In the integral double integral over D z(x,y) dx dy
In the integral $\iint_D z(x,y)\,dx\,dy$, the region $D$ is the quarter disk $x^2+y^2\leqslant a^2$, $x>0$, $y>0$.
$1^{\circ}$ Map this region to a rectangle by a change of variables.
$2^{\circ}$ Map this region to an isosceles right triangle by a change of variables.
Explanation of Method
Polar coordinates $(r,\theta)$ immediately turn the quarter disk into the rectangle $[0,a]\times\left[0,\frac\pi2\right]$, since the curved boundary $r=a$ and the two straight edges $\theta=0,\frac\pi2$ are each a single coordinate line. To get a triangle instead, a further substitution is needed that keeps $r$ but replaces $\theta$ with a quantity that scales linearly with $r$ — this way the two straight edges of the quarter disk (which lie at fixed angle but variable radius) become two edges meeting at the origin, and the arc becomes the third (straight) edge.
- Part $1^\circ$: use ordinary polar coordinates $(r,\theta)$.
- Part $2^\circ$: starting from $(r,\theta)$, substitute $q=\dfrac{2r\theta}{\pi}$ in place of $\theta$, keeping $p=r$.
- Compute the Jacobian of each transformation.
Part $1^\circ$ · Mapping to a rectangle
Step 1 · Choose the transformation
Use polar coordinates: $u=r=\sqrt{x^2+y^2}$, $v=\theta=\arctan\dfrac yx$, i.e. $x=r\cos\theta$, $y=r\sin\theta$.
Step 2 · Transform the region
$x^2+y^2\leqslant a^2 \Rightarrow 0\leqslant r\leqslant a$; and $x>0,y>0 \Rightarrow 0<\theta<\frac\pi2$. So $D$ maps to the rectangle:
$$ 0\leqslant r\leqslant a, \qquad 0\leqslant\theta\leqslant\frac{\pi}{2}. $$
Step 3 · Jacobian
$$ \frac{\partial(x,y)}{\partial(r,\theta)} = \begin{vmatrix}\cos\theta&-r\sin\theta\\\sin\theta&r\cos\theta\end{vmatrix} = r, \qquad dx\,dy=r\,dr\,d\theta. $$
Step 4 · Transformed integral
$$ \iint_D z(x,y)\,dx\,dy = \int_0^{\pi/2}\int_0^a z(r\cos\theta,r\sin\theta)\cdot r\,dr\,d\theta. $$
Part $2^\circ$ · Mapping to an isosceles right triangle
Step 1 · Choose the transformation
Starting from the polar rectangle of Part $1^\circ$, apply the further substitution:
$$ p=r, \qquad q=\frac{2r\theta}{\pi} \quad\Longleftrightarrow\quad r=p,\ \ \theta=\frac{\pi q}{2p}. $$
In terms of the original $x,y$: $p=\sqrt{x^2+y^2}$, $q=\dfrac{2}{\pi}\sqrt{x^2+y^2}\arctan\dfrac yx$.
Step 2 · Transform the region
For each fixed $r=p\in[0,a]$, as $\theta$ runs from $0$ to $\frac\pi2$, $q=\frac{2r\theta}{\pi}$ runs from $0$ to $r=p$. So the rectangle $[0,a]\times\left[0,\frac\pi2\right]$ in $(r,\theta)$ becomes:
$$ 0\leqslant q\leqslant p\leqslant a, $$
which is the isosceles right triangle with vertices $(0,0)$, $(a,0)$, $(a,a)$ (legs of equal length $a$ along $q=0$ and $p=a$, hypotenuse along $q=p$).
Step 3 · Jacobian
First, in terms of $(r,\theta)$:
$$ \frac{\partial(p,q)}{\partial(r,\theta)} = \begin{vmatrix}1&0\\[2pt]\frac{2\theta}{\pi}&\frac{2r}{\pi}\end{vmatrix} = \frac{2r}{\pi}, \qquad dr\,d\theta = \frac{\pi}{2r}\,dp\,dq = \frac{\pi}{2p}\,dp\,dq. $$
Combining with $dx\,dy=r\,dr\,d\theta$ from Part $1^\circ$ (and $r=p$):
$$ dx\,dy = p\cdot\frac{\pi}{2p}\,dp\,dq = \frac{\pi}{2}\,dp\,dq. $$
Remarkably, the combined Jacobian is constant.
Step 4 · Transformed integral
$$ \iint_D z(x,y)\,dx\,dy = \frac{\pi}{2}\int_0^a\int_0^p z\!\left(p\cos\frac{\pi q}{2p},\,p\sin\frac{\pi q}{2p}\right)dq\,dp. $$
Additional Notes
- The constant Jacobian $\frac\pi2$ is a useful consistency check: the triangle has area $\frac{a^2}{2}$, so $\frac\pi2\cdot\frac{a^2}{2}=\frac{\pi a^2}{4}$, exactly the area of the original quarter disk.
- This is one of infinitely many valid transformations achieving the requested mapping; any substitution that linearly reparametrizes the angle $\theta$ in proportion to $r$ works equally well.
- Verified symbolically: computing $\dfrac{\partial(p,q)}{\partial(x,y)}$ directly from $p=\sqrt{x^2+y^2}$, $q=\frac2\pi\sqrt{x^2+y^2}\arctan\frac yx$ gives the constant value $\frac2\pi$, confirming $\left|\dfrac{\partial(x,y)}{\partial(p,q)}\right|=\dfrac\pi2$.
Result
The transformations are: