How to find the integral by changing the coordinates?

Question

Let R be the region in the first quadrant where $$3 \geq y-x \geq 0$$ $$5 \geq xy \geq2$$ Compute $$\int_A (x^2-y^2)\,dx\,dy.$$

I tried to use $ u= y-x, v= xy$ as my change of coordinates, but then I don't know how to solve it. Can someone help me?

Answer 1

Hint: try the tranformation

$$v:=x+y,~~u:=x-y$$

or $x=\frac{u+v}{2}$, $y=\frac{v-u}{2}$. Now the integrand becomes the product $uv$ (up to scalar coming from the determinant of the Jacobian transformation) and the domain of integration is

$$A=\{(u,v): -u\geq 0, -u\leq 3, v^2-u^2\geq 8, v^2-u^2\leq 20 \}= \{(u,v): -3\leq u\leq 0, v^2-u^2\geq 8, v^2-u^2\leq 20 \} $$

The integral can now be computed.

Answer 2

For the Jacobian, use this fact that:

$$\frac{\partial(x,y)}{\partial(u,v)}=\frac{1}{\frac{\partial(u,v)}{\partial(x,y)}}$$ provided $\frac{\partial(x,y)}{\partial(u,v)}\neq 0$.

Answer 3

$$ \mbox{With}\quad x \equiv u + v\quad\mbox{and}\quad y \equiv u - v \quad\Longrightarrow\quad {\partial\left(x,y\right) \over \partial\left(u,v\right)} = -2 $$

\begin{align} & \\[6mm]?& = \left.\vphantom{\Huge A} \int_{0}^{\infty}{\rm d}x\int_{0}^{\infty}{\rm d}y\,\left(x^{2} - y^{2}\right) \right\vert_{3\ \geq\ y - x\ >\ 0 \atop \vphantom{\Large A}5\ \geq\ xy\ >\ 2} = -2\left.\vphantom{\Huge A} \int_{-\infty}^{\infty}{\rm d}v\int_{-\infty}^{\infty}{\rm d}u\,4uv \right\vert_{3\ \geq\ -2v\ >\ 0 \atop {\vphantom{\Large A^{A}}5\ \geq\ u^{2}\ -\ v^{2}\ >\ 2\atop u\ >\ \left\vert v\right\vert}} \\[4mm]&= -8\int_{-3/2}^{0}{\rm d}v\ v\int_{\sqrt{v^{2} + 2}}^{\sqrt{v^{2} + 5}}{\rm d}u\ u = -8\int_{-3/2}^{0}{\rm d}v\ v\left({v^{2} + 5 \over 2} - {v^{2} + 2 \over 2}\right) = -12\,\left\lbrack -\,{\left(-3/2\right)^{2} \over 2}\right\rbrack \\[4mm]&={\Large {27 \over 2}} \end{align}