Calculating $ \iint_D y\,dx\,dy$

Question

Calculating the integral: $$ \iint_D y\,dx\,dy$$
$D$ is a closed domain whose boundary is $x=0,\ y=0,\ \left(\frac{x}{a}+\frac{y}{b}\right)^3=\frac{xy}{c^2}.\quad (x,y\geq0)$

Fisrt it's easy to transform the D into $\left(x+y\right)^3=\frac{ab}{c^2}xy$, then I want to change $x+y$ into $m$, $xy$ into $n$, but in that case, this integral turns into an improper integral. And also, under this change, I'm not sure the $y$ should be turned into $\frac{m+\sqrt{m^2-4n}}{2}$ or $\frac{m-\sqrt{m^2-4n}}{2}$. Hence I'm stucked. If there is any better method of element changing?

Answer 1

Set $a > 0,\,b > 0,\,c > 0$ and assigned the domain:

$$ D := \left\{ (x,\,y) \in \mathbb{R}^2 : x \ge 0, \, y \ge 0, \, \left(\frac{x}{a} + \frac{y}{b}\right)^3 \le \frac{x\,y}{c^2} \right\} $$

writing the following code in Wolfram Mathematica 12.2:

{a, b, c} = {6, 6, 6};

RegionPlot[x > 0 && y > 0 && (x/a + y/b)^3 < x y/c^2,
           {x, 0, 1}, {y, 0, 1}, FrameLabel -> {x, y}]

we obtain:

from which we can get a more concrete idea of what it's, even if it isn't strictly necessary.

At this point, if we want to calculate the double integral:

$$ I := \iint_D y\,\text{d}x\,\text{d}y $$

there are essentially two ways:

• calculate it directly through a substitution of the type $u = \frac{x}{a} + \frac{y}{b}$ and $v = \frac{x\,y}{c^2}$, which by presenting two pairs of $x,\,y$ implies having to break the double integral into the algebraic sum of two double integrals;

• calculate it indirectly by applying the Gauss-Green theorem in the plane, i.e. through a line integral along the boundary of the domain $\partial D$ parameterized with equations of the type $x = \rho(\theta)\,\cos\theta$ and $y = \rho(\theta)\,\sin\theta$, where $\rho(\theta) > 0$ is computable by substituting these parametric equations in the respective Cartesian equation of the lobe.

Regardless of the chosen path, some calculations have to be made, although personally I would prefer the second path, and the result is:

$$ I = \frac{a^4\,b^5}{840\,c^6}\,. $$

I hope it's clear enough, good luck! ^_^