By introducing new variables u and v in place of x and y

By introducing new variables $u$ and $v$ in place of $x$ and $y$, determine the limits of integration for: $$ \iint_D z(x,y)\,dx\,dy, $$ where $D$ is the triangle bounded by $x=0$, $y=0$, $x+y=a$, using the substitution $x=\dfrac{u(a-v)}{a}$, $y=\dfrac{uv}{a}$.

Explanation of Method

Here $u$ plays the role of "distance along the ray from the origin toward the hypotenuse," and $v$ parametrizes where along that ray's endpoint on the hypotenuse the point sits, since $x+y=\frac{u(a-v)}{a}+\frac{uv}{a}=u$ regardless of $v$. So the hypotenuse $x+y=a$ becomes simply $u=a$, while the two legs $x=0$ and $y=0$ become $v=a$ and $v=0$ — turning the triangle into a square.

Transform each of the three boundaries into $(u,v)$-coordinates.
Compute the Jacobian of $x=\frac{u(a-v)}{a}$, $y=\frac{uv}{a}$.
Write the transformed integral over the resulting square.

Step 1 · Transform the boundaries

$$ x=0 \Rightarrow u(a-v)=0 \Rightarrow v=a \ (\text{for }u\neq0), \qquad y=0 \Rightarrow uv=0 \Rightarrow v=0 \ (\text{for }u\neq0). $$

For the hypotenuse:

$$ x+y = \frac{u(a-v)}{a}+\frac{uv}{a} = \frac{u(a-v)+uv}{a} = \frac{ua}{a} = u = a. $$

Checking the vertices confirms this: $(0,0)\mapsto u=0$; $(a,0)\mapsto(u,v)=(a,0)$; $(0,a)\mapsto(u,v)=(a,a)$. So the region $D'$ is the square $0\leqslant u\leqslant a$, $0\leqslant v\leqslant a$.

Step 2 · Compute the Jacobian

$$ \frac{\partial x}{\partial u}=\frac{a-v}{a},\quad \frac{\partial x}{\partial v}=-\frac ua, \qquad \frac{\partial y}{\partial u}=\frac va,\quad \frac{\partial y}{\partial v}=\frac ua. $$

$$ \frac{\partial(x,y)}{\partial(u,v)} = \frac{a-v}{a}\cdot\frac ua - \left(-\frac ua\right)\cdot\frac va = \frac{u(a-v)}{a^2}+\frac{uv}{a^2} = \frac{ua}{a^2}=\frac ua. $$

$$ dx\,dy = \frac ua\,du\,dv. $$

Step 3 · Write the transformed integral

$$ \iint_D z(x,y)\,dx\,dy = \int_0^a\int_0^a \frac ua\,z\!\left(\frac{u(a-v)}{a},\frac{uv}{a}\right)dv\,du. $$

Additional Notes

Here $v$ can be read as playing the role of a "$y$-coordinate along the hypotenuse" and $u$ as the scaling factor from the origin out to that point — a barycentric-style parametrization of the triangle, which is why it converts a triangle into a perfect square rather than a slanted parallelogram.

Result

The limits of integration and transformed integral are:

$$ 0\leqslant u\leqslant a, \qquad 0\leqslant v\leqslant a, $$ $$ \iint_D z(x,y)\,dx\,dy = \int_0^a\int_0^a \frac ua\,z\!\left(\frac{u(a-v)}{a},\frac{uv}{a}\right)dv\,du. $$