1
$\begingroup$

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

1 Answers 1

4

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

  • 1
    +1. Note that under this change of variables, you have $\left(10x^2+39x y+14 y^2\right)^2=u^2v^2$.2012-12-30
  • 0
    You need the Jacobian of $(x,y)$ with respect to $(u,v)$.2012-12-30
  • 0
    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? /Alexander2012-12-30
  • 0
    Yes. $\ddot\smile$2012-12-30
  • 0
    Nice smiley, and ++++answer!2013-02-27