Change of variables leading to a messy Jacobian

Question

I'm trying to solve the following problem:

$$\int_E \frac{y}{x}d(x,y)$$

Where E is bounded by $x^2-y^2=1$, $x^2-y^2=4$, $y=0$, and $y=\frac{x}{2}$. I've tried using the transformation $u=x^2-y^2$ and $v=\frac{y}{x}$ but this leads to a really messy Jacobian. Am I missing a better transformation?

Answer 1

I decided to use a different transformation, letting $u=\sqrt{x^2-y^2}$ and keeping $v=\frac{y}{x}$. The bounds are $1 \leq u^2 \leq 4$ and $0 \leq v \leq \frac{1}{2}$. Then the bounds for $u$ are $1 \leq u \leq 2$.

This gives $x=\sqrt{\frac{u^2}{1-v^2}}$ and $y=v\sqrt{\frac{u^2}{1-v^2}}$, while the integrand simplifies nicely to $v$.

The partials are then: $$\frac{\delta x}{\delta u}=\frac{1}{\sqrt{1-v^2}}, \frac{\delta x}{\delta v}=\frac{uv}{\sqrt{1-v^2}^3}, \frac{\delta y}{\delta u}=\frac{v}{\sqrt{1-v^2}}, \frac{\delta y}{\delta v}=\frac{u}{\sqrt{1-v^2}^3}$$ The Jacobian comes out to $\frac{u}{1-v^2}$, so the transformed integral is:

$$\int_0^{\frac{1}{2}}\int_1^2 \frac{uv}{1-v^2}dudv$$

If my math is right, this comes out to $-\frac{3}{4}ln\frac{3}{4}$.

Answer 2

Your process looks right: define the change of variables, determine new bounds of integration and calculate the Jacobian. But of course I didn't verify the details of the calculation.

I still maintain that your initial choice for the change of variables is the better one, with no square roots etc. Here's how I'd approach the problem. We let $u = x^2 - y^2$ and $v = \dfrac{y}{x}$. Then, $1 \leq u \leq 4$ and $0 \leq v \leq \dfrac{1}{2}$. Now, you're supposed to calculate the determinant: \begin{align} \det \dfrac{\partial(x,y)}{\partial(u,v)} \end{align} (this is simply the $2 \times 2$ matrix of partial derivatives, evaluated at the appropriate point). Now, by the inverse function theorem, this is in fact equal to \begin{align} \dfrac{1}{\det \dfrac{\partial(u,v)}{\partial(x,y)}} \end{align} (again, you have to interpret the notation correctly and evaluate at the right points). Well, let's calculate this: \begin{align} \left( \det \dfrac{\partial(u,v)}{\partial(x,y)} \right)^{-1} &= \left( \det \begin{pmatrix} 2x & -2y \\ -\frac{y}{x^2} & \frac{1}{x}\end{pmatrix} \right)^{-1} \\ &= \left( 2 - \dfrac{2y^2}{x^2}\right)^{-1} \\ &= \dfrac{1}{2(1-v^2)}. \end{align}

So, the integral now becomes: \begin{align} \int_E \dfrac{y}{x} \, d(x,y) &= \int_{[1,4] \times [0, \frac{1}{2}]} v \cdot \left| \det \dfrac{\partial(x,y)}{\partial(u,v)} \right|\, d(u,v) \\ &= \int_{[1,4] \times [0, \frac{1}{2}]} \dfrac{v}{2(1-v^2)} \, d(u,v) \\\\ &= \quad \ddot{\smile} \text{ you already have the correct answer} \end{align}

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It is much quicker to go with the original change of variables, and then apply the inverse function theorem. Because as in this example (and many others) it is usually easier to calculate the determinant's reciprocal rather than calculating the inverse of your change of variables, then differentiating, and then calculating the determinant.