# $\iint 1/(x+y)$ in region bounded by $x=0, y=0, x+y =1, x+y = 4$

$$\iint 1/(x+y)$$ 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.

• 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. Oct 16, 2018 at 20:34
• Yes, the method is totally correct.
– Hugo
Oct 16, 2018 at 20:46

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 $$1$$ 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

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