How can this multiple integral be evaluated?












3












$begingroup$


I am stuck trying to solve the following integral:



$$int_R (y+2x^2)(y-x^2) dA$$ where $R$ is defined by the following equations: $xy=1$, $xy=2$, $y=x^2$, $y=x^2-1$ with $x$ and $y$ positives.



I've tried several changes of variables for example: $u=xy$, $v=y-x^2$ or $u=y-x^2$, $v=x^2$ but I get stuck because for the Jacobian ot for the limits of integration I have to solve a third degree equation. I know that I could solve it using Cardano's formula but it has to be an easy way to do it.



Thank you very much.
Merry Christmas.






integration multivariable-calculus change-of-variable







share|cite|improve this question











$endgroup$












  • $begingroup$
    On behalf of @math15: It seems that there is an error in the integral. Are you sure that it is $(y+2x)$ and not $(y+2x^2)$ ?
    $endgroup$
    – dantopa
    Dec 26 '18 at 10:05










  • $begingroup$
    Yes sorry, you are right!. I've already corrected it
    $endgroup$
    – Alonso Quijano
    Dec 26 '18 at 23:00








  • 3




    $begingroup$
    The first idea, $u=xy$ and $v=y-x^2$, looks good with $frac{partial (u,v)}{partial (x,y)}=y+2x^2$ and therefore $frac{partial (x,y)}{partial (u,v)}=frac 1{y+2x^2}$.
    $endgroup$
    – random
    Dec 27 '18 at 10:02










  • $begingroup$
    Thank you very much! I've checked the inverse function theorem and yes, it works!
    $endgroup$
    – Alonso Quijano
    Dec 27 '18 at 12:55
















3












$begingroup$


I am stuck trying to solve the following integral:



$$int_R (y+2x^2)(y-x^2) dA$$ where $R$ is defined by the following equations: $xy=1$, $xy=2$, $y=x^2$, $y=x^2-1$ with $x$ and $y$ positives.



I've tried several changes of variables for example: $u=xy$, $v=y-x^2$ or $u=y-x^2$, $v=x^2$ but I get stuck because for the Jacobian ot for the limits of integration I have to solve a third degree equation. I know that I could solve it using Cardano's formula but it has to be an easy way to do it.



Thank you very much.
Merry Christmas.






integration multivariable-calculus change-of-variable







share|cite|improve this question











$endgroup$












  • $begingroup$
    On behalf of @math15: It seems that there is an error in the integral. Are you sure that it is $(y+2x)$ and not $(y+2x^2)$ ?
    $endgroup$
    – dantopa
    Dec 26 '18 at 10:05










  • $begingroup$
    Yes sorry, you are right!. I've already corrected it
    $endgroup$
    – Alonso Quijano
    Dec 26 '18 at 23:00








  • 3




    $begingroup$
    The first idea, $u=xy$ and $v=y-x^2$, looks good with $frac{partial (u,v)}{partial (x,y)}=y+2x^2$ and therefore $frac{partial (x,y)}{partial (u,v)}=frac 1{y+2x^2}$.
    $endgroup$
    – random
    Dec 27 '18 at 10:02










  • $begingroup$
    Thank you very much! I've checked the inverse function theorem and yes, it works!
    $endgroup$
    – Alonso Quijano
    Dec 27 '18 at 12:55














3












3








3


1



$begingroup$


I am stuck trying to solve the following integral:



$$int_R (y+2x^2)(y-x^2) dA$$ where $R$ is defined by the following equations: $xy=1$, $xy=2$, $y=x^2$, $y=x^2-1$ with $x$ and $y$ positives.



I've tried several changes of variables for example: $u=xy$, $v=y-x^2$ or $u=y-x^2$, $v=x^2$ but I get stuck because for the Jacobian ot for the limits of integration I have to solve a third degree equation. I know that I could solve it using Cardano's formula but it has to be an easy way to do it.



Thank you very much.
Merry Christmas.






integration multivariable-calculus change-of-variable







share|cite|improve this question











$endgroup$




I am stuck trying to solve the following integral:



$$int_R (y+2x^2)(y-x^2) dA$$ where $R$ is defined by the following equations: $xy=1$, $xy=2$, $y=x^2$, $y=x^2-1$ with $x$ and $y$ positives.



I've tried several changes of variables for example: $u=xy$, $v=y-x^2$ or $u=y-x^2$, $v=x^2$ but I get stuck because for the Jacobian ot for the limits of integration I have to solve a third degree equation. I know that I could solve it using Cardano's formula but it has to be an easy way to do it.



Thank you very much.
Merry Christmas.






integration multivariable-calculus change-of-variable




integration multivariable-calculus change-of-variable






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 27 '18 at 0:06









rafa11111

1,1952417




1,1952417










asked Dec 26 '18 at 8:22









Alonso QuijanoAlonso Quijano

445




445












  • $begingroup$
    On behalf of @math15: It seems that there is an error in the integral. Are you sure that it is $(y+2x)$ and not $(y+2x^2)$ ?
    $endgroup$
    – dantopa
    Dec 26 '18 at 10:05










  • $begingroup$
    Yes sorry, you are right!. I've already corrected it
    $endgroup$
    – Alonso Quijano
    Dec 26 '18 at 23:00








  • 3




    $begingroup$
    The first idea, $u=xy$ and $v=y-x^2$, looks good with $frac{partial (u,v)}{partial (x,y)}=y+2x^2$ and therefore $frac{partial (x,y)}{partial (u,v)}=frac 1{y+2x^2}$.
    $endgroup$
    – random
    Dec 27 '18 at 10:02










  • $begingroup$
    Thank you very much! I've checked the inverse function theorem and yes, it works!
    $endgroup$
    – Alonso Quijano
    Dec 27 '18 at 12:55


















  • $begingroup$
    On behalf of @math15: It seems that there is an error in the integral. Are you sure that it is $(y+2x)$ and not $(y+2x^2)$ ?
    $endgroup$
    – dantopa
    Dec 26 '18 at 10:05










  • $begingroup$
    Yes sorry, you are right!. I've already corrected it
    $endgroup$
    – Alonso Quijano
    Dec 26 '18 at 23:00








  • 3




    $begingroup$
    The first idea, $u=xy$ and $v=y-x^2$, looks good with $frac{partial (u,v)}{partial (x,y)}=y+2x^2$ and therefore $frac{partial (x,y)}{partial (u,v)}=frac 1{y+2x^2}$.
    $endgroup$
    – random
    Dec 27 '18 at 10:02










  • $begingroup$
    Thank you very much! I've checked the inverse function theorem and yes, it works!
    $endgroup$
    – Alonso Quijano
    Dec 27 '18 at 12:55
















$begingroup$
On behalf of @math15: It seems that there is an error in the integral. Are you sure that it is $(y+2x)$ and not $(y+2x^2)$ ?
$endgroup$
– dantopa
Dec 26 '18 at 10:05




$begingroup$
On behalf of @math15: It seems that there is an error in the integral. Are you sure that it is $(y+2x)$ and not $(y+2x^2)$ ?
$endgroup$
– dantopa
Dec 26 '18 at 10:05












$begingroup$
Yes sorry, you are right!. I've already corrected it
$endgroup$
– Alonso Quijano
Dec 26 '18 at 23:00






$begingroup$
Yes sorry, you are right!. I've already corrected it
$endgroup$
– Alonso Quijano
Dec 26 '18 at 23:00






3




3




$begingroup$
The first idea, $u=xy$ and $v=y-x^2$, looks good with $frac{partial (u,v)}{partial (x,y)}=y+2x^2$ and therefore $frac{partial (x,y)}{partial (u,v)}=frac 1{y+2x^2}$.
$endgroup$
– random
Dec 27 '18 at 10:02




$begingroup$
The first idea, $u=xy$ and $v=y-x^2$, looks good with $frac{partial (u,v)}{partial (x,y)}=y+2x^2$ and therefore $frac{partial (x,y)}{partial (u,v)}=frac 1{y+2x^2}$.
$endgroup$
– random
Dec 27 '18 at 10:02












$begingroup$
Thank you very much! I've checked the inverse function theorem and yes, it works!
$endgroup$
– Alonso Quijano
Dec 27 '18 at 12:55




$begingroup$
Thank you very much! I've checked the inverse function theorem and yes, it works!
$endgroup$
– Alonso Quijano
Dec 27 '18 at 12:55










1 Answer
1






active

oldest

votes


















0












$begingroup$

As a first step it is better to represent the region of integration, as in the figure.



enter image description here



From this we see that, to integrate, we have to divide the domain in three subregions as:



1) $ x_Ale xle x_B$ where we have $frac{1}{x}le yle x^2$



2) $ x_Ble xle x_C$ where we have $frac{1}{x}le yle frac{2}{x}$



3) $ x_Cle xle x_D$ where we have $x^2-1le yle frac{2}{x}$



This gives us the limits of integration and the integration is simple for the given function.
The problem is to find the coordinates of the points $A,B,C,D$ where we find some difficulty in solving equations of third degree.



Can you do from there?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Well, actually my problem is to compute C and D. I was told that there is a way to solve the integral without solving a third degree equation with a change of of variables but I cannot find one which avoid this problem.
    $endgroup$
    – Alonso Quijano
    Dec 26 '18 at 23:08











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

As a first step it is better to represent the region of integration, as in the figure.



enter image description here



From this we see that, to integrate, we have to divide the domain in three subregions as:



1) $ x_Ale xle x_B$ where we have $frac{1}{x}le yle x^2$



2) $ x_Ble xle x_C$ where we have $frac{1}{x}le yle frac{2}{x}$



3) $ x_Cle xle x_D$ where we have $x^2-1le yle frac{2}{x}$



This gives us the limits of integration and the integration is simple for the given function.
The problem is to find the coordinates of the points $A,B,C,D$ where we find some difficulty in solving equations of third degree.



Can you do from there?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Well, actually my problem is to compute C and D. I was told that there is a way to solve the integral without solving a third degree equation with a change of of variables but I cannot find one which avoid this problem.
    $endgroup$
    – Alonso Quijano
    Dec 26 '18 at 23:08
















0












$begingroup$

As a first step it is better to represent the region of integration, as in the figure.



enter image description here



From this we see that, to integrate, we have to divide the domain in three subregions as:



1) $ x_Ale xle x_B$ where we have $frac{1}{x}le yle x^2$



2) $ x_Ble xle x_C$ where we have $frac{1}{x}le yle frac{2}{x}$



3) $ x_Cle xle x_D$ where we have $x^2-1le yle frac{2}{x}$



This gives us the limits of integration and the integration is simple for the given function.
The problem is to find the coordinates of the points $A,B,C,D$ where we find some difficulty in solving equations of third degree.



Can you do from there?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Well, actually my problem is to compute C and D. I was told that there is a way to solve the integral without solving a third degree equation with a change of of variables but I cannot find one which avoid this problem.
    $endgroup$
    – Alonso Quijano
    Dec 26 '18 at 23:08














0












0








0





$begingroup$

As a first step it is better to represent the region of integration, as in the figure.



enter image description here



From this we see that, to integrate, we have to divide the domain in three subregions as:



1) $ x_Ale xle x_B$ where we have $frac{1}{x}le yle x^2$



2) $ x_Ble xle x_C$ where we have $frac{1}{x}le yle frac{2}{x}$



3) $ x_Cle xle x_D$ where we have $x^2-1le yle frac{2}{x}$



This gives us the limits of integration and the integration is simple for the given function.
The problem is to find the coordinates of the points $A,B,C,D$ where we find some difficulty in solving equations of third degree.



Can you do from there?






share|cite|improve this answer









$endgroup$



As a first step it is better to represent the region of integration, as in the figure.



enter image description here



From this we see that, to integrate, we have to divide the domain in three subregions as:



1) $ x_Ale xle x_B$ where we have $frac{1}{x}le yle x^2$



2) $ x_Ble xle x_C$ where we have $frac{1}{x}le yle frac{2}{x}$



3) $ x_Cle xle x_D$ where we have $x^2-1le yle frac{2}{x}$



This gives us the limits of integration and the integration is simple for the given function.
The problem is to find the coordinates of the points $A,B,C,D$ where we find some difficulty in solving equations of third degree.



Can you do from there?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 26 '18 at 9:23









Emilio NovatiEmilio Novati

52.2k43474




52.2k43474












  • $begingroup$
    Well, actually my problem is to compute C and D. I was told that there is a way to solve the integral without solving a third degree equation with a change of of variables but I cannot find one which avoid this problem.
    $endgroup$
    – Alonso Quijano
    Dec 26 '18 at 23:08


















  • $begingroup$
    Well, actually my problem is to compute C and D. I was told that there is a way to solve the integral without solving a third degree equation with a change of of variables but I cannot find one which avoid this problem.
    $endgroup$
    – Alonso Quijano
    Dec 26 '18 at 23:08
















$begingroup$
Well, actually my problem is to compute C and D. I was told that there is a way to solve the integral without solving a third degree equation with a change of of variables but I cannot find one which avoid this problem.
$endgroup$
– Alonso Quijano
Dec 26 '18 at 23:08




$begingroup$
Well, actually my problem is to compute C and D. I was told that there is a way to solve the integral without solving a third degree equation with a change of of variables but I cannot find one which avoid this problem.
$endgroup$
– Alonso Quijano
Dec 26 '18 at 23:08


















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