Evaluate $iint_D x sin (y -x^2) ,dA$.

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Let $D$ be the region, in the first quadrant of the $x,y$ plane, bounded by the curves $y = x^2$, $y = x^2+1$, $x+y=1$ and $x+y=2$. Using an appropriate change of variables, compute the integral



$$iint_D x sin (y -x^2) ,dA.$$



I've been reviewing for an upcoming test and this problem was recommended to do for study -- I just can't get it. I've tried many changes of variables and nothing has worked. I would really appreciate a hint or a solution. Thanks in advance!




calculus integration multivariable-calculus change-of-variable




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  • 2




    Try u=x+y and v= y-x^2. The new limits will be 1<u<2 and 0<v<1. Don't forget to compute the jacobian
    – zokomoko
    yesterday










  • The Jacobian ends up being $frac12x + 1$, leaving the $x$ term in the integral -- and solving for $x$ here in terms of $u$ and $v$ is horrid and leaves you with a terrible looking integral. :/
    – SAWblade
    yesterday










  • Just guessing, but I don't think this integral has a nice closed form. It could be a typo in the text.
    – Dylan
    yesterday














up vote
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Let $D$ be the region, in the first quadrant of the $x,y$ plane, bounded by the curves $y = x^2$, $y = x^2+1$, $x+y=1$ and $x+y=2$. Using an appropriate change of variables, compute the integral



$$iint_D x sin (y -x^2) ,dA.$$



I've been reviewing for an upcoming test and this problem was recommended to do for study -- I just can't get it. I've tried many changes of variables and nothing has worked. I would really appreciate a hint or a solution. Thanks in advance!




calculus integration multivariable-calculus change-of-variable




share|cite|improve this question

















  • 2




    Try u=x+y and v= y-x^2. The new limits will be 1<u<2 and 0<v<1. Don't forget to compute the jacobian
    – zokomoko
    yesterday










  • The Jacobian ends up being $frac12x + 1$, leaving the $x$ term in the integral -- and solving for $x$ here in terms of $u$ and $v$ is horrid and leaves you with a terrible looking integral. :/
    – SAWblade
    yesterday










  • Just guessing, but I don't think this integral has a nice closed form. It could be a typo in the text.
    – Dylan
    yesterday












up vote
1
down vote

favorite
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up vote
1
down vote

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Let $D$ be the region, in the first quadrant of the $x,y$ plane, bounded by the curves $y = x^2$, $y = x^2+1$, $x+y=1$ and $x+y=2$. Using an appropriate change of variables, compute the integral



$$iint_D x sin (y -x^2) ,dA.$$



I've been reviewing for an upcoming test and this problem was recommended to do for study -- I just can't get it. I've tried many changes of variables and nothing has worked. I would really appreciate a hint or a solution. Thanks in advance!




calculus integration multivariable-calculus change-of-variable




share|cite|improve this question













Let $D$ be the region, in the first quadrant of the $x,y$ plane, bounded by the curves $y = x^2$, $y = x^2+1$, $x+y=1$ and $x+y=2$. Using an appropriate change of variables, compute the integral



$$iint_D x sin (y -x^2) ,dA.$$



I've been reviewing for an upcoming test and this problem was recommended to do for study -- I just can't get it. I've tried many changes of variables and nothing has worked. I would really appreciate a hint or a solution. Thanks in advance!





calculus integration multivariable-calculus change-of-variable





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited yesterday









Michael Hardy

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204k23185460









asked yesterday









SAWblade

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  • 2




    Try u=x+y and v= y-x^2. The new limits will be 1<u<2 and 0<v<1. Don't forget to compute the jacobian
    – zokomoko
    yesterday










  • The Jacobian ends up being $frac12x + 1$, leaving the $x$ term in the integral -- and solving for $x$ here in terms of $u$ and $v$ is horrid and leaves you with a terrible looking integral. :/
    – SAWblade
    yesterday










  • Just guessing, but I don't think this integral has a nice closed form. It could be a typo in the text.
    – Dylan
    yesterday












  • 2




    Try u=x+y and v= y-x^2. The new limits will be 1<u<2 and 0<v<1. Don't forget to compute the jacobian
    – zokomoko
    yesterday










  • The Jacobian ends up being $frac12x + 1$, leaving the $x$ term in the integral -- and solving for $x$ here in terms of $u$ and $v$ is horrid and leaves you with a terrible looking integral. :/
    – SAWblade
    yesterday










  • Just guessing, but I don't think this integral has a nice closed form. It could be a typo in the text.
    – Dylan
    yesterday







2




2




Try u=x+y and v= y-x^2. The new limits will be 1<u<2 and 0<v<1. Don't forget to compute the jacobian
– zokomoko
yesterday




Try u=x+y and v= y-x^2. The new limits will be 1<u<2 and 0<v<1. Don't forget to compute the jacobian
– zokomoko
yesterday












The Jacobian ends up being $frac12x + 1$, leaving the $x$ term in the integral -- and solving for $x$ here in terms of $u$ and $v$ is horrid and leaves you with a terrible looking integral. :/
– SAWblade
yesterday




The Jacobian ends up being $frac12x + 1$, leaving the $x$ term in the integral -- and solving for $x$ here in terms of $u$ and $v$ is horrid and leaves you with a terrible looking integral. :/
– SAWblade
yesterday












Just guessing, but I don't think this integral has a nice closed form. It could be a typo in the text.
– Dylan
yesterday




Just guessing, but I don't think this integral has a nice closed form. It could be a typo in the text.
– Dylan
yesterday















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