$\iint 1/(x+y)$ in region bounded by $x=0, y=0, x+y =1, x+y = 4$ using following transformation: $T(u,v) = (u - uv, uv)$. I want to make sure that my method is correct

Calculating the jacobian, I get $u$ since $x+y = u$, i get that $1 < u < 4$ Additionally, $v = y/y+x$ and thus at $x = 0$, we have $v = 0$ and at $y = 0$ we have $v = 1$.

So this the double integral is equivalent to $\iint(1/u)u dudv$ over the area $[1,4]$x$[0,1]$ which is easily solved

I am not sure about the method I employed, any help would be greatly appreciated.

  • $\begingroup$ Whats your region? {(x,y) | x > 0, y>0, x+y=1}? Or {(x,y) | x > 0, y>0, x+y=4}? Or [1,4]x[0,1] ? Please specify. $\endgroup$
    – Stockfish
    Oct 16, 2018 at 20:34
  • $\begingroup$ Yes, the method is totally correct. $\endgroup$
    – Hugo
    Oct 16, 2018 at 20:46

1 Answer 1


Yes the method is correct, indeed we are considering the following change of coordinates

  • $u=x+y \implies 1\le u \le 4$

  • $v=\frac{y}{x+y}\implies 0\le v \le 1$

indeed for any fixed value for $u=x+y\,$ we have that $y$ varies form $0$ to $u$ and therefore $v$ varies from $0$ to $1$

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the jacobian is

$$du\,dv=|J|dx\,dy=\begin{vmatrix}1&1\\-\frac{y}{(x+y)^2}&\frac{x}{(x+y)^2}\end{vmatrix}dx\,dy=\frac1udxdy \implies dx\,dy=u\,du\,dv$$

and therefore

$$\iint_D \frac1{x+y}dxdy=\int_1^4\int_0^1\frac1u\cdot u\,du\,dv$$


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