Evaluate double integral $\iint_R\sqrt \frac{x+y}{x-2y}dA$ using change of variables

Question

Evaluate the integral $$\iint_R\sqrt \frac{x+y}{x-2y}dA$$ where $R$ is the region bounded by $ y - \frac{x}{2} = 0, y = 0,x+y = 1.$

I know I need to change the variable to $u$ and $v$, since that is the related section to this problem, and also use the Jacobian, but I am not sure where to start. Any help is appreciated.

Answer 1

Here's probably how they wanted you to do it: let's use the transformation $u = x + y, v = x - 2y.$ Solving the system of equations for $x$ and $y$ gives us the reverse transformation $x = \frac13(2u + v), y = \frac13(u-v).$

In order to change variables like this, we have to transform the domain of integration as well. I'm going to do this by choosing $v$ to be bounded by two constants and $u$ be bounded by functions of $v.$ So $v$ will go from $0$ ($x - 2y = v = 0$) to $1.$ (at $(1,0)$) $u$ will now go from the line $y = \frac13(u-v) = 0$ up to $x + y = u = 1.$ So, rewriting our bounds of integration should be

$$\bar{R} = \{(u,v): v \leq u \leq 1, 0 \leq v \leq 1\}$$

Now the next thing we have to do to is calculate the determinant of the Jacobian of our transformation. We can do this as follows:

$$|J| = \begin{vmatrix}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{vmatrix} = \begin{vmatrix}\frac23 & \frac13 \\ \frac13 & -\frac13\end{vmatrix} = -\frac13$$

So, we can now rewrite our integral as follows:

$$\iint_R \sqrt{\frac{x+y}{x-2y}} dA = \iint_{\bar{R}} \sqrt{\frac{u}{v}} \cdot \left|-\frac{1}{3}\right| du dv = \frac13\int_0^1\int_v^1 \sqrt{\frac{u}{v}} du dv$$

We can now integrate as follows:

$$\frac13\int_0^1 v^{-\frac12} \int_v^1 u^{\frac12} du dv = \frac13\cdot\frac23\int_0^1 v^{-\frac12}\left(1 - v^{\frac32}\right) dv = \frac29 \int_0^1 v^{-1/2} - v dv$$

Noting that our integral is improper: (it always has been)

$$\lim_{a\to 0} \frac29 \int_a^1 v^{-\frac12} - v dv = \frac29 \lim_{a \to 0} 2\sqrt{v} - \frac12v^2\Bigg|_a^1 = \frac29 \lim_{a \to 0} \left(\frac32 - 2\sqrt{a} + \frac12a^2\right) = \boxed{\frac13}$$

Answer 2

Change the base to polar coordinates, i.e. $x=r\cos t$ and $ y= r\sin t$

\begin{align} &\iint_R\sqrt \frac{x+y}{x-2y}dxdy =\int_0^{\tan^{-1}\frac12}\int_0^{\frac1{\cos t +\sin t}} \sqrt \frac{\cos t+\sin t}{\cos t-2\sin t}\>rdr dt\\ &\hspace{15mm}=\int_0^{\tan^{-1}\frac12} \frac{\frac1{2(\cos t+\sin t)^2}} {\sqrt \frac {\cos t-2\sin t} {\cos t+\sin t}}dt =-\int_0^{\tan^{-1}\frac12} \frac {d\left(\frac{\cos t-2\sin t} {\cos t+\sin t}\right)}{6\sqrt \frac{\cos t-2\sin t} {\cos t+\sin t}} =\frac13 \end{align}