Double Integration with change of variables

I am having trouble with the following double integral:

$$\iint\limits_D(x^2+y^2) \;dA$$

where $D$ is given by the region enclosed by the curves

• $xy=1$
• $xy=2$
• $x^2-y^2 =1$
• $x^2-y^2 =2$

I have tried changing variables to $u=x^2+y^2$, $v=x^2-y^2$ (and other changes of variables) but the Jacobian looks grim.

The change of variables $u = xy$, $v = x^2 - y^2$ works. Check that it's one-to-one in the first quadrant. (I assume you want only the first quadrant region bounded by those curves.)

EDIT:

First we prove that the the transformation is one-to-one on the first quadrant. Let $x, y > 0$, and define $u$, $v$ as above. $x^2$ and $-y^2$ are the roots of the polynomial $f(T) = T^2 -vT - u^2$. Since one is positive and the other negative, $x^2$ and $-y^2$, hence $x$ and $y$, are uniquely determined by $u$ and $v$.

Moreover the image of the transformation is the entire half-plane $u > 0$. For given $u$ and $v$ with $u > 0$, write the polynomial $f(T)$ as above. Since $f(0) < 0$, $f$ has one negative root $-y^2$ and one positive root $x^2$ (for some $x, y > 0$), and we have $v = x^2 - y^2$ and $- u^2 = -x^2 y^2$. It follows that $u = xy$.

(Here is an alternative proof: the mapping $x + iy \mapsto v + 2iu$ is really the mapping $z \mapsto z^2$ from the first quadrant to the upper half-plane.)

The Jacobian is calculated as $$\frac{\partial(u,v)}{\partial(x,y)} = \left| \matrix{y & x \\ 2x & -2y} \right| = -2(x^2 + y^2).$$ Thus $du \, dv = 2(x^2 + y^2) \, dx \, dy$.

Now the image of $D$ under the transformation is the set defined by the inequalities $1 \leq u \leq 2$ and $1 \leq v \leq 2$, so $$\iint_D (x^2 + y^2) \, dx \, dy = \int_1^2 \int_1^2 \frac{1}{2} \, du \, dv = \frac{1}{2}.$$

Consider the variables \begin{align*} u = xy, && v = x^2-y^2. \end{align*} We obtain that $x=\frac{u}{y}$. Therefore: $$y^2=x^2-v,$$ $$y^2 -\frac{u^2}{y^2}=-v,$$ $$\frac{y^4-u^2}{y^2} =-v,$$ $$y^4+vy^2-u^2=0,$$ $$y^2 = \frac{-v\pm \sqrt{v^2-4(-u^2)}}{2} = \frac{-v \pm \sqrt{4u^2+v^2}}{2}.$$ From which we obtain: $$x^2=\frac{-v \pm \sqrt{4u^2+v^2}}{2}+v.$$ Thinking now on the integrand: $$x^2+y^2 = \frac{-v\pm \sqrt{4u^2+v^2}}{2}+\frac{-v\pm \sqrt{4u^2+v^2}}{2}+v = \pm\sqrt{4u^2+v^2}.$$ Last equation gives us 2 options for the integrand

$i)$$\displaystyle \frac{v+\sqrt{4u^2+v^2}}{2}+\frac{v+\sqrt{4u^2+v^2}}{2}-v=\sqrt{4u^2+v^2}.$

$ii)$$\displaystyle \frac{v-\sqrt{4u^2+v^2}}{2}+\frac{v-\sqrt{4u^2+v^2}}{2}-v= -\sqrt{4u^2+v^2}.$

Of course, the volume of the solid must be positive, hence we take the first case. Then we have that: $$\iint\limits_{D}(x^2+y^2)\;dA=\iint\limits_{S}\sqrt{4u^2+v^2}\;|\textbf{J}(T(u,v))|d\hat{A},$$ where $T(u,v)$ is the transform that takes us from $S$ to $D$. It is clear that $S$ is the rectangle: \begin{align*} S:\begin{cases} u=1\\ u=2\\ v=1\\ v=2 \end{cases}. \end{align*}

Now let's calculate the Jacobian: $$\textbf{J}(T)= \left| {\begin{array}{cc} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{array}}\right|. $$ Para calcular $\displaystyle \frac{\partial x}{\partial u}$ considere las variables $u=xy$ y $v=x^2-y^2$, de donde se obtiene que: \begin{align*} 0=\frac{\partial x}{\partial u}y+ \frac{\partial y}{\partial u}x, && 1 = 2x\frac{\partial x}{\partial u}-2y\frac{\partial y}{\partial u}.\\ \end{align*}

Multiplying the first equation times $2y$ and the second one times $x$ and adding the equations, we obtain: \begin{align*} 2y=2y^2\frac{\partial x}{\partial u} +2xy \frac{\partial y}{\partial u}, && 0=2x^2\frac{\partial x}{\partial u}-2xy\frac{\partial x}{\partial u}, \end{align*} $$2y = \frac{\partial x}{\partial u}(2y^2+2x^2),$$ $$\frac{\partial x}{\partial u} = \frac{2y}{2y^2+2x^2}.$$ In a similar fashion we obtain the rest of the partial derivatives: $$\textbf{J}(T)= \left| {\begin{array}{cc} \frac{y}{y^2+x^2} & \frac{x}{2x^2+2y^2}\\ \frac{x}{x^2+y^2} & \frac{-y}{2x^2+2y^2} \end{array}}\right| = -\frac{y^2}{2(y^2+x^2)^2}-\frac{x^2}{2(y^2+x^2)^2} = -\frac{y^2+x^2}{2(y^2+x^2)^2} = -\frac{1}{2\sqrt{4u^2+v^2}}. $$ Therefore, the original integral is now: $$\iint\limits_{S}\sqrt{4u^2+v^2}\left(\frac{1}{2\sqrt{4u^2+v^2}}\right)\;d\hat{A} = \iint\limits_{S}\frac{1}{2}\;d\hat{A}.$$ Because of Fubini's Theorem we can write: $$\int_1^2 \left(\int_1^2 \frac{1}{2}\;du\right)dv=\int_1^2 \left[\frac{1}{2}u\right]^2_1dv =\int_1^2 \frac{1}{2}\;dv =\boxed{\frac{1}{2}}.$$