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calculation double integral by transformation
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By using the transformation $x+y=u, ; y=uv$, evaluate the integral $int int sqrt(xy(1-x-y)) dxdy$ taken over the area enclosed by the straight lines $x=0, ; y=0$ and $x+y=1$.
First I have calculated the Jacobian, it came as $u$.
By using the transformation, integrand is changed to $usqrt(v(1-u)(1-v))$. I am unable to find the new limits of the integration. Please help me with this.
calculus integration
edited Jul 26 at 17:32
Joseph Eck
570212
asked Jul 26 at 17:07
Balaji
6617
add a comment |Â
up vote
0
down vote
favorite
By using the transformation $x+y=u, ; y=uv$, evaluate the integral $int int sqrt(xy(1-x-y)) dxdy$ taken over the area enclosed by the straight lines $x=0, ; y=0$ and $x+y=1$.
First I have calculated the Jacobian, it came as $u$.
By using the transformation, integrand is changed to $usqrt(v(1-u)(1-v))$. I am unable to find the new limits of the integration. Please help me with this.
calculus integration
edited Jul 26 at 17:32
Joseph Eck
570212
asked Jul 26 at 17:07
Balaji
6617
2
You are a member of this forum for over two years. It's better if you learn how to write mathematical expressions in Mathjax.
– Math Lover
Jul 26 at 17:09
How about looking at the old limits of the integration first, but not in terms of straight lines, but in terms of $x$ and $y$.
– mvw
Jul 26 at 17:24
math.meta.stackexchange.com/questions/5020/…
– John Wayland Bales
Jul 26 at 17:24
$u = x+y, v = frac yu = frac yx+y$ the line $x+y = 1$ transforms to $u = 1$ the line $y = 0 implies v = 0$ and $x = 0 implies v = 1$ along those respective boundaries.
– Doug M
Jul 26 at 17:28
@DougM thanks a lot.
– Balaji
Jul 26 at 17:45
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
By using the transformation $x+y=u, ; y=uv$, evaluate the integral $int int sqrt(xy(1-x-y)) dxdy$ taken over the area enclosed by the straight lines $x=0, ; y=0$ and $x+y=1$.
First I have calculated the Jacobian, it came as $u$.
By using the transformation, integrand is changed to $usqrt(v(1-u)(1-v))$. I am unable to find the new limits of the integration. Please help me with this.
calculus integration
edited Jul 26 at 17:32
Joseph Eck
570212
asked Jul 26 at 17:07
Balaji
6617
By using the transformation $x+y=u, ; y=uv$, evaluate the integral $int int sqrt(xy(1-x-y)) dxdy$ taken over the area enclosed by the straight lines $x=0, ; y=0$ and $x+y=1$.
First I have calculated the Jacobian, it came as $u$.
By using the transformation, integrand is changed to $usqrt(v(1-u)(1-v))$. I am unable to find the new limits of the integration. Please help me with this.
calculus integration
edited Jul 26 at 17:32
Joseph Eck
570212
asked Jul 26 at 17:07
Balaji
6617
edited Jul 26 at 17:32
Joseph Eck
570212
edited Jul 26 at 17:32
Joseph Eck
570212
edited Jul 26 at 17:32
Joseph Eck
570212
570212
asked Jul 26 at 17:07
Balaji
6617
asked Jul 26 at 17:07
Balaji
6617
asked Jul 26 at 17:07
Balaji
6617
6617
2
You are a member of this forum for over two years. It's better if you learn how to write mathematical expressions in Mathjax.
– Math Lover
Jul 26 at 17:09
How about looking at the old limits of the integration first, but not in terms of straight lines, but in terms of $x$ and $y$.
– mvw
Jul 26 at 17:24
math.meta.stackexchange.com/questions/5020/…
– John Wayland Bales
Jul 26 at 17:24
$u = x+y, v = frac yu = frac yx+y$ the line $x+y = 1$ transforms to $u = 1$ the line $y = 0 implies v = 0$ and $x = 0 implies v = 1$ along those respective boundaries.
– Doug M
Jul 26 at 17:28
@DougM thanks a lot.
– Balaji
Jul 26 at 17:45
add a comment |Â
2
You are a member of this forum for over two years. It's better if you learn how to write mathematical expressions in Mathjax.
– Math Lover
Jul 26 at 17:09
How about looking at the old limits of the integration first, but not in terms of straight lines, but in terms of $x$ and $y$.
– mvw
Jul 26 at 17:24
math.meta.stackexchange.com/questions/5020/…
– John Wayland Bales
Jul 26 at 17:24
$u = x+y, v = frac yu = frac yx+y$ the line $x+y = 1$ transforms to $u = 1$ the line $y = 0 implies v = 0$ and $x = 0 implies v = 1$ along those respective boundaries.
– Doug M
Jul 26 at 17:28
@DougM thanks a lot.
– Balaji
Jul 26 at 17:45
2
2
You are a member of this forum for over two years. It's better if you learn how to write mathematical expressions in Mathjax.
– Math Lover
Jul 26 at 17:09
You are a member of this forum for over two years. It's better if you learn how to write mathematical expressions in Mathjax.
– Math Lover
Jul 26 at 17:09
How about looking at the old limits of the integration first, but not in terms of straight lines, but in terms of $x$ and $y$.
– mvw
Jul 26 at 17:24
How about looking at the old limits of the integration first, but not in terms of straight lines, but in terms of $x$ and $y$.
– mvw
Jul 26 at 17:24
math.meta.stackexchange.com/questions/5020/…
– John Wayland Bales
Jul 26 at 17:24
math.meta.stackexchange.com/questions/5020/…
– John Wayland Bales
Jul 26 at 17:24
$u = x+y, v = frac yu = frac yx+y$ the line $x+y = 1$ transforms to $u = 1$ the line $y = 0 implies v = 0$ and $x = 0 implies v = 1$ along those respective boundaries.
– Doug M
Jul 26 at 17:28
$u = x+y, v = frac yu = frac yx+y$ the line $x+y = 1$ transforms to $u = 1$ the line $y = 0 implies v = 0$ and $x = 0 implies v = 1$ along those respective boundaries.
– Doug M
Jul 26 at 17:28
@DougM thanks a lot.
– Balaji
Jul 26 at 17:45
@DougM thanks a lot.
– Balaji
Jul 26 at 17:45
add a comment |Â
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2
You are a member of this forum for over two years. It's better if you learn how to write mathematical expressions in Mathjax.
– Math Lover
Jul 26 at 17:09
How about looking at the old limits of the integration first, but not in terms of straight lines, but in terms of $x$ and $y$.
– mvw
Jul 26 at 17:24
math.meta.stackexchange.com/questions/5020/…
– John Wayland Bales
Jul 26 at 17:24
$u = x+y, v = frac yu = frac yx+y$ the line $x+y = 1$ transforms to $u = 1$ the line $y = 0 implies v = 0$ and $x = 0 implies v = 1$ along those respective boundaries.
– Doug M
Jul 26 at 17:28
@DougM thanks a lot.
– Balaji
Jul 26 at 17:45