Mixing
The double integral $int_{1}^{2}int_{x}^{2x}f(x,y)dxdy$ under the tranformation $x=u-uv $ and $y=uv$...
$begingroup$
The double integral $int_{1}^{2}int_{x}^{2x}f(x,y)dxdy$ under the tranformation $x=u-uv $ and $y=uv$ is__________________
I have calculated jacobian as u but I am not able to find out the limits of integration for u and v .
u and v in terms of x and y are $u=x+y$ and $v=frac{y}{x+y}$.
x goes from 1 to 2 and y goes from x to 2x .
calculus multivariable-calculus multiple-integral
asked Jan 19 at 7:53
sejysejy
1539
$endgroup$
add a comment |
$begingroup$
The double integral $int_{1}^{2}int_{x}^{2x}f(x,y)dxdy$ under the tranformation $x=u-uv $ and $y=uv$ is__________________
I have calculated jacobian as u but I am not able to find out the limits of integration for u and v .
u and v in terms of x and y are $u=x+y$ and $v=frac{y}{x+y}$.
x goes from 1 to 2 and y goes from x to 2x .
calculus multivariable-calculus multiple-integral
asked Jan 19 at 7:53
sejysejy
1539
$endgroup$
2
$begingroup$
Is the integral supposed to be $int_1^2 int_x^{2x} f(x,y),dy,dx$ ?
$endgroup$
– StubbornAtom
Jan 19 at 12:40
add a comment |
$begingroup$
The double integral $int_{1}^{2}int_{x}^{2x}f(x,y)dxdy$ under the tranformation $x=u-uv $ and $y=uv$ is__________________
I have calculated jacobian as u but I am not able to find out the limits of integration for u and v .
u and v in terms of x and y are $u=x+y$ and $v=frac{y}{x+y}$.
x goes from 1 to 2 and y goes from x to 2x .
calculus multivariable-calculus multiple-integral
asked Jan 19 at 7:53
sejysejy
1539
$endgroup$
The double integral $int_{1}^{2}int_{x}^{2x}f(x,y)dxdy$ under the tranformation $x=u-uv $ and $y=uv$ is__________________
I have calculated jacobian as u but I am not able to find out the limits of integration for u and v .
u and v in terms of x and y are $u=x+y$ and $v=frac{y}{x+y}$.
x goes from 1 to 2 and y goes from x to 2x .
calculus multivariable-calculus multiple-integral
calculus multivariable-calculus multiple-integral
asked Jan 19 at 7:53
sejysejy
1539
asked Jan 19 at 7:53
sejysejy
1539
asked Jan 19 at 7:53
sejysejy
1539
asked Jan 19 at 7:53
sejysejy
1539
asked Jan 19 at 7:53
sejysejy
1539
1539
2
$begingroup$
Is the integral supposed to be $int_1^2 int_x^{2x} f(x,y),dy,dx$ ?
$endgroup$
– StubbornAtom
Jan 19 at 12:40
add a comment |
2
$begingroup$
Is the integral supposed to be $int_1^2 int_x^{2x} f(x,y),dy,dx$ ?
$endgroup$
– StubbornAtom
Jan 19 at 12:40
2
2
$begingroup$
Is the integral supposed to be $int_1^2 int_x^{2x} f(x,y),dy,dx$ ?
$endgroup$
– StubbornAtom
Jan 19 at 12:40
$begingroup$
Is the integral supposed to be $int_1^2 int_x^{2x} f(x,y),dy,dx$ ?
$endgroup$
– StubbornAtom
Jan 19 at 12:40
add a comment |
1 Answer
1
active
oldest
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$begingroup$
The region of integration is the set of points $(x,y)$ such that $ 1le xle 2$ and $x le y le 2x$. Notice that under the change of variables, the bottom line $y=x$ can be written as $uv = u - uv$, or $v = 1/2$. Along the top line $y=2x$, we have $uv/2 = u - uv$, or $v= 2/3$. More generally, along any of the lines $y=tx$ for $tin[1,2]$, we have $uv/t = u - uv$, or $（1+1/t)v = 1$ i.e. $v = frac{t}{t+1} = 1-frac1{t+1}$. We also need to describe the vertical lines $x=1,2$. For fixed $u$ and $x$, $v = 1- x/u$. A graph of the level sets of $u,v$:
Thus the region is
$$ 2 le u le 6,quad max(1/2, 1-2/u) le v le min(2/3, 1-1/u).$$
Alternatively, by describing the lines $x=x_0$ as $u=x_0/(1-v)$, we obtain
$$ frac12 le v le frac23 , quad frac1{1-v} le u le frac2{1-v}.$$
answered Jan 20 at 12:35
Calvin KhorCalvin Khor
12.2k21438
$endgroup$
2
$begingroup$
One could add, that it's probably best to integrate first over $u$ i.e. $$ int_{frac{1}{2}}^{frac{2}{3}} {rm d}v int_{frac{1}{1-v}}^{frac{2}{1-v}} {rm d}u , left| frac{partial (x,y)}{partial (u,v)} right| , fleft(x(u,v),y(u,v)right) , . $$
$endgroup$
– Diger
Jan 20 at 14:38
1
$begingroup$
@Diger indeed, that looks nicer if we don't know anything else about the problem. I've added it to the answer(thanks)
$endgroup$
– Calvin Khor
Jan 20 at 14:54
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
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active
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active
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votes
$begingroup$
The region of integration is the set of points $(x,y)$ such that $ 1le xle 2$ and $x le y le 2x$. Notice that under the change of variables, the bottom line $y=x$ can be written as $uv = u - uv$, or $v = 1/2$. Along the top line $y=2x$, we have $uv/2 = u - uv$, or $v= 2/3$. More generally, along any of the lines $y=tx$ for $tin[1,2]$, we have $uv/t = u - uv$, or $（1+1/t)v = 1$ i.e. $v = frac{t}{t+1} = 1-frac1{t+1}$. We also need to describe the vertical lines $x=1,2$. For fixed $u$ and $x$, $v = 1- x/u$. A graph of the level sets of $u,v$:
Thus the region is
$$ 2 le u le 6,quad max(1/2, 1-2/u) le v le min(2/3, 1-1/u).$$
Alternatively, by describing the lines $x=x_0$ as $u=x_0/(1-v)$, we obtain
$$ frac12 le v le frac23 , quad frac1{1-v} le u le frac2{1-v}.$$
answered Jan 20 at 12:35
Calvin KhorCalvin Khor
12.2k21438
$endgroup$
2
$begingroup$
One could add, that it's probably best to integrate first over $u$ i.e. $$ int_{frac{1}{2}}^{frac{2}{3}} {rm d}v int_{frac{1}{1-v}}^{frac{2}{1-v}} {rm d}u , left| frac{partial (x,y)}{partial (u,v)} right| , fleft(x(u,v),y(u,v)right) , . $$
$endgroup$
– Diger
Jan 20 at 14:38
1
$begingroup$
@Diger indeed, that looks nicer if we don't know anything else about the problem. I've added it to the answer(thanks)
$endgroup$
– Calvin Khor
Jan 20 at 14:54
add a comment |
$begingroup$
The region of integration is the set of points $(x,y)$ such that $ 1le xle 2$ and $x le y le 2x$. Notice that under the change of variables, the bottom line $y=x$ can be written as $uv = u - uv$, or $v = 1/2$. Along the top line $y=2x$, we have $uv/2 = u - uv$, or $v= 2/3$. More generally, along any of the lines $y=tx$ for $tin[1,2]$, we have $uv/t = u - uv$, or $（1+1/t)v = 1$ i.e. $v = frac{t}{t+1} = 1-frac1{t+1}$. We also need to describe the vertical lines $x=1,2$. For fixed $u$ and $x$, $v = 1- x/u$. A graph of the level sets of $u,v$:
Thus the region is
$$ 2 le u le 6,quad max(1/2, 1-2/u) le v le min(2/3, 1-1/u).$$
Alternatively, by describing the lines $x=x_0$ as $u=x_0/(1-v)$, we obtain
$$ frac12 le v le frac23 , quad frac1{1-v} le u le frac2{1-v}.$$
answered Jan 20 at 12:35
Calvin KhorCalvin Khor
12.2k21438
$endgroup$
2
$begingroup$
One could add, that it's probably best to integrate first over $u$ i.e. $$ int_{frac{1}{2}}^{frac{2}{3}} {rm d}v int_{frac{1}{1-v}}^{frac{2}{1-v}} {rm d}u , left| frac{partial (x,y)}{partial (u,v)} right| , fleft(x(u,v),y(u,v)right) , . $$
$endgroup$
– Diger
Jan 20 at 14:38
1
$begingroup$
@Diger indeed, that looks nicer if we don't know anything else about the problem. I've added it to the answer(thanks)
$endgroup$
– Calvin Khor
Jan 20 at 14:54
add a comment |
$begingroup$
The region of integration is the set of points $(x,y)$ such that $ 1le xle 2$ and $x le y le 2x$. Notice that under the change of variables, the bottom line $y=x$ can be written as $uv = u - uv$, or $v = 1/2$. Along the top line $y=2x$, we have $uv/2 = u - uv$, or $v= 2/3$. More generally, along any of the lines $y=tx$ for $tin[1,2]$, we have $uv/t = u - uv$, or $（1+1/t)v = 1$ i.e. $v = frac{t}{t+1} = 1-frac1{t+1}$. We also need to describe the vertical lines $x=1,2$. For fixed $u$ and $x$, $v = 1- x/u$. A graph of the level sets of $u,v$:
Thus the region is
$$ 2 le u le 6,quad max(1/2, 1-2/u) le v le min(2/3, 1-1/u).$$
Alternatively, by describing the lines $x=x_0$ as $u=x_0/(1-v)$, we obtain
$$ frac12 le v le frac23 , quad frac1{1-v} le u le frac2{1-v}.$$
answered Jan 20 at 12:35
Calvin KhorCalvin Khor
12.2k21438
$endgroup$
The region of integration is the set of points $(x,y)$ such that $ 1le xle 2$ and $x le y le 2x$. Notice that under the change of variables, the bottom line $y=x$ can be written as $uv = u - uv$, or $v = 1/2$. Along the top line $y=2x$, we have $uv/2 = u - uv$, or $v= 2/3$. More generally, along any of the lines $y=tx$ for $tin[1,2]$, we have $uv/t = u - uv$, or $（1+1/t)v = 1$ i.e. $v = frac{t}{t+1} = 1-frac1{t+1}$. We also need to describe the vertical lines $x=1,2$. For fixed $u$ and $x$, $v = 1- x/u$. A graph of the level sets of $u,v$:
Thus the region is
$$ 2 le u le 6,quad max(1/2, 1-2/u) le v le min(2/3, 1-1/u).$$
Alternatively, by describing the lines $x=x_0$ as $u=x_0/(1-v)$, we obtain
$$ frac12 le v le frac23 , quad frac1{1-v} le u le frac2{1-v}.$$
answered Jan 20 at 12:35
Calvin KhorCalvin Khor
12.2k21438
edited Jan 20 at 14:54
answered Jan 20 at 12:35
Calvin KhorCalvin Khor
12.2k21438
answered Jan 20 at 12:35
Calvin KhorCalvin Khor
12.2k21438
answered Jan 20 at 12:35
Calvin KhorCalvin Khor
12.2k21438
12.2k21438
2
$begingroup$
One could add, that it's probably best to integrate first over $u$ i.e. $$ int_{frac{1}{2}}^{frac{2}{3}} {rm d}v int_{frac{1}{1-v}}^{frac{2}{1-v}} {rm d}u , left| frac{partial (x,y)}{partial (u,v)} right| , fleft(x(u,v),y(u,v)right) , . $$
$endgroup$
– Diger
Jan 20 at 14:38
1
$begingroup$
@Diger indeed, that looks nicer if we don't know anything else about the problem. I've added it to the answer(thanks)
$endgroup$
– Calvin Khor
Jan 20 at 14:54
add a comment |
2
$begingroup$
One could add, that it's probably best to integrate first over $u$ i.e. $$ int_{frac{1}{2}}^{frac{2}{3}} {rm d}v int_{frac{1}{1-v}}^{frac{2}{1-v}} {rm d}u , left| frac{partial (x,y)}{partial (u,v)} right| , fleft(x(u,v),y(u,v)right) , . $$
$endgroup$
– Diger
Jan 20 at 14:38
1
$begingroup$
@Diger indeed, that looks nicer if we don't know anything else about the problem. I've added it to the answer(thanks)
$endgroup$
– Calvin Khor
Jan 20 at 14:54
2
2
$begingroup$
One could add, that it's probably best to integrate first over $u$ i.e. $$ int_{frac{1}{2}}^{frac{2}{3}} {rm d}v int_{frac{1}{1-v}}^{frac{2}{1-v}} {rm d}u , left| frac{partial (x,y)}{partial (u,v)} right| , fleft(x(u,v),y(u,v)right) , . $$
$endgroup$
– Diger
Jan 20 at 14:38
$begingroup$
One could add, that it's probably best to integrate first over $u$ i.e. $$ int_{frac{1}{2}}^{frac{2}{3}} {rm d}v int_{frac{1}{1-v}}^{frac{2}{1-v}} {rm d}u , left| frac{partial (x,y)}{partial (u,v)} right| , fleft(x(u,v),y(u,v)right) , . $$
$endgroup$
– Diger
Jan 20 at 14:38
1
1
$begingroup$
@Diger indeed, that looks nicer if we don't know anything else about the problem. I've added it to the answer(thanks)
$endgroup$
– Calvin Khor
Jan 20 at 14:54
$begingroup$
@Diger indeed, that looks nicer if we don't know anything else about the problem. I've added it to the answer(thanks)
$endgroup$
– Calvin Khor
Jan 20 at 14:54
add a comment |
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$begingroup$
Is the integral supposed to be $int_1^2 int_x^{2x} f(x,y),dy,dx$ ?
$endgroup$
– StubbornAtom
Jan 19 at 12:40