calculating an integral with green's theorem
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
add a comment |
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
add a comment |
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
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
edited Nov 28 '18 at 19:28
asked Nov 27 '18 at 15:17
wondering1123
10011
10011
add a comment |
add a comment |
2 Answers
2
active
oldest
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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*}
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 '18 at 20:54
@wondering1123 The formula for area,i.e.,I(M) has been derived from Green's theorem
– Dikshit Gautam
Nov 28 '18 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 '18 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 '18 at 19:38
add a comment |
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.
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 '18 at 21:27
I'll add a few more details. Wait a bit.
– saulspatz
Nov 28 '18 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 '18 at 22:01
add a comment |
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2 Answers
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2 Answers
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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*}
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 '18 at 20:54
@wondering1123 The formula for area,i.e.,I(M) has been derived from Green's theorem
– Dikshit Gautam
Nov 28 '18 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 '18 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 '18 at 19:38
add a comment |
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*}
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 '18 at 20:54
@wondering1123 The formula for area,i.e.,I(M) has been derived from Green's theorem
– Dikshit Gautam
Nov 28 '18 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 '18 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 '18 at 19:38
add a comment |
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*}
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 '18 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 '18 at 20:54
@wondering1123 The formula for area,i.e.,I(M) has been derived from Green's theorem
– Dikshit Gautam
Nov 28 '18 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 '18 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 '18 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 '18 at 20:54
@wondering1123 The formula for area,i.e.,I(M) has been derived from Green's theorem
– Dikshit Gautam
Nov 28 '18 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 '18 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 '18 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 '18 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 '18 at 20:54
@wondering1123 The formula for area,i.e.,I(M) has been derived from Green's theorem
– Dikshit Gautam
Nov 28 '18 at 4:53
@wondering1123 The formula for area,i.e.,I(M) has been derived from Green's theorem
– Dikshit Gautam
Nov 28 '18 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 '18 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 '18 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 '18 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 '18 at 19:38
add a comment |
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.
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 '18 at 21:27
I'll add a few more details. Wait a bit.
– saulspatz
Nov 28 '18 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 '18 at 22:01
add a comment |
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.
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 '18 at 21:27
I'll add a few more details. Wait a bit.
– saulspatz
Nov 28 '18 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 '18 at 22:01
add a comment |
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.
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.
edited Nov 28 '18 at 21:58
answered Nov 28 '18 at 20:10
saulspatz
14k21329
14k21329
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 '18 at 21:27
I'll add a few more details. Wait a bit.
– saulspatz
Nov 28 '18 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 '18 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 '18 at 21:27
I'll add a few more details. Wait a bit.
– saulspatz
Nov 28 '18 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 '18 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 '18 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 '18 at 21:27
I'll add a few more details. Wait a bit.
– saulspatz
Nov 28 '18 at 21:48
I'll add a few more details. Wait a bit.
– saulspatz
Nov 28 '18 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 '18 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 '18 at 22:01
add a comment |
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