calculating an integral with green's theorem












0














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







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    0














    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







    share|cite|improve this question



























      0












      0








      0







      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







      share|cite|improve this question















      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 28 '18 at 19:28







      wondering1123

















      asked Nov 27 '18 at 15:17









      wondering1123wondering1123

      11011




      11011






















          2 Answers
          2






          active

          oldest

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          2














          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



          enter image description here



          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*}






          share|cite|improve this answer





















          • 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



















          1














          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.enter image description here



          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.






          share|cite|improve this answer























          • 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











          Your Answer





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          2 Answers
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          active

          oldest

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2














          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



          enter image description here



          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*}






          share|cite|improve this answer





















          • 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
















          2














          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



          enter image description here



          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*}






          share|cite|improve this answer





















          • 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














          2












          2








          2






          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



          enter image description here



          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*}






          share|cite|improve this answer












          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



          enter image description here



          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*}







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 27 '18 at 16:14









          Dikshit GautamDikshit 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


















          • 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











          1














          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.enter image description here



          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.






          share|cite|improve this answer























          • 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
















          1














          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.enter image description here



          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.






          share|cite|improve this answer























          • 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














          1












          1








          1






          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.enter image description here



          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.






          share|cite|improve this answer














          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.enter image description here



          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 28 '18 at 21:58

























          answered Nov 28 '18 at 20:10









          saulspatzsaulspatz

          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


















          • 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


















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