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 bounded by $\sqrt x+\sqrt y=\sqrt a$, $x=0$, $y=0$ (with $a>0$), using the substitution $x=u\cos^4v$, $y=u\sin^4v$.

Explanation of Method

The curved boundary $\sqrt x+\sqrt y=\sqrt a$ is awkward in Cartesian form, but under this substitution, $\sqrt x+\sqrt y=\sqrt u\,(\cos^2v+\sin^2v)=\sqrt u$, so the curve collapses to the single condition $u=a$. The two axes $x=0$ and $y=0$ become $v=\frac\pi2$ and $v=0$ respectively, so the whole region becomes a rectangle in $(u,v)$.

Transform each of the three boundaries into $(u,v)$-coordinates.
Compute the Jacobian of $x=u\cos^4v$, $y=u\sin^4v$.
Write the transformed integral over the resulting rectangle.

Step 1 · Transform the boundaries

For $u>0$ in the first quadrant: $x=0 \Rightarrow \cos v=0 \Rightarrow v=\frac\pi2$; $y=0 \Rightarrow \sin v=0 \Rightarrow v=0$.

For the curved boundary:

$$ \sqrt x+\sqrt y = \sqrt u\cos^2v+\sqrt u\sin^2v = \sqrt u\left(\cos^2v+\sin^2v\right) = \sqrt u = \sqrt a \ \Longrightarrow\ u=a. $$

So the region $D'$ is the rectangle $0\leqslant u\leqslant a$, $0\leqslant v\leqslant\frac\pi2$.

Step 2 · Compute the Jacobian

$$ \frac{\partial x}{\partial u}=\cos^4v,\quad \frac{\partial x}{\partial v}=-4u\cos^3v\sin v, \quad \frac{\partial y}{\partial u}=\sin^4v,\quad \frac{\partial y}{\partial v}=4u\sin^3v\cos v. $$

$$ \frac{\partial(x,y)}{\partial(u,v)} = \cos^4v\cdot4u\sin^3v\cos v - \sin^4v\cdot(-4u\cos^3v\sin v) = 4u\cos^3v\sin^3v\left(\cos^2v+\sin^2v\right) = 4u\cos^3v\sin^3v. $$

$$ dx\,dy = 4u\cos^3v\sin^3v\,du\,dv. $$

Step 3 · Write the transformed integral

$$ \iint_D z(x,y)\,dx\,dy = \int_0^{\pi/2}\int_0^a 4u\cos^3v\sin^3v\cdot z\!\left(u\cos^4v,\,u\sin^4v\right)du\,dv. $$

Additional Notes

This substitution is the natural analogue of polar coordinates for the "astroid-like" curve $\sqrt x+\sqrt y=\sqrt a$: just as $x=r\cos^2\theta$, $y=r\sin^2\theta$ would collapse $\sqrt x+\sqrt y=\sqrt r$, using fourth powers here collapses the constraint into the single condition $u=a$ cleanly, at the cost of a slightly more complex (but still fully explicit) Jacobian.

Result

The limits of integration and transformed integral are:

$$ 0\leqslant u\leqslant a, \qquad 0\leqslant v\leqslant \frac\pi2, $$ $$ \iint_D z(x,y)\,dx\,dy = \int_0^{\pi/2}\int_0^a 4u\cos^3v\sin^3v\cdot z\!\left(u\cos^4v,u\sin^4v\right)du\,dv. $$