Calculate $\iint_D \frac{y}{x} $

Question

I am trying to calculate $\iint_D \frac{y}{x} $ where D is restricted by 3<_3x+y<_5 and by 4x<_y<_7x. I started by switching variables so that s=3x+y and t=$\frac{y}{x}$. However that gives me a very weird Jacobian determinant; $\frac{x^3}{3x^3+y}$, so maybe I shouldn't set t to $\frac{y}{x}$? Does anyone have any suggestion on how to go about this? Here are my calculations this far:

Thanks in advance!

/Nick

Answer 1

What about defining $s = 3x + y$ and $t = x$? You don't get constant integration limits, but it is simple enough, as the integral becomes

$$ \int_3^5 \int_{s/10}^{s/7} |J| \dfrac{s- 3t}{t} dt ds, $$

with $J=\begin{pmatrix} 1 & 0 \\ -3 & 1 \\ \end{pmatrix}$. The integral then becomes

$$ \int_3^5 \int_{s/10}^{s/7} \left(\frac st -3 \right) dt\,ds = 8 \left(\log \left(\frac{10}{7}\right)-\frac{9}{70}\right). $$

obs: Final result computed with wolfram, but the primitive functions are immediate.

Answer 2

Pierre's answer is a good one, but for completeness I'll continue on the track you started. From your picture you have the derivatives in the wrong entries of your matrix so the Jacobian is not correct. Instead it should be

$$J = \frac{x^2}{3x+y}$$

To isolate $x$ substitute $y=xt$ into $s$

$$3x+xt = s \implies x =\frac{s}{3+t}$$

which means the Jacobian is

$$\frac{\frac{s^2}{(3+t)^2}}{s} = \frac{s}{(3+t)^2} $$

and the integral becomes

$$\int_3^5\int_4^7 \frac{st}{(3+t)^2}dtds = \left(\int_3^5 sds\right)\left(\int_4^7\frac{1}{3+t}-\frac{3}{(3+t)^2}dt\right)$$

$$=8\cdot \left(\log\frac{10}{7} -\frac{9}{70}\right)$$