I have the following double integral

$$ \iint_D xdxdy $$

where D is given by the inequalities

$$ x^2+xy+y^2 \le 4, x\ge 0 $$

In the solution given to the problem they apply this variable substitution when solving the integral:

$$ u = \frac{\sqrt{3}}{2}x $$ $$ v = \frac{x}2 + y $$

My question is how they arrive at this substitution? Is there a method for finding substitutions like these?

While I understand that this substitution works, there is no explanation for how they arrive at this substitution at all in my solution.

  • $\begingroup$ In my opinion, this change of variable doesn't really ease the resolution. It is more direct to use $y\in[\frac{-x-\sqrt{16-3x^2}}2, \frac{-x+\sqrt{16-3x^2}}2]$. $\endgroup$ – Yves Daoust Jun 2 '15 at 10:28

You can transform the region D by completing the square

$$x^2+xy+y^2=\frac{3}{4}x^2+\frac{1}{4}x^2+xy+y^2=(\frac{1}{2}x+y)^2+(\frac{\sqrt{3}}{2}x)^2\leq 4$$

Let $u = \frac{\sqrt{3}}{2}x, v=\frac{x}{2}+y$. And notice that $x\geq 0$ gives $u\geq 0$ too.

  • $\begingroup$ This is not the given change of variable. $\endgroup$ – Yves Daoust Jun 2 '15 at 10:16
  • $\begingroup$ @YvesDaoust: I said starting from $y$ would give the one the OP has. I am just giving the general way to do it. Thanks for telling me the reason though. $\endgroup$ – KittyL Jun 2 '15 at 10:16
  • $\begingroup$ Right, but it would have been nice to take into account the constraint $x\ge0$. $\endgroup$ – Yves Daoust Jun 2 '15 at 10:19
  • $\begingroup$ @YvesDaoust: You are right. Thanks. $\endgroup$ – KittyL Jun 2 '15 at 10:20
  • $\begingroup$ @KittyL: Thanks for the fast answer. I see how this works, but how do you get on the train of thought for finding this substitution? My first idea was to use polar coordinates, but that didn't get me anywhere... Is it just experience or is there a kind of method for it? Thanks! $\endgroup$ – haeger Jun 2 '15 at 10:43

We want to represent x as a function of y for the first constraint. So we complete to square with respect to x:

$$y^2+2(y)(x/2)+x^2/4+\frac 3 4x^2=(y+x/2)^2+\frac 3 4x^2<4$$

You don't have to use any change of variables, you could simply write y in terms of x


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