How can this multiple integral be evaluated?
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
edited Dec 27 at 0:06
rafa11111
1,115417
asked Dec 26 at 8:22
Alonso Quijano
445
New contributor
Alonso Quijano is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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
edited Dec 27 at 0:06
rafa11111
1,115417
asked Dec 26 at 8:22
Alonso Quijano
445
New contributor
Alonso Quijano is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
On behalf of @math15: It seems that there is an error in the integral. Are you sure that it is $(y+2x)$ and not $(y+2x^2)$ ?
– dantopa
Dec 26 at 10:05
Yes sorry, you are right!. I've already corrected it
– Alonso Quijano
Dec 26 at 23:00
3
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 $frac{partial (x,y)}{partial (u,v)}=frac 1{y+2x^2}$.
– random
2 days ago
Thank you very much! I've checked the inverse function theorem and yes, it works!
– Alonso Quijano
2 days ago
add a comment |
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
edited Dec 27 at 0:06
rafa11111
1,115417
asked Dec 26 at 8:22
Alonso Quijano
445
New contributor
Alonso Quijano is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
edited Dec 27 at 0:06
rafa11111
1,115417
asked Dec 26 at 8:22
Alonso Quijano
445
New contributor
Alonso Quijano is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 27 at 0:06
rafa11111
1,115417
asked Dec 26 at 8:22
Alonso Quijano
445
New contributor
Alonso Quijano is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 27 at 0:06
rafa11111
1,115417
edited Dec 27 at 0:06
rafa11111
1,115417
edited Dec 27 at 0:06
rafa11111
1,115417
1,115417
asked Dec 26 at 8:22
Alonso Quijano
445
New contributor
Alonso Quijano is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 26 at 8:22
Alonso Quijano
445
asked Dec 26 at 8:22
Alonso Quijano
445
445
New contributor
Alonso Quijano is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Alonso Quijano is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Alonso Quijano is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
On behalf of @math15: It seems that there is an error in the integral. Are you sure that it is $(y+2x)$ and not $(y+2x^2)$ ?
– dantopa
Dec 26 at 10:05
Yes sorry, you are right!. I've already corrected it
– Alonso Quijano
Dec 26 at 23:00
3
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 $frac{partial (x,y)}{partial (u,v)}=frac 1{y+2x^2}$.
– random
2 days ago
Thank you very much! I've checked the inverse function theorem and yes, it works!
– Alonso Quijano
2 days ago
add a comment |
On behalf of @math15: It seems that there is an error in the integral. Are you sure that it is $(y+2x)$ and not $(y+2x^2)$ ?
– dantopa
Dec 26 at 10:05
Yes sorry, you are right!. I've already corrected it
– Alonso Quijano
Dec 26 at 23:00
3
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 $frac{partial (x,y)}{partial (u,v)}=frac 1{y+2x^2}$.
– random
2 days ago
Thank you very much! I've checked the inverse function theorem and yes, it works!
– Alonso Quijano
2 days ago
On behalf of @math15: It seems that there is an error in the integral. Are you sure that it is $(y+2x)$ and not $(y+2x^2)$ ?
– dantopa
Dec 26 at 10:05
On behalf of @math15: It seems that there is an error in the integral. Are you sure that it is $(y+2x)$ and not $(y+2x^2)$ ?
– dantopa
Dec 26 at 10:05
Yes sorry, you are right!. I've already corrected it
– Alonso Quijano
Dec 26 at 23:00
Yes sorry, you are right!. I've already corrected it
– Alonso Quijano
Dec 26 at 23:00
3
3
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 $frac{partial (x,y)}{partial (u,v)}=frac 1{y+2x^2}$.
– random
2 days ago
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 $frac{partial (x,y)}{partial (u,v)}=frac 1{y+2x^2}$.
– random
2 days ago
Thank you very much! I've checked the inverse function theorem and yes, it works!
– Alonso Quijano
2 days ago
Thank you very much! I've checked the inverse function theorem and yes, it works!
– Alonso Quijano
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
As a first step it is better to represent the region of integration, as in the figure.
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?
answered Dec 26 at 9:23
Emilio Novati
51.5k43472
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.
– Alonso Quijano
Dec 26 at 23:08
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
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oldest
votes
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oldest
votes
As a first step it is better to represent the region of integration, as in the figure.
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?
answered Dec 26 at 9:23
Emilio Novati
51.5k43472
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.
– Alonso Quijano
Dec 26 at 23:08
add a comment |
As a first step it is better to represent the region of integration, as in the figure.
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?
answered Dec 26 at 9:23
Emilio Novati
51.5k43472
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.
– Alonso Quijano
Dec 26 at 23:08
add a comment |
As a first step it is better to represent the region of integration, as in the figure.
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?
answered Dec 26 at 9:23
Emilio Novati
51.5k43472
As a first step it is better to represent the region of integration, as in the figure.
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?
answered Dec 26 at 9:23
Emilio Novati
51.5k43472
answered Dec 26 at 9:23
Emilio Novati
51.5k43472
answered Dec 26 at 9:23
Emilio Novati
51.5k43472
answered Dec 26 at 9:23
Emilio Novati
51.5k43472
51.5k43472
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.
– Alonso Quijano
Dec 26 at 23:08
add a comment |
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.
– Alonso Quijano
Dec 26 at 23:08
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.
– Alonso Quijano
Dec 26 at 23:08
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.
– Alonso Quijano
Dec 26 at 23:08
add a comment |
Alonso Quijano is a new contributor. Be nice, and check out our Code of Conduct.
Alonso Quijano is a new contributor. Be nice, and check out our Code of Conduct.
Alonso Quijano is a new contributor. Be nice, and check out our Code of Conduct.
Alonso Quijano is a new contributor. Be nice, and check out our Code of Conduct.
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On behalf of @math15: It seems that there is an error in the integral. Are you sure that it is $(y+2x)$ and not $(y+2x^2)$ ?
– dantopa
Dec 26 at 10:05
Yes sorry, you are right!. I've already corrected it
– Alonso Quijano
Dec 26 at 23:00
3
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 $frac{partial (x,y)}{partial (u,v)}=frac 1{y+2x^2}$.
– random
2 days ago
Thank you very much! I've checked the inverse function theorem and yes, it works!
– Alonso Quijano
2 days ago