i'd like to know how can i change the following domain to a simple one $$D=\left\{(x,y)\mid \frac12\le x+y\le1 ;x^3\le y\le 4x^3\right\}$$ regarding the integral: $$\iint\limits_D\frac{x+3y}{x^4}e^{y/x^3} \,dx \, dy$$ It's pretty obvious that i should use $u=\frac{y}{x^3}$, so that $1\le u\le4$, but i'm not so sure about the other one. Can anyone help?
Parameter change
2
$\begingroup$
integration
integral-domain
jacobian
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0If you think that your $u$ is obvious, then why is not the $v=x+y$ obvious? If you explain that, then I think you will get better help. – 2017-02-04
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2I added proper spacing before $dx$ and $dy$ and changed $\displaystyle\int\int$ to $\displaystyle\iint$ and $|$ to $\mid$ and $[ \cdots ]$ to $\{ \cdots \},$ etc., but one thing in the MathJax code was really weird: $x^3{\le}y$ rather than $x^3\le y$ and $1{\le}u$ rather than $1\le u.$ Putting those braces around the binary relation symbol prevented proper spacing. – 2017-02-04
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0@mickep Because i don't see what would be the inverse function in that case. – 2017-02-04
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1But maybe (hint hint) you will not need that. Try to calculate the jacobian, and you will see how the integrand was chosen. – 2017-02-04
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0Right, thanks.. – 2017-02-04