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




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    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














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




share|cite|improve this question

















  • 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












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




share|cite|improve this question













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





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 26 at 17:32









Joseph Eck

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asked Jul 26 at 17:07









Balaji

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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












  • 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















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