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I'm trying to solve the following integral: $\iint\limits_D (10x^2+39xy+14y^2)^2 dxdy$ bounded by the lines: $2x+7y=1$, $2x+7y=-1$, $5x+2y=3$ and $5x+2y=-3$.

Now I'm new to calculus and do not know how to solve this kind of problem other than when say $x=a+by, x=c+by$, $y=0$ and $y=2$.

It doesn't look that advanced but I cannot find any similar examples. I'm expected to use Mathematica for the calculation.

Can I rotate the axes to simplify it? Or where do I start?

Thanks! Alexander

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1 Answer 1

Hint: Take $u=2x+7y,v=5x+2y$ so you have the range of them as $$u|_{-1}^{1}, v|_{-3}^3$$ Now change the integrand according to $u$ and $v$ and calculate the Jacobian of them respect to $x$ and $y$ as well. See this link for more http://www.math24.net/change-of-variables-in-double-integrals.html

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+1. Note that under this change of variables, you have $\left(10x^2+39x y+14 y^2\right)^2=u^2v^2$. –  yohBS Dec 30 '12 at 8:49
    
You need the Jacobian of $(x,y)$ with respect to $(u,v)$. –  Christian Blatter Dec 30 '12 at 9:00
    
Thank you guys!, I calculate the Jacobian as $\begin{vmatrix}du/dx& du/dy\\dv/dx& dv/dy\end{vmatrix} = \begin{vmatrix}2& 7\\5& 2\end{vmatrix} = -31$. Thus I have the integral: $\int_{-3}^3 \!\int_{-1}^1 \! u^2v^2 \, 31 dudv$ which I evaluate to $372$, is that correct? /Alexander –  Alexander Dec 30 '12 at 16:04
    
Yes. $\ddot\smile$ –  Babak S. Dec 30 '12 at 16:07
    
Nice smiley, and ++++answer! –  amWhy Feb 27 '13 at 0:46

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