Integral in hyperbolic coordinates all.
My homework problem is the following:
Define $D=\{(x,y)\mid x,y>0, 1\leq x^2-y^2\leq 9, 2\leq xy\leq4 \}$.. For a continuous function $f:D\rightarrow\mathbb{R}$, use the hyperbolic coordinates from Exercise 7 to show that
$$
\int_D[x^2+y^2]dxdy=8
$$
In exercise 7, the function we defined was, for $x,y>0$,
$$
\Phi(x,y)=(x^2-y^2,xy)
$$
And we proved that this is a smooth change of variables with
$$
\det{D\Phi}=\begin{vmatrix}2x&y\\-2y&x\end{vmatrix}=2x^2+2y^2
$$
So what I've tried is, by the change of variables theorem,
$$
\int_D[x^2+y^2]dxdy=\int_1^9\int_2^4\left(\left((x^2-y^2)^2+(xy)^2\right)(2x^2+2y^2) \right)dxdy=\frac{19391968}{7}\neq 8
$$
I know why this is wrong, but I'm not sure how to do it. I should have some $u=x^2-y^2$, $v=xy$ and integrate from $u=1$ to $u=9$ and $v=2$ to $v=4$, but I'm not quite sure how to fill in all the details, including how to get my integral and $\det(D\Phi$ in terms of $u,v$.
Thanks
 A: You are applying the change of variables theorem backwards. (It may help to imagine the one-variable case: if you want to compute
$$
\int_0^3 x\sin(x^2) dx
$$
Then if you let $u = x^2$, then $du =2x dx$ and the integral transforms into
$$
\int_0^9 \sin(u)* x\text{ }du = \frac{1}{2}\int_0^9du.
$$
Let's call your new coordinates $u$ and $v$, so $u = x^2 + y^2$ and $v = xy$. Your determinant calculation shows that
$$
du dv = (2x^2+2y^2)dxdy.
$$
(again, look at the one-variable case to convince yourself that this is what it's showing you and not the other way around).
So your integrand should simplify to
$$
\frac{du dv}{2}
$$
which is not so bad!
A: Let $\cal{O}=\{(x,y)\in\Bbb{R}: x>0, \,y>0\}$ and $\Phi:\mathcal{O}\to\Bbb{R}^2$ defined by $\Phi(x,y)=(x^2-y^2,xy)$ for $(x,y)\in\cal{O}$. It is clear that the mapping $\Phi(x,y)=(u,v)$ with $u=x^2-y^2$ and $v=xy$ is both continuously differentiable and one-to-one. Also, at each point $(x, y)$ in $\cal{O}$
$$
\operatorname{det}D\Phi(x,y)=
\operatorname{det}\pmatrix{2x & -2y\\ y &x}=2(x^2+y^2)\neq0
$$
so that the derivative matrix $D\Phi(x,y)$ is invertible. Thus, the mapping  $\Phi$ is a smooth change of variables.
Using the Inverse Function Theorem it follows that the inverse mapping $\Phi^{-1}:\Phi(\cal{O})\to\Bbb{R}^2$ is a smooth change of variables on the open subset $\Phi(\cal{O})$ of $\Bbb{R}^2$.
Define $$D=\{(x,y)\in\Bbb{R}^2∣x,y>0,1≤x^2−y^2≤9,\;2≤xy≤4\}$$ and also define $$K=\Phi(D)=\{(u,v)\in\Bbb{R}^2 ∣1≤u≤9,\;2≤v≤4\}\subset{\cal{O}}.$$
Then $\Phi^{-1}(K)$ is a Jordan domain with the property that for the continuous function $f:D=\Phi^{-1}(K)\to\Bbb{R}$ defined by $f(x,y)=x^2+y^2$ the following integral transformation formula holds:
$$
\int_{D=\Phi^{-1}(K)} f(x,y)\operatorname{d}x\operatorname{d}y=\int_{K}f\left(\Phi^{-1}(u,v)\right) \left|\det\Phi^{-1}(u,v)\right|\operatorname{d}u\operatorname{d}v
$$
with $\left|\det\Phi^{-1}(u,v)\right|=\left|\det\Phi(x,y)\right|^{-1}=\frac{1}{2(x^2+y^2)}$.
Observing that $$\left(\Phi^{-1}(u,v)\right) \left|\det\Phi^{-1}(x,y)\right|=(x^2+y^2)\cdot\frac{1}{2(x^2+y^2)}=\frac{1}{2}$$ we have
$$
\int_{D} (x^2+y^2)\operatorname{d}x\operatorname{d}y=\int_1^9\int_2^4\frac{1}{2}\operatorname{d}u\operatorname{d}v=\frac{1}{2}(9-1)(4-2)=8.
$$
