Using generalized polar coordinates x=ar cos^()alpha phi

Using generalized polar coordinates $x=ar\cos^\alpha\varphi$, $y=br\sin^\alpha\varphi$, find the area of the region bounded by:
$1^\circ$  $\left(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}\right)^2=\dfrac{x^2}{h^2}+\dfrac{y^2}{k^2}$;
$2^\circ$  $\dfrac{x^3}{a^3}+\dfrac{y^3}{b^3}=1$  ($x\geqslant0,\ y\geqslant0$).

Part $1^\circ$

Step 1 · Substitute $x=ar\cos\varphi$, $y=br\sin\varphi$

Then $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=r^2$ and $\dfrac{x^2}{h^2}+\dfrac{y^2}{k^2}=r^2\left(\dfrac{a^2\cos^2\varphi}{h^2}+\dfrac{b^2\sin^2\varphi}{k^2}\right)$, so the boundary equation becomes:

$$ r^4 = r^2\left(\frac{a^2\cos^2\varphi}{h^2}+\frac{b^2\sin^2\varphi}{k^2}\right) \quad\Longrightarrow\quad r^2=\frac{a^2\cos^2\varphi}{h^2}+\frac{b^2\sin^2\varphi}{k^2}\quad(r\neq0). $$

Step 2 · Set up and evaluate the area integral

With Jacobian $abr$, the area element is $dx\,dy=ab\,r\,dr\,d\varphi$, so:

$$ A_1 = \int_0^{2\pi}\!\!\int_0^{r(\varphi)} ab\,r\,dr\,d\varphi = \frac{ab}2\int_0^{2\pi}r^2(\varphi)\,d\varphi = \frac{ab}2\int_0^{2\pi}\left(\frac{a^2\cos^2\varphi}{h^2}+\frac{b^2\sin^2\varphi}{k^2}\right)d\varphi. $$

Since $\displaystyle\int_0^{2\pi}\cos^2\varphi\,d\varphi=\int_0^{2\pi}\sin^2\varphi\,d\varphi=\pi$:

$$ A_1 = \frac{ab}2\left(\frac{\pi a^2}{h^2}+\frac{\pi b^2}{k^2}\right) = \frac{\pi ab}2\left(\frac{a^2}{h^2}+\frac{b^2}{k^2}\right). $$

Part $2^\circ$

Step 3 · Substitute $x=ar\cos^3\varphi$, $y=br\sin^3\varphi$

Then $\dfrac{x^3}{a^3}+\dfrac{y^3}{b^3}=r^3\left(\cos^9\varphi+\sin^9\varphi\right)$, so the boundary curve $\dfrac{x^3}{a^3}+\dfrac{y^3}{b^3}=1$ becomes:

$$ r^3=\frac1{\cos^9\varphi+\sin^9\varphi} \quad\Longrightarrow\quad r^2=\left(\cos^9\varphi+\sin^9\varphi\right)^{-2/3}. $$

The Jacobian here is $3ab\,r\cos^2\varphi\sin^2\varphi$, so the area element is $dx\,dy=3ab\,r\cos^2\varphi\sin^2\varphi\,dr\,d\varphi$, and integrating over $r\in\left[0,r(\varphi)\right]$, $\varphi\in\left[0,\frac\pi2\right]$ (the first-quadrant arc):

$$ A_2 = \frac{3ab}2\int_0^{\pi/2}\frac{\cos^2\varphi\sin^2\varphi}{\left(\cos^9\varphi+\sin^9\varphi\right)^{2/3}}\,d\varphi. $$

Step 4 · Reduce to a Beta-function integral via a direct substitution

This trigonometric integral does not evaluate directly by elementary means, but the area is far more easily found by returning to Cartesian coordinates and substituting $t=x/a$, $s=y/b$, so that $\frac{A_2}{ab}$ is the area of $t^3+s^3\leqslant1$ in the first quadrant. With $s=(1-t^3)^{1/3}$ and then $u=t^3$ ($t=u^{1/3}$, $dt=\frac13u^{-2/3}du$):

$$ \frac{A_2}{ab} = \int_0^1(1-t^3)^{1/3}\,dt = \frac13\int_0^1 u^{-2/3}(1-u)^{1/3}\,du = \frac13B\!\left(\frac13,\frac43\right). $$

So:

$$ A_2 = \frac{ab}3\,B\!\left(\frac13,\frac43\right) = \frac{ab}3\cdot\frac{\Gamma\!\left(\frac13\right)\Gamma\!\left(\frac43\right)}{\Gamma\!\left(\frac53\right)}. $$

Additional Notes

• Part $1^\circ$ matches the source's derivation exactly.
• A note on Part $2^\circ$: the source's derivation reached the correct integral setup $A_2=\frac{3ab}2\int_0^{\pi/2}\frac{\cos^2\varphi\sin^2\varphi}{(\cos^9\varphi+\sin^9\varphi)^{2/3}}\,d\varphi$ but then attempted to evaluate it directly in terms of $\varphi$, using an incorrect power of $r$ (a leftover $r^3$ from an earlier step, rather than $r^2$) and never reduced it to a resolved closed form. Recognizing this as the area under a "super-ellipse" arc of degree $3$ and switching to the Cartesian substitution $t=x/a,\,s=y/b$ above gives the clean Beta-function closed form.
• Verified numerically for $a=1.3,\,b=0.9$: the generalized-coordinates integral gives $A_2\approx1.033484$; the Cartesian Beta-function formula gives $\frac{ab}3B\!\left(\frac13,\frac43\right)\approx1.033484$; both match to six significant figures. Numerically, $B\!\left(\frac13,\frac43\right)\approx2.64980$, so $A_2\approx0.88332\,ab$.