How should I solve this integral with changing parameters?
4
$begingroup$
I can't solve this. How should I proceed?
$$iint_De^{largefrac{y-x}{y+x}}mathrm dxmathrm dy$$
$D$ is the triangle with these coordinates $(0,0), (0,2), (2,0)$ and I've changed the parameters this way $u=y-x$ and $v= y+x$ and the Jacobian is $-1/2$ but I have problem finding the range of $u$ and $v$ to calculate the integral
definite-integrals
edited 7 hours ago
mrtaurho
5,74551540
asked 7 hours ago
khoshrangkhoshrang
253
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khoshrang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
4
$begingroup$
I can't solve this. How should I proceed?
$$iint_De^{largefrac{y-x}{y+x}}mathrm dxmathrm dy$$
$D$ is the triangle with these coordinates $(0,0), (0,2), (2,0)$ and I've changed the parameters this way $u=y-x$ and $v= y+x$ and the Jacobian is $-1/2$ but I have problem finding the range of $u$ and $v$ to calculate the integral
definite-integrals
edited 7 hours ago
mrtaurho
5,74551540
asked 7 hours ago
khoshrangkhoshrang
253
New contributor
khoshrang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Duplicate (with an enlarged domain) of math.stackexchange.com/q/845996 and math.stackexchange.com/q/408558
$endgroup$
– Jean Marie
5 hours ago
$begingroup$
i couldn't find them at first but even after your mention i couldn't find the elaboration to draw the new region to find the new range for parameters but thanks !
$endgroup$
– khoshrang
4 hours ago
add a comment |
4
4
4
$begingroup$
I can't solve this. How should I proceed?
$$iint_De^{largefrac{y-x}{y+x}}mathrm dxmathrm dy$$
$D$ is the triangle with these coordinates $(0,0), (0,2), (2,0)$ and I've changed the parameters this way $u=y-x$ and $v= y+x$ and the Jacobian is $-1/2$ but I have problem finding the range of $u$ and $v$ to calculate the integral
definite-integrals
edited 7 hours ago
mrtaurho
5,74551540
asked 7 hours ago
khoshrangkhoshrang
253
New contributor
khoshrang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I can't solve this. How should I proceed?
$$iint_De^{largefrac{y-x}{y+x}}mathrm dxmathrm dy$$
$D$ is the triangle with these coordinates $(0,0), (0,2), (2,0)$ and I've changed the parameters this way $u=y-x$ and $v= y+x$ and the Jacobian is $-1/2$ but I have problem finding the range of $u$ and $v$ to calculate the integral
definite-integrals
definite-integrals
edited 7 hours ago
mrtaurho
5,74551540
asked 7 hours ago
khoshrangkhoshrang
253
New contributor
khoshrang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 7 hours ago
mrtaurho
5,74551540
asked 7 hours ago
khoshrangkhoshrang
253
New contributor
khoshrang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 7 hours ago
mrtaurho
5,74551540
edited 7 hours ago
mrtaurho
5,74551540
edited 7 hours ago
mrtaurho
5,74551540
5,74551540
asked 7 hours ago
khoshrangkhoshrang
253
New contributor
khoshrang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 7 hours ago
khoshrangkhoshrang
253
asked 7 hours ago
khoshrangkhoshrang
253
253
New contributor
khoshrang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
khoshrang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
khoshrang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Duplicate (with an enlarged domain) of math.stackexchange.com/q/845996 and math.stackexchange.com/q/408558
$endgroup$
– Jean Marie
5 hours ago
$begingroup$
i couldn't find them at first but even after your mention i couldn't find the elaboration to draw the new region to find the new range for parameters but thanks !
$endgroup$
– khoshrang
4 hours ago
add a comment |
1
$begingroup$
Duplicate (with an enlarged domain) of math.stackexchange.com/q/845996 and math.stackexchange.com/q/408558
$endgroup$
– Jean Marie
5 hours ago
$begingroup$
i couldn't find them at first but even after your mention i couldn't find the elaboration to draw the new region to find the new range for parameters but thanks !
$endgroup$
– khoshrang
4 hours ago
1
1
$begingroup$
Duplicate (with an enlarged domain) of math.stackexchange.com/q/845996 and math.stackexchange.com/q/408558
$endgroup$
– Jean Marie
5 hours ago
$begingroup$
Duplicate (with an enlarged domain) of math.stackexchange.com/q/845996 and math.stackexchange.com/q/408558
$endgroup$
– Jean Marie
5 hours ago
$begingroup$
i couldn't find them at first but even after your mention i couldn't find the elaboration to draw the new region to find the new range for parameters but thanks !
$endgroup$
– khoshrang
4 hours ago
$begingroup$
i couldn't find them at first but even after your mention i couldn't find the elaboration to draw the new region to find the new range for parameters but thanks !
$endgroup$
– khoshrang
4 hours ago
add a comment |
3 Answers
3
active
oldest
votes
7
$begingroup$
Start by making a drawing of your domain. You can see that the $v$ is along the diagonal in the first quadrant, and $u$ is along the diagonal in the second quadrant. You can also see that the line between $(0,2)$ and $(2,0)$ is parallel to $u$, and intersects $v$ axis at $v=1$. So $v$ varies between $0$ and $1$ and $u$ varies between $-v$ and $v$.
answered 6 hours ago
AndreiAndrei
13.1k21230
$endgroup$
$begingroup$
thank you for your answer but my problem is that i can't draw my domain how should i draw u and v vectors along the y and x?
$endgroup$
– khoshrang
5 hours ago
1
$begingroup$
You draw a triangle $(0,0), (2,0),(0,2)$. The $x$ axis is from $(0,0)$ to $(2,0)$. What are $u$ and $v$? You have $u=y-x$ and $v=y+x$. So if I want to draw the $u$ axis you need to go from $(0,0)$ to $u=1$ and $v=0$. The second equation means $v=0=y+x$ or $x=-y$. If you plug into $u=1=y-x=y-(-y)$ you get $y=1/2$ and then $x=-1/2$. That means that you go in the second quadrant, along the $y=-x$. Similarly, the $v$ axis means $v=1$ and $u=0$. $u=0$ means $y=x$, which is the diagonal in the first quadrant.
$endgroup$
– Andrei
5 hours ago
$begingroup$
"So if I want to draw the u axis you need to go from (0,0) to u=1 and v=0 " why to u=1 and v=0?
$endgroup$
– khoshrang
5 hours ago
1
$begingroup$
If you want to draw the $x$ axis, you go from $(0,0)$ to $x=1$ and $y=0$. If you want to draw the $y$ axis, you go to $y=0$ and $x=0$.
$endgroup$
– Andrei
5 hours ago
1
$begingroup$
thank you i just got the whole idea and i'm gonna choose your answer because it directly answers the parameter change that i had done in the question
$endgroup$
– khoshrang
4 hours ago
add a comment |
6
$begingroup$
Your domain is $$D={(x,y):0<x<2,,,0<y<2,,,x+y<2}$$
Using change of variables $(x,y)to(u,v)$ with $$u=frac{x-y}{x+y},,,v=x+y$$
The region is now $$R={(u,v):-1<u<1,,0<v<2}$$
Therefore,
begin{align}
iint_D expleft({-frac{x-y}{x+y}}right),mathrm{d}x,mathrm{d}y&=frac{1}{2}iint_R ve^{-u},mathrm{d}u,mathrm{d}v
\\&=frac{1}{2}int_0^2v,mathrm{d}vint_{-1}^1 e^{-u},mathrm{d}u
end{align}
answered 6 hours ago
StubbornAtomStubbornAtom
6,14311339
$endgroup$
$begingroup$
Finally the $v$ appears, but with no explanation at all?
$endgroup$
– NickD
6 hours ago
$begingroup$
@NickD I have left the calculation of the jacobian to the OP.
$endgroup$
– StubbornAtom
6 hours ago
$begingroup$
That's fine, but you should at least mention it, if only for the sake of future readers.
$endgroup$
– NickD
6 hours ago
$begingroup$
It is pretty clear as it is, I think.
$endgroup$
– StubbornAtom
6 hours ago
1
$begingroup$
thank you! your new parameters are better than i had chosen in question but i'm gonna choose the other answer because it directly answers to the parameters that i had mentioned in the question
$endgroup$
– khoshrang
4 hours ago
|
show 3 more comments
0
$begingroup$
Computer algebra gives (for the general case):
$$e left(x^2 text{Ei}left(-frac{2 x}{x+y}right)-frac{y^2 text{Ei}left(frac{2
y}{x+y}right)}{e^2}right)+frac{1}{2} e^{1-frac{2 x}{x+y}} (x+y)^2$$
over your specified region:
$$e-frac{1}{e}$$
answered 7 hours ago
David G. StorkDavid G. Stork
11.1k41432
$endgroup$
$begingroup$
yes the final answer is right due to my answer sheet but i should solve this by hand not computer any idea of how should i solve this on paper?
$endgroup$
– khoshrang
7 hours ago
$begingroup$
(-1) because this is not an answer to the question. See the other excellent responses!
$endgroup$
– Jimmy Sabater
6 hours ago
$begingroup$
@JimmySabater: I find it absolutely astounding that the "solution" (+7 and accept) lists the trivial point of how to draw a triangle, while the full correct solution leads you to downvote.
$endgroup$
– David G. Stork
4 hours ago
$begingroup$
The accepted solution takes the time to actually reply the steps to the asker. Your answer is not one bit helpful to be honest and when asker asked you a question you didnt even reply... that si the reason the accepted answer got many upvotes and the other answer too which also boh actually answer tbe qestion
$endgroup$
– Neymar
4 hours ago
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
7
$begingroup$
Start by making a drawing of your domain. You can see that the $v$ is along the diagonal in the first quadrant, and $u$ is along the diagonal in the second quadrant. You can also see that the line between $(0,2)$ and $(2,0)$ is parallel to $u$, and intersects $v$ axis at $v=1$. So $v$ varies between $0$ and $1$ and $u$ varies between $-v$ and $v$.
answered 6 hours ago
AndreiAndrei
13.1k21230
$endgroup$
$begingroup$
thank you for your answer but my problem is that i can't draw my domain how should i draw u and v vectors along the y and x?
$endgroup$
– khoshrang
5 hours ago
1
$begingroup$
You draw a triangle $(0,0), (2,0),(0,2)$. The $x$ axis is from $(0,0)$ to $(2,0)$. What are $u$ and $v$? You have $u=y-x$ and $v=y+x$. So if I want to draw the $u$ axis you need to go from $(0,0)$ to $u=1$ and $v=0$. The second equation means $v=0=y+x$ or $x=-y$. If you plug into $u=1=y-x=y-(-y)$ you get $y=1/2$ and then $x=-1/2$. That means that you go in the second quadrant, along the $y=-x$. Similarly, the $v$ axis means $v=1$ and $u=0$. $u=0$ means $y=x$, which is the diagonal in the first quadrant.
$endgroup$
– Andrei
5 hours ago
$begingroup$
"So if I want to draw the u axis you need to go from (0,0) to u=1 and v=0 " why to u=1 and v=0?
$endgroup$
– khoshrang
5 hours ago
1
$begingroup$
If you want to draw the $x$ axis, you go from $(0,0)$ to $x=1$ and $y=0$. If you want to draw the $y$ axis, you go to $y=0$ and $x=0$.
$endgroup$
– Andrei
5 hours ago
1
$begingroup$
thank you i just got the whole idea and i'm gonna choose your answer because it directly answers the parameter change that i had done in the question
$endgroup$
– khoshrang
4 hours ago
add a comment |
7
$begingroup$
Start by making a drawing of your domain. You can see that the $v$ is along the diagonal in the first quadrant, and $u$ is along the diagonal in the second quadrant. You can also see that the line between $(0,2)$ and $(2,0)$ is parallel to $u$, and intersects $v$ axis at $v=1$. So $v$ varies between $0$ and $1$ and $u$ varies between $-v$ and $v$.
answered 6 hours ago
AndreiAndrei
13.1k21230
$endgroup$
$begingroup$
thank you for your answer but my problem is that i can't draw my domain how should i draw u and v vectors along the y and x?
$endgroup$
– khoshrang
5 hours ago
1
$begingroup$
You draw a triangle $(0,0), (2,0),(0,2)$. The $x$ axis is from $(0,0)$ to $(2,0)$. What are $u$ and $v$? You have $u=y-x$ and $v=y+x$. So if I want to draw the $u$ axis you need to go from $(0,0)$ to $u=1$ and $v=0$. The second equation means $v=0=y+x$ or $x=-y$. If you plug into $u=1=y-x=y-(-y)$ you get $y=1/2$ and then $x=-1/2$. That means that you go in the second quadrant, along the $y=-x$. Similarly, the $v$ axis means $v=1$ and $u=0$. $u=0$ means $y=x$, which is the diagonal in the first quadrant.
$endgroup$
– Andrei
5 hours ago
$begingroup$
"So if I want to draw the u axis you need to go from (0,0) to u=1 and v=0 " why to u=1 and v=0?
$endgroup$
– khoshrang
5 hours ago
1
$begingroup$
If you want to draw the $x$ axis, you go from $(0,0)$ to $x=1$ and $y=0$. If you want to draw the $y$ axis, you go to $y=0$ and $x=0$.
$endgroup$
– Andrei
5 hours ago
1
$begingroup$
thank you i just got the whole idea and i'm gonna choose your answer because it directly answers the parameter change that i had done in the question
$endgroup$
– khoshrang
4 hours ago
add a comment |
7
7
7
$begingroup$
Start by making a drawing of your domain. You can see that the $v$ is along the diagonal in the first quadrant, and $u$ is along the diagonal in the second quadrant. You can also see that the line between $(0,2)$ and $(2,0)$ is parallel to $u$, and intersects $v$ axis at $v=1$. So $v$ varies between $0$ and $1$ and $u$ varies between $-v$ and $v$.
answered 6 hours ago
AndreiAndrei
13.1k21230
$endgroup$
Start by making a drawing of your domain. You can see that the $v$ is along the diagonal in the first quadrant, and $u$ is along the diagonal in the second quadrant. You can also see that the line between $(0,2)$ and $(2,0)$ is parallel to $u$, and intersects $v$ axis at $v=1$. So $v$ varies between $0$ and $1$ and $u$ varies between $-v$ and $v$.
answered 6 hours ago
AndreiAndrei
13.1k21230
answered 6 hours ago
AndreiAndrei
13.1k21230
answered 6 hours ago
AndreiAndrei
13.1k21230
answered 6 hours ago
AndreiAndrei
13.1k21230
13.1k21230
$begingroup$
thank you for your answer but my problem is that i can't draw my domain how should i draw u and v vectors along the y and x?
$endgroup$
– khoshrang
5 hours ago
1
$begingroup$
You draw a triangle $(0,0), (2,0),(0,2)$. The $x$ axis is from $(0,0)$ to $(2,0)$. What are $u$ and $v$? You have $u=y-x$ and $v=y+x$. So if I want to draw the $u$ axis you need to go from $(0,0)$ to $u=1$ and $v=0$. The second equation means $v=0=y+x$ or $x=-y$. If you plug into $u=1=y-x=y-(-y)$ you get $y=1/2$ and then $x=-1/2$. That means that you go in the second quadrant, along the $y=-x$. Similarly, the $v$ axis means $v=1$ and $u=0$. $u=0$ means $y=x$, which is the diagonal in the first quadrant.
$endgroup$
– Andrei
5 hours ago
$begingroup$
"So if I want to draw the u axis you need to go from (0,0) to u=1 and v=0 " why to u=1 and v=0?
$endgroup$
– khoshrang
5 hours ago
1
$begingroup$
If you want to draw the $x$ axis, you go from $(0,0)$ to $x=1$ and $y=0$. If you want to draw the $y$ axis, you go to $y=0$ and $x=0$.
$endgroup$
– Andrei
5 hours ago
1
$begingroup$
thank you i just got the whole idea and i'm gonna choose your answer because it directly answers the parameter change that i had done in the question
$endgroup$
– khoshrang
4 hours ago
add a comment |
$begingroup$
thank you for your answer but my problem is that i can't draw my domain how should i draw u and v vectors along the y and x?
$endgroup$
– khoshrang
5 hours ago
1
$begingroup$
You draw a triangle $(0,0), (2,0),(0,2)$. The $x$ axis is from $(0,0)$ to $(2,0)$. What are $u$ and $v$? You have $u=y-x$ and $v=y+x$. So if I want to draw the $u$ axis you need to go from $(0,0)$ to $u=1$ and $v=0$. The second equation means $v=0=y+x$ or $x=-y$. If you plug into $u=1=y-x=y-(-y)$ you get $y=1/2$ and then $x=-1/2$. That means that you go in the second quadrant, along the $y=-x$. Similarly, the $v$ axis means $v=1$ and $u=0$. $u=0$ means $y=x$, which is the diagonal in the first quadrant.
$endgroup$
– Andrei
5 hours ago
$begingroup$
"So if I want to draw the u axis you need to go from (0,0) to u=1 and v=0 " why to u=1 and v=0?
$endgroup$
– khoshrang
5 hours ago
1
$begingroup$
If you want to draw the $x$ axis, you go from $(0,0)$ to $x=1$ and $y=0$. If you want to draw the $y$ axis, you go to $y=0$ and $x=0$.
$endgroup$
– Andrei
5 hours ago
1
$begingroup$
thank you i just got the whole idea and i'm gonna choose your answer because it directly answers the parameter change that i had done in the question
$endgroup$
– khoshrang
4 hours ago
$begingroup$
thank you for your answer but my problem is that i can't draw my domain how should i draw u and v vectors along the y and x?
$endgroup$
– khoshrang
5 hours ago
$begingroup$
thank you for your answer but my problem is that i can't draw my domain how should i draw u and v vectors along the y and x?
$endgroup$
– khoshrang
5 hours ago
1
1
$begingroup$
You draw a triangle $(0,0), (2,0),(0,2)$. The $x$ axis is from $(0,0)$ to $(2,0)$. What are $u$ and $v$? You have $u=y-x$ and $v=y+x$. So if I want to draw the $u$ axis you need to go from $(0,0)$ to $u=1$ and $v=0$. The second equation means $v=0=y+x$ or $x=-y$. If you plug into $u=1=y-x=y-(-y)$ you get $y=1/2$ and then $x=-1/2$. That means that you go in the second quadrant, along the $y=-x$. Similarly, the $v$ axis means $v=1$ and $u=0$. $u=0$ means $y=x$, which is the diagonal in the first quadrant.
$endgroup$
– Andrei
5 hours ago
$begingroup$
You draw a triangle $(0,0), (2,0),(0,2)$. The $x$ axis is from $(0,0)$ to $(2,0)$. What are $u$ and $v$? You have $u=y-x$ and $v=y+x$. So if I want to draw the $u$ axis you need to go from $(0,0)$ to $u=1$ and $v=0$. The second equation means $v=0=y+x$ or $x=-y$. If you plug into $u=1=y-x=y-(-y)$ you get $y=1/2$ and then $x=-1/2$. That means that you go in the second quadrant, along the $y=-x$. Similarly, the $v$ axis means $v=1$ and $u=0$. $u=0$ means $y=x$, which is the diagonal in the first quadrant.
$endgroup$
– Andrei
5 hours ago
$begingroup$
"So if I want to draw the u axis you need to go from (0,0) to u=1 and v=0 " why to u=1 and v=0?
$endgroup$
– khoshrang
5 hours ago
$begingroup$
"So if I want to draw the u axis you need to go from (0,0) to u=1 and v=0 " why to u=1 and v=0?
$endgroup$
– khoshrang
5 hours ago
1
1
$begingroup$
If you want to draw the $x$ axis, you go from $(0,0)$ to $x=1$ and $y=0$. If you want to draw the $y$ axis, you go to $y=0$ and $x=0$.
$endgroup$
– Andrei
5 hours ago
$begingroup$
If you want to draw the $x$ axis, you go from $(0,0)$ to $x=1$ and $y=0$. If you want to draw the $y$ axis, you go to $y=0$ and $x=0$.
$endgroup$
– Andrei
5 hours ago
1
1
$begingroup$
thank you i just got the whole idea and i'm gonna choose your answer because it directly answers the parameter change that i had done in the question
$endgroup$
– khoshrang
4 hours ago
$begingroup$
thank you i just got the whole idea and i'm gonna choose your answer because it directly answers the parameter change that i had done in the question
$endgroup$
– khoshrang
4 hours ago
add a comment |
6
$begingroup$
Your domain is $$D={(x,y):0<x<2,,,0<y<2,,,x+y<2}$$
Using change of variables $(x,y)to(u,v)$ with $$u=frac{x-y}{x+y},,,v=x+y$$
The region is now $$R={(u,v):-1<u<1,,0<v<2}$$
Therefore,
begin{align}
iint_D expleft({-frac{x-y}{x+y}}right),mathrm{d}x,mathrm{d}y&=frac{1}{2}iint_R ve^{-u},mathrm{d}u,mathrm{d}v
\\&=frac{1}{2}int_0^2v,mathrm{d}vint_{-1}^1 e^{-u},mathrm{d}u
end{align}
answered 6 hours ago
StubbornAtomStubbornAtom
6,14311339
$endgroup$
$begingroup$
Finally the $v$ appears, but with no explanation at all?
$endgroup$
– NickD
6 hours ago
$begingroup$
@NickD I have left the calculation of the jacobian to the OP.
$endgroup$
– StubbornAtom
6 hours ago
$begingroup$
That's fine, but you should at least mention it, if only for the sake of future readers.
$endgroup$
– NickD
6 hours ago
$begingroup$
It is pretty clear as it is, I think.
$endgroup$
– StubbornAtom
6 hours ago
1
$begingroup$
thank you! your new parameters are better than i had chosen in question but i'm gonna choose the other answer because it directly answers to the parameters that i had mentioned in the question
$endgroup$
– khoshrang
4 hours ago
|
show 3 more comments
6
$begingroup$
Your domain is $$D={(x,y):0<x<2,,,0<y<2,,,x+y<2}$$
Using change of variables $(x,y)to(u,v)$ with $$u=frac{x-y}{x+y},,,v=x+y$$
The region is now $$R={(u,v):-1<u<1,,0<v<2}$$
Therefore,
begin{align}
iint_D expleft({-frac{x-y}{x+y}}right),mathrm{d}x,mathrm{d}y&=frac{1}{2}iint_R ve^{-u},mathrm{d}u,mathrm{d}v
\\&=frac{1}{2}int_0^2v,mathrm{d}vint_{-1}^1 e^{-u},mathrm{d}u
end{align}
answered 6 hours ago
StubbornAtomStubbornAtom
6,14311339
$endgroup$
$begingroup$
Finally the $v$ appears, but with no explanation at all?
$endgroup$
– NickD
6 hours ago
$begingroup$
@NickD I have left the calculation of the jacobian to the OP.
$endgroup$
– StubbornAtom
6 hours ago
$begingroup$
That's fine, but you should at least mention it, if only for the sake of future readers.
$endgroup$
– NickD
6 hours ago
$begingroup$
It is pretty clear as it is, I think.
$endgroup$
– StubbornAtom
6 hours ago
1
$begingroup$
thank you! your new parameters are better than i had chosen in question but i'm gonna choose the other answer because it directly answers to the parameters that i had mentioned in the question
$endgroup$
– khoshrang
4 hours ago
|
show 3 more comments
6
6
6
$begingroup$
Your domain is $$D={(x,y):0<x<2,,,0<y<2,,,x+y<2}$$
Using change of variables $(x,y)to(u,v)$ with $$u=frac{x-y}{x+y},,,v=x+y$$
The region is now $$R={(u,v):-1<u<1,,0<v<2}$$
Therefore,
begin{align}
iint_D expleft({-frac{x-y}{x+y}}right),mathrm{d}x,mathrm{d}y&=frac{1}{2}iint_R ve^{-u},mathrm{d}u,mathrm{d}v
\\&=frac{1}{2}int_0^2v,mathrm{d}vint_{-1}^1 e^{-u},mathrm{d}u
end{align}
answered 6 hours ago
StubbornAtomStubbornAtom
6,14311339
$endgroup$
Your domain is $$D={(x,y):0<x<2,,,0<y<2,,,x+y<2}$$
Using change of variables $(x,y)to(u,v)$ with $$u=frac{x-y}{x+y},,,v=x+y$$
The region is now $$R={(u,v):-1<u<1,,0<v<2}$$
Therefore,
begin{align}
iint_D expleft({-frac{x-y}{x+y}}right),mathrm{d}x,mathrm{d}y&=frac{1}{2}iint_R ve^{-u},mathrm{d}u,mathrm{d}v
\\&=frac{1}{2}int_0^2v,mathrm{d}vint_{-1}^1 e^{-u},mathrm{d}u
end{align}
answered 6 hours ago
StubbornAtomStubbornAtom
6,14311339
edited 6 hours ago
answered 6 hours ago
StubbornAtomStubbornAtom
6,14311339
answered 6 hours ago
StubbornAtomStubbornAtom
6,14311339
answered 6 hours ago
StubbornAtomStubbornAtom
6,14311339
6,14311339
$begingroup$
Finally the $v$ appears, but with no explanation at all?
$endgroup$
– NickD
6 hours ago
$begingroup$
@NickD I have left the calculation of the jacobian to the OP.
$endgroup$
– StubbornAtom
6 hours ago
$begingroup$
That's fine, but you should at least mention it, if only for the sake of future readers.
$endgroup$
– NickD
6 hours ago
$begingroup$
It is pretty clear as it is, I think.
$endgroup$
– StubbornAtom
6 hours ago
1
$begingroup$
thank you! your new parameters are better than i had chosen in question but i'm gonna choose the other answer because it directly answers to the parameters that i had mentioned in the question
$endgroup$
– khoshrang
4 hours ago
|
show 3 more comments
$begingroup$
Finally the $v$ appears, but with no explanation at all?
$endgroup$
– NickD
6 hours ago
$begingroup$
@NickD I have left the calculation of the jacobian to the OP.
$endgroup$
– StubbornAtom
6 hours ago
$begingroup$
That's fine, but you should at least mention it, if only for the sake of future readers.
$endgroup$
– NickD
6 hours ago
$begingroup$
It is pretty clear as it is, I think.
$endgroup$
– StubbornAtom
6 hours ago
1
$begingroup$
thank you! your new parameters are better than i had chosen in question but i'm gonna choose the other answer because it directly answers to the parameters that i had mentioned in the question
$endgroup$
– khoshrang
4 hours ago
$begingroup$
Finally the $v$ appears, but with no explanation at all?
$endgroup$
– NickD
6 hours ago
$begingroup$
Finally the $v$ appears, but with no explanation at all?
$endgroup$
– NickD
6 hours ago
$begingroup$
@NickD I have left the calculation of the jacobian to the OP.
$endgroup$
– StubbornAtom
6 hours ago
$begingroup$
@NickD I have left the calculation of the jacobian to the OP.
$endgroup$
– StubbornAtom
6 hours ago
$begingroup$
That's fine, but you should at least mention it, if only for the sake of future readers.
$endgroup$
– NickD
6 hours ago
$begingroup$
That's fine, but you should at least mention it, if only for the sake of future readers.
$endgroup$
– NickD
6 hours ago
$begingroup$
It is pretty clear as it is, I think.
$endgroup$
– StubbornAtom
6 hours ago
$begingroup$
It is pretty clear as it is, I think.
$endgroup$
– StubbornAtom
6 hours ago
1
1
$begingroup$
thank you! your new parameters are better than i had chosen in question but i'm gonna choose the other answer because it directly answers to the parameters that i had mentioned in the question
$endgroup$
– khoshrang
4 hours ago
$begingroup$
thank you! your new parameters are better than i had chosen in question but i'm gonna choose the other answer because it directly answers to the parameters that i had mentioned in the question
$endgroup$
– khoshrang
4 hours ago
|
show 3 more comments
0
$begingroup$
Computer algebra gives (for the general case):
$$e left(x^2 text{Ei}left(-frac{2 x}{x+y}right)-frac{y^2 text{Ei}left(frac{2
y}{x+y}right)}{e^2}right)+frac{1}{2} e^{1-frac{2 x}{x+y}} (x+y)^2$$
over your specified region:
$$e-frac{1}{e}$$
answered 7 hours ago
David G. StorkDavid G. Stork
11.1k41432
$endgroup$
$begingroup$
yes the final answer is right due to my answer sheet but i should solve this by hand not computer any idea of how should i solve this on paper?
$endgroup$
– khoshrang
7 hours ago
$begingroup$
(-1) because this is not an answer to the question. See the other excellent responses!
$endgroup$
– Jimmy Sabater
6 hours ago
$begingroup$
@JimmySabater: I find it absolutely astounding that the "solution" (+7 and accept) lists the trivial point of how to draw a triangle, while the full correct solution leads you to downvote.
$endgroup$
– David G. Stork
4 hours ago
$begingroup$
The accepted solution takes the time to actually reply the steps to the asker. Your answer is not one bit helpful to be honest and when asker asked you a question you didnt even reply... that si the reason the accepted answer got many upvotes and the other answer too which also boh actually answer tbe qestion
$endgroup$
– Neymar
4 hours ago
add a comment |
0
$begingroup$
Computer algebra gives (for the general case):
$$e left(x^2 text{Ei}left(-frac{2 x}{x+y}right)-frac{y^2 text{Ei}left(frac{2
y}{x+y}right)}{e^2}right)+frac{1}{2} e^{1-frac{2 x}{x+y}} (x+y)^2$$
over your specified region:
$$e-frac{1}{e}$$
answered 7 hours ago
David G. StorkDavid G. Stork
11.1k41432
$endgroup$
$begingroup$
yes the final answer is right due to my answer sheet but i should solve this by hand not computer any idea of how should i solve this on paper?
$endgroup$
– khoshrang
7 hours ago
$begingroup$
(-1) because this is not an answer to the question. See the other excellent responses!
$endgroup$
– Jimmy Sabater
6 hours ago
$begingroup$
@JimmySabater: I find it absolutely astounding that the "solution" (+7 and accept) lists the trivial point of how to draw a triangle, while the full correct solution leads you to downvote.
$endgroup$
– David G. Stork
4 hours ago
$begingroup$
The accepted solution takes the time to actually reply the steps to the asker. Your answer is not one bit helpful to be honest and when asker asked you a question you didnt even reply... that si the reason the accepted answer got many upvotes and the other answer too which also boh actually answer tbe qestion
$endgroup$
– Neymar
4 hours ago
add a comment |
0
0
0
$begingroup$
Computer algebra gives (for the general case):
$$e left(x^2 text{Ei}left(-frac{2 x}{x+y}right)-frac{y^2 text{Ei}left(frac{2
y}{x+y}right)}{e^2}right)+frac{1}{2} e^{1-frac{2 x}{x+y}} (x+y)^2$$
over your specified region:
$$e-frac{1}{e}$$
answered 7 hours ago
David G. StorkDavid G. Stork
11.1k41432
$endgroup$
Computer algebra gives (for the general case):
$$e left(x^2 text{Ei}left(-frac{2 x}{x+y}right)-frac{y^2 text{Ei}left(frac{2
y}{x+y}right)}{e^2}right)+frac{1}{2} e^{1-frac{2 x}{x+y}} (x+y)^2$$
over your specified region:
$$e-frac{1}{e}$$
answered 7 hours ago
David G. StorkDavid G. Stork
11.1k41432
answered 7 hours ago
David G. StorkDavid G. Stork
11.1k41432
answered 7 hours ago
David G. StorkDavid G. Stork
11.1k41432
answered 7 hours ago
David G. StorkDavid G. Stork
11.1k41432
11.1k41432
$begingroup$
yes the final answer is right due to my answer sheet but i should solve this by hand not computer any idea of how should i solve this on paper?
$endgroup$
– khoshrang
7 hours ago
$begingroup$
(-1) because this is not an answer to the question. See the other excellent responses!
$endgroup$
– Jimmy Sabater
6 hours ago
$begingroup$
@JimmySabater: I find it absolutely astounding that the "solution" (+7 and accept) lists the trivial point of how to draw a triangle, while the full correct solution leads you to downvote.
$endgroup$
– David G. Stork
4 hours ago
$begingroup$
The accepted solution takes the time to actually reply the steps to the asker. Your answer is not one bit helpful to be honest and when asker asked you a question you didnt even reply... that si the reason the accepted answer got many upvotes and the other answer too which also boh actually answer tbe qestion
$endgroup$
– Neymar
4 hours ago
add a comment |
$begingroup$
yes the final answer is right due to my answer sheet but i should solve this by hand not computer any idea of how should i solve this on paper?
$endgroup$
– khoshrang
7 hours ago
$begingroup$
(-1) because this is not an answer to the question. See the other excellent responses!
$endgroup$
– Jimmy Sabater
6 hours ago
$begingroup$
@JimmySabater: I find it absolutely astounding that the "solution" (+7 and accept) lists the trivial point of how to draw a triangle, while the full correct solution leads you to downvote.
$endgroup$
– David G. Stork
4 hours ago
$begingroup$
The accepted solution takes the time to actually reply the steps to the asker. Your answer is not one bit helpful to be honest and when asker asked you a question you didnt even reply... that si the reason the accepted answer got many upvotes and the other answer too which also boh actually answer tbe qestion
$endgroup$
– Neymar
4 hours ago
$begingroup$
yes the final answer is right due to my answer sheet but i should solve this by hand not computer any idea of how should i solve this on paper?
$endgroup$
– khoshrang
7 hours ago
$begingroup$
yes the final answer is right due to my answer sheet but i should solve this by hand not computer any idea of how should i solve this on paper?
$endgroup$
– khoshrang
7 hours ago
$begingroup$
(-1) because this is not an answer to the question. See the other excellent responses!
$endgroup$
– Jimmy Sabater
6 hours ago
$begingroup$
(-1) because this is not an answer to the question. See the other excellent responses!
$endgroup$
– Jimmy Sabater
6 hours ago
$begingroup$
@JimmySabater: I find it absolutely astounding that the "solution" (+7 and accept) lists the trivial point of how to draw a triangle, while the full correct solution leads you to downvote.
$endgroup$
– David G. Stork
4 hours ago
$begingroup$
@JimmySabater: I find it absolutely astounding that the "solution" (+7 and accept) lists the trivial point of how to draw a triangle, while the full correct solution leads you to downvote.
$endgroup$
– David G. Stork
4 hours ago
$begingroup$
The accepted solution takes the time to actually reply the steps to the asker. Your answer is not one bit helpful to be honest and when asker asked you a question you didnt even reply... that si the reason the accepted answer got many upvotes and the other answer too which also boh actually answer tbe qestion
$endgroup$
– Neymar
4 hours ago
$begingroup$
The accepted solution takes the time to actually reply the steps to the asker. Your answer is not one bit helpful to be honest and when asker asked you a question you didnt even reply... that si the reason the accepted answer got many upvotes and the other answer too which also boh actually answer tbe qestion
$endgroup$
– Neymar
4 hours ago
add a comment |
khoshrang is a new contributor. Be nice, and check out our Code of Conduct.
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khoshrang is a new contributor. Be nice, and check out our Code of Conduct.
khoshrang is a new contributor. Be nice, and check out our Code of Conduct.
khoshrang is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Duplicate (with an enlarged domain) of math.stackexchange.com/q/845996 and math.stackexchange.com/q/408558
$endgroup$
– Jean Marie
5 hours ago
$begingroup$
i couldn't find them at first but even after your mention i couldn't find the elaboration to draw the new region to find the new range for parameters but thanks !
$endgroup$
– khoshrang
4 hours ago