 4:53
@wondering1123 The formula for area,i.e.,I(M) has been derived from Green's theorem
– Dikshit Gautam
Nov 28 at 4:53
I am sorry, but thats the way i would normally integrate it..I thought about it and I still can not explain it with the formular $I(M)$ above. would you mind explaining it?
– wondering1123
Nov 28 at 19:27
I am sorry, but thats the way i would normally integrate it..I thought about it and I still can not explain it with the formular $I(M)$ above. would you mind explaining it?
– wondering1123
Nov 28 at 19:27
You haven't used Green's theorem to evaluate the area. If you use Green's theorem, you will evaluate a line integral, not a double integral.
– saulspatz
Nov 28 at 19:38
You haven't used Green's theorem to evaluate the area. If you use Green's theorem, you will evaluate a line integral, not a double integral.
– saulspatz
Nov 28 at 19:38
add a comment |
up vote
1
down vote
Green's theorem says that $$int_{partial S}{Pdx+Qdy}=intint_S(Q_y-P_x)dxdy$$ with suitable hypotheses on $P,Q,S$. So if we set $P(x,y)=0, Q(x,y)=y,$ we see that the area of $S$ is $$int_{partial S}{y dy}$$ You simply have to parametrize the curves bounding $S$ and integrate the expression for $ydy$.
$S$ breaks into two congruent regions, so it's enough to compute the area of one of them.
The region on the right is bounded below by a segment the x-axis. If we parameterize $y$ on this axis, we will have $y=0$ of course, so the line integral will be $0$ and we can ignore it. The blue boundary curve runs from $(2,0)$ to the red dot at $(sqrt{1.6},sqrt{.4})$ and we can parameterize along this curve as $y$ as $$
y=sqrt{{4-x^2over4}}implies dy= {-xdxover 2sqrt{4-x^2}},$$ so you have to integrate ${-xover4}dx$ from $x=2$ to x=$sqrt{1.6}.$ The direction of the integral is determined because in Green's theorem, the contour of the line integral is oriented counter-clockwise.
We have $ydy={-xover8}dx,$ so we compute $$int_2^{sqrt{.6}}{-xover8}dx={-x^2over16}biggrrvert_2^{sqrt{.6}}={3.4over16}$$
Do the same thing for the arc of the hyperbola that bounds the region on the right.
answered Nov 28 at 20:10
saulspatz
13.7k21328
thanks for the detailled explanation! It is still quite hard for me to get it..and I need to integrate only for positive $x,y$
– wondering1123
Nov 28 at 21:27
I'll add a few more details. Wait a bit.
– saulspatz
Nov 28 at 21:48
I added some details. Note that the contour satisfies $xge0, yge0$ so we are considering only positive $x$ and $y$. What makes you think we aren't?
– saulspatz
Nov 28 at 22:01
add a comment |
up vote
1
down vote
Green's theorem says that $$int_{partial S}{Pdx+Qdy}=intint_S(Q_y-P_x)dxdy$$ with suitable hypotheses on $P,Q,S$. So if we set $P(x,y)=0, Q(x,y)=y,$ we see that the area of $S$ is $$int_{partial S}{y dy}$$ You simply have to parametrize the curves bounding $S$ and integrate the expression for $ydy$.
$S$ breaks into two congruent regions, so it's enough to compute the area of one of them.
The region on the right is bounded below by a segment the x-axis. If we parameterize $y$ on this axis, we will have $y=0$ of course, so the line integral will be $0$ and we can ignore it. The blue boundary curve runs from $(2,0)$ to the red dot at $(sqrt{1.6},sqrt{.4})$ and we can parameterize along this curve as $y$ as $$
y=sqrt{{4-x^2over4}}implies dy= {-xdxover 2sqrt{4-x^2}},$$ so you have to integrate ${-xover4}dx$ from $x=2$ to x=$sqrt{1.6}.$ The direction of the integral is determined because in Green's theorem, the contour of the line integral is oriented counter-clockwise.
We have $ydy={-xover8}dx,$ so we compute $$int_2^{sqrt{.6}}{-xover8}dx={-x^2over16}biggrrvert_2^{sqrt{.6}}={3.4over16}$$
Do the same thing for the arc of the hyperbola that bounds the region on the right.
answered Nov 28 at 20:10
saulspatz
13.7k21328
thanks for the detailled explanation! It is still quite hard for me to get it..and I need to integrate only for positive $x,y$
– wondering1123
Nov 28 at 21:27
I'll add a few more details. Wait a bit.
– saulspatz
Nov 28 at 21:48
I added some details. Note that the contour satisfies $xge0, yge0$ so we are considering only positive $x$ and $y$. What makes you think we aren't?
– saulspatz
Nov 28 at 22:01
add a comment |
up vote
1
down vote
up vote
1
down vote
Green's theorem says that $$int_{partial S}{Pdx+Qdy}=intint_S(Q_y-P_x)dxdy$$ with suitable hypotheses on $P,Q,S$. So if we set $P(x,y)=0, Q(x,y)=y,$ we see that the area of $S$ is $$int_{partial S}{y dy}$$ You simply have to parametrize the curves bounding $S$ and integrate the expression for $ydy$.
$S$ breaks into two congruent regions, so it's enough to compute the area of one of them.
The region on the right is bounded below by a segment the x-axis. If we parameterize $y$ on this axis, we will have $y=0$ of course, so the line integral will be $0$ and we can ignore it. The blue boundary curve runs from $(2,0)$ to the red dot at $(sqrt{1.6},sqrt{.4})$ and we can parameterize along this curve as $y$ as $$
y=sqrt{{4-x^2over4}}implies dy= {-xdxover 2sqrt{4-x^2}},$$ so you have to integrate ${-xover4}dx$ from $x=2$ to x=$sqrt{1.6}.$ The direction of the integral is determined because in Green's theorem, the contour of the line integral is oriented counter-clockwise.
We have $ydy={-xover8}dx,$ so we compute $$int_2^{sqrt{.6}}{-xover8}dx={-x^2over16}biggrrvert_2^{sqrt{.6}}={3.4over16}$$
Do the same thing for the arc of the hyperbola that bounds the region on the right.
answered Nov 28 at 20:10
saulspatz
13.7k21328
Green's theorem says that $$int_{partial S}{Pdx+Qdy}=intint_S(Q_y-P_x)dxdy$$ with suitable hypotheses on $P,Q,S$. So if we set $P(x,y)=0, Q(x,y)=y,$ we see that the area of $S$ is $$int_{partial S}{y dy}$$ You simply have to parametrize the curves bounding $S$ and integrate the expression for $ydy$.
$S$ breaks into two congruent regions, so it's enough to compute the area of one of them.
The region on the right is bounded below by a segment the x-axis. If we parameterize $y$ on this axis, we will have $y=0$ of course, so the line integral will be $0$ and we can ignore it. The blue boundary curve runs from $(2,0)$ to the red dot at $(sqrt{1.6},sqrt{.4})$ and we can parameterize along this curve as $y$ as $$
y=sqrt{{4-x^2over4}}implies dy= {-xdxover 2sqrt{4-x^2}},$$ so you have to integrate ${-xover4}dx$ from $x=2$ to x=$sqrt{1.6}.$ The direction of the integral is determined because in Green's theorem, the contour of the line integral is oriented counter-clockwise.
We have $ydy={-xover8}dx,$ so we compute $$int_2^{sqrt{.6}}{-xover8}dx={-x^2over16}biggrrvert_2^{sqrt{.6}}={3.4over16}$$
Do the same thing for the arc of the hyperbola that bounds the region on the right.
answered Nov 28 at 20:10
saulspatz
13.7k21328
edited Nov 28 at 21:58
answered Nov 28 at 20:10
saulspatz
13.7k21328
answered Nov 28 at 20:10
saulspatz
13.7k21328
answered Nov 28 at 20:10
saulspatz
13.7k21328
13.7k21328
thanks for the detailled explanation! It is still quite hard for me to get it..and I need to integrate only for positive $x,y$
– wondering1123
Nov 28 at 21:27
I'll add a few more details. Wait a bit.
– saulspatz
Nov 28 at 21:48
I added some details. Note that the contour satisfies $xge0, yge0$ so we are considering only positive $x$ and $y$. What makes you think we aren't?
– saulspatz
Nov 28 at 22:01
add a comment |
thanks for the detailled explanation! It is still quite hard for me to get it..and I need to integrate only for positive $x,y$
– wondering1123
Nov 28 at 21:27
I'll add a few more details. Wait a bit.
– saulspatz
Nov 28 at 21:48
I added some details. Note that the contour satisfies $xge0, yge0$ so we are considering only positive $x$ and $y$. What makes you think we aren't?
– saulspatz
Nov 28 at 22:01
thanks for the detailled explanation! It is still quite hard for me to get it..and I need to integrate only for positive $x,y$
– wondering1123
Nov 28 at 21:27
thanks for the detailled explanation! It is still quite hard for me to get it..and I need to integrate only for positive $x,y$
– wondering1123
Nov 28 at 21:27
I'll add a few more details. Wait a bit.
– saulspatz
Nov 28 at 21:48
I'll add a few more details. Wait a bit.
– saulspatz
Nov 28 at 21:48
I added some details. Note that the contour satisfies $xge0, yge0$ so we are considering only positive $x$ and $y$. What makes you think we aren't?
– saulspatz
Nov 28 at 22:01
I added some details. Note that the contour satisfies $xge0, yge0$ so we are considering only positive $x$ and $y$. What makes you think we aren't?
– saulspatz
Nov 28 at 22:01
add a comment |
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