I am having a bit of trouble with the following question:
Given a region $D$ in the first quadrant bounded by $y = \sqrt{x}$, $y=2\sqrt{x}$, $x^2 + y^2 = 1$ and $x^2 + y^2 = 4$, evaluate:
$$\iint_D \frac{2x^2 + y^2}{xy}dA$$
I first set $u = \frac{y^2}{x}$ and $v = x^2 + y^2$, then tried expressing $x$ and $y$ in terms of $u$ and $v$, but this made computing the Jacobian very difficult. I thus resorted to implicit differentiation to calculate the Jacobian:
\begin{align*} u &= \frac{y^2}{x} \\ v &= x^2 + y^2 \\ \\ u \frac{\partial x}{\partial u} &= 2y \frac{\partial y}{\partial u} \\ u \frac{\partial x}{\partial v} &= 2y \frac{\partial y}{\partial v} \\ 0 &= 2x \frac{\partial x}{\partial u} + 2y \frac{\partial y}{\partial u} \\ 1 &= 2x \frac{\partial x}{\partial v} + 2y \frac{\partial y}{\partial v} \\ \\ \frac{\partial x}{\partial u} &= 0 \\ \frac{\partial y}{\partial u} &= 0 \\ \frac{\partial x}{\partial v} &= \frac{1}{2x + \frac{y^2}{x}} \\ \frac{\partial y}{\partial v} &= \frac{y}{4x^2 + y^2} \\ \end{align*}
I then found the Jacobian to be $\frac{\partial x}{\partial u}\frac{\partial y}{\partial v} - \frac{\partial x}{\partial v}\frac{\partial y}{\partial u} = 0$ which seems very odd considering the entire double integral would become $0$.
Is there something wrong with the implicit differentation, or my approach in general?
Thank you!
You can compute \begin{equation} \renewcommand{\arraystretch}{1.5} \frac{d \left(u , v\right)}{d \left(x , y\right)} = \left|\begin{array}{cc}{u}_{x}&{u}_{y}\\ {v}_{x}&{v}_{y} \end{array}\right| = \left|\begin{array}{cc}\displaystyle -\frac{{y}^{2}}{{x}^{2}}&\frac{2 y}{x}\\ \displaystyle 2 x&2 y \end{array}\right| =-2 y \frac{{y}^{2}+2 {x}^{2}}{{x}^{2}}\end{equation}
Hence
\begin{equation}\renewcommand{\arraystretch}{1.5} \frac{d \left(x , y\right)}{d \left(u , v\right)} = \frac{1}{\frac{d \left(u , v\right)}{d \left(x , y\right)}} = \frac{{x}^{2}}{{-2} y \left({y}^{2}+2 {x}^{2}\right)}\end{equation}
This is the jacobian that you need.