$\iint_D(x^2-y^2)dxdy$ with D enclosed by $y=\frac2x$, $y=\frac4x$, $y=x$, $y=x-3$?

Question

I have been presented with the following problem: Calculate the double integral $$\iint_D(x^2-y^2)dxdy$$ where D is the area enclosed by the curves $y=\frac2x$, $y=\frac4x$, $y=x$, and $y=x-3$.

Here's a visualisation of the area D: https://i.stack.imgur.com/Y7ewp.png

I've been trying to use various substitutions, but I always fail in finding the new area or what x and y should be substituted with.

I'm kind of stuck and would like a pointer in what I should substitute (or if I should even you substitution to begin with). Thanks in advance!

Answer 1

Start with a graph of the region $D$

The region $D$ is closed by $$y=x, \quad y=x-3,\quad y=\frac{4}{x},\quad y=\frac{2}{x}$$ Rewriting with the change of variables (this is a transformation one-to-one with Jacobian not null) $$\underbrace{x-y}_{u}=0,\quad \underbrace{x-y}_{u}=3,\quad \underbrace{yx}_{v}=4,\quad \underbrace{yx}_{v}=2, \quad x\geqslant 0, y\geqslant 0$$ we have $$D^{*}=\{(u,v)\in \mathbb{R}^{2}:0\leqslant u\leqslant 3, 2\leqslant v\leqslant 4 \}$$

The Jacobian of the change of variables is given by,

$$\boxed{\left|\frac{\partial (x,y)}{\partial (u,v)}\right|=\begin{vmatrix}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix}}$$ But in this case first we can calculate $$\left|\frac{\partial (u,v)}{\partial (x,y)}\right|=\begin{vmatrix}\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{vmatrix}=\begin{vmatrix} 1 & -1 \\ y & x\end{vmatrix}=x+y$$ Using the fact $$\boxed{\left| \frac{\partial (x,y)}{\partial (u,v)}\frac{\partial (u,v)}{\partial (x,y)}\right|=1 \implies \left|\frac{\partial (x,y)}{\partial (u,v)}\right|=\frac{1}{\left| \frac{\partial (u,v)}{\partial (x,y)}\right|}}$$

Hence, $$\left|\frac{\partial(x,y)}{\partial (u,v)}\right|=\frac{1}{x+y}$$

Using the change of variables theorem in a double integral $$\boxed{\iint_{D}f(x,y)\, {\rm d}A=\iint_{D^{*}}f(x(u,v),y(u,v))\left|\frac{\partial (x,y)}{\partial (u,v)}\right|{\rm d}A^{*}}$$ Therefore, \begin{align*} \iint_{D}(x^{2}-y^{2})\, {\rm d}A&=\iint_{D}(x-y)(x+y)\, {\rm d}A,\\ &=\iint_{D^{*}}u\cdot (x+y)\frac{1}{x+y}{\rm d}A^{*},\\ &=\iint_{D^{*}}u\, {\rm d}A^{*},\\ &=\int_{2}^{4}\int_{0}^{3}u{\rm d}u{\rm d}v,\\ &=9. \end{align*}

Answer 2

The domain you require to integrate into is quasi-cartesian in the sense you may change coordinates and convert it to a rectangle. Since you have a line and a hyperbolic functions, lets call $y - x$ as $u$ and $xy$ as v. From here, you can follow the steps given up section "Change of Variables for a Double Integral" on the link https://tutorial.math.lamar.edu/classes/calciii/changeofvariables.aspx . I hope it suffices.

Answer 3

Let $u=y-x$ and $v=xy $. Then, the integrand becomes $x^2-y^2=u\sqrt{u^2+4v}$ with the Jacobian $ \frac1{\sqrt{u^2+4v}}$. The integral is thus \begin{align}\int_2^4 \int_0^3 u \ du dv= 9 \end{align}