Change of Variables — Question 10

Question 10 – Solution

Strategy. Use total distance along the coordinate-sum direction and the fraction allocated to $x$.

Step 1: Inverse and geometry Since $u=x+y>0$, $$x=uv,\qquad y=u(1-v).$$ Nonnegativity gives $0\le v\le 1$, while the two diagonal boundaries give $1\le u\le 3$.

Step 2: Jacobian $$\frac{\partial(x,y)}{\partial(u,v)} =\begin{vmatrix}v&u\\1-v&-u\end{vmatrix}=-u, \qquad dA=u\,du\,dv.$$ Hence the integrand $1/(x+y)=1/u$ cancels the area scale: $$I=\int_1^3\int_0^1\frac 1u(u)\,dv\,du =\boxed{2}.$$

Step 3: Interpretation and check On a segment $x+y=u$, the quantity $v=x/u$ is the fraction of the total coordinate sum assigned to $x$; it runs once from the $y$-axis to the $x$-axis. Thus the inverse is one-to-one in the interior, and $u>0$ confirms the absolute Jacobian used above.