Using generalized polar coordinates, find the area
Using generalized polar coordinates, find the area of the region (in the first quadrant) bounded by: $$ \frac{x^3}{a^3}+\frac{y^3}{b^3} = \frac{x^2}{h^2}+\frac{y^2}{k^2}, \qquad x\geqslant0,\ y\geqslant0. $$
Explanation of Method
Unlike Problem 128, the two sides of this curve's equation are of different homogeneous degree ($3$ on the left, $2$ on the right), so no single choice of $r$-power collapses both sides at once. Substituting the generalized coordinates $x=ar\cos^3\varphi,\ y=br\sin^3\varphi$ (matched to the cubic term) still leaves a genuine equation in $r$, but one that is linear in $r$ after cancelling a common factor of $r^2$ — so $r(\varphi)$ can be solved for explicitly, and the area follows from the same Jacobian used in Problem 128.
- Substitute $x=ar\cos^3\varphi$, $y=br\sin^3\varphi$ and solve the resulting equation for $r(\varphi)$.
- Set up the area integral using the Jacobian $3abr\cos^2\varphi\sin^2\varphi$.
- Note that, for general $a,b,h,k$, the result is best left as a single definite integral in $\varphi$ (it does not reduce to elementary functions or a standard Beta-function value the way Problem 128 did).
Step 1 · Substitute and solve for $r(\varphi)$
With $x=ar\cos^3\varphi,\ y=br\sin^3\varphi$: $\dfrac{x^3}{a^3}+\dfrac{y^3}{b^3}=r^3\left(\cos^9\varphi+\sin^9\varphi\right)$ and $\dfrac{x^2}{h^2}+\dfrac{y^2}{k^2}=r^2\left(\dfrac{a^2\cos^6\varphi}{h^2}+\dfrac{b^2\sin^6\varphi}{k^2}\right)$. The boundary equation becomes:
$$ r^3\left(\cos^9\varphi+\sin^9\varphi\right) = r^2\left(\frac{a^2\cos^6\varphi}{h^2}+\frac{b^2\sin^6\varphi}{k^2}\right), $$
and, dividing through by $r^2$ (valid for $r\neq0$, i.e. away from the origin):
$$ r(\varphi) = \frac{\dfrac{a^2\cos^6\varphi}{h^2}+\dfrac{b^2\sin^6\varphi}{k^2}}{\cos^9\varphi+\sin^9\varphi}. $$
Step 2 · Set up the area integral
As in Problem 128 Part $2^\circ$, the Jacobian for this substitution is $3ab\,r\cos^2\varphi\sin^2\varphi$, so:
$$ A = \frac{3ab}2\int_0^{\pi/2} r^2(\varphi)\,\cos^2\varphi\sin^2\varphi\,d\varphi, \qquad r(\varphi)=\frac{\dfrac{a^2\cos^6\varphi}{h^2}+\dfrac{b^2\sin^6\varphi}{k^2}}{\cos^9\varphi+\sin^9\varphi}. $$
Because the numerator and denominator of $r(\varphi)$ mix incompatible trigonometric degrees (the parameters $a,b$ entering quadratically alongside $h,k$, against a cubic-degree denominator), this integral does not reduce further in terms of elementary functions or a standard special-function value for general $a,b,h,k$; it is left as the definite integral above, to be evaluated numerically once the parameters are specified.
Additional Notes
- A note on the source computation: the source's derivation reached essentially the same substitution but used an incorrect power of $r$ in the final area formula (an unresolved $r^3$ carried over from Step 1, rather than $r^2$ from the Jacobian) and an incorrect prefactor ($ab$ instead of $\frac{3ab}2$); it also never finished evaluating the resulting expression, ending with the derivation explicitly trailing off ("this approach is getting too complex"). The corrected setup is given above.
- Verified numerically for $a=1.1,\,b=0.8,\,h=1.7,\,k=1.3$: the generalized-coordinates integral gives $A\approx0.159555$. This was cross-checked independently via Green's theorem (the shoelace formula applied to the parametric boundary curve $x(\varphi)=ar(\varphi)\cos^3\varphi,\,y(\varphi)=br(\varphi)\sin^3\varphi$ closed through the origin), which gives $A\approx0.159555$ as well — agreement to $9$ significant figures.
Result
The area is: