Using an appropriate change of variables, calculate

Using an appropriate change of variables, calculate the area of the loop bounded by the curve: $$ \left(\frac xa+\frac yb\right)^3 = \frac{xy}{c^2}. $$

Step 1 · Substitute and transform the curve

With $u=\dfrac xa+\dfrac yb,\ v=\dfrac xa-\dfrac yb$ (so $x=\frac{a(u+v)}2,\ y=\frac{b(u-v)}2$, Jacobian $\frac{ab}2$, exactly as in Problem 131), $xy=\dfrac{ab(u^2-v^2)}4$, so the curve $\left(\frac xa+\frac yb\right)^3=\dfrac{xy}{c^2}$ becomes:

$$ u^3 = \frac{ab(u^2-v^2)}{4c^2} \quad\Longrightarrow\quad 4c^2u^3 = ab(u^2-v^2). $$

Step 2 · Solve for $v^2$ and find the loop's range

Solving for $v^2$:

$$ v^2 = u^2-\frac{4c^2}{ab}u^3 = u^2\left(1-\frac{4c^2}{ab}u\right). $$

Writing $k=\dfrac{4c^2}{ab}$, real $v$ requires $1-ku\geqslant0$, i.e. $0\leqslant u\leqslant\dfrac1k$ (taking the branch through the origin, which is the physically relevant loop since $u=\frac xa+\frac yb\geqslant0$ in the first quadrant). At $u=0$, $v=0$; at $u=\frac1k$, $v=0$ again — so the curve $v=\pm u\sqrt{1-ku}$ traces a closed loop over $u\in\left[0,\frac1k\right]$, symmetric about $v=0$.

Step 3 · Set up and evaluate the area integral

By symmetry in $v$:

$$ A = 2\int_0^{1/k}\int_0^{u\sqrt{1-ku}}\frac{ab}2\,dv\,du = ab\int_0^{1/k} u\sqrt{1-ku}\,du. $$

Substituting $w=ku$ ($u=w/k,\,du=dw/k$):

$$ A = \frac{ab}{k^2}\int_0^1 w\sqrt{1-w}\,dw = \frac{ab}{k^2}\,B\!\left(2,\frac32\right) = \frac{ab}{k^2}\cdot\frac{\Gamma(2)\Gamma\left(\frac32\right)}{\Gamma\left(\frac72\right)} = \frac{ab}{k^2}\cdot\frac4{15}. $$

Since $k=\dfrac{4c^2}{ab}$, $k^2=\dfrac{16c^4}{a^2b^2}$, so:

$$ A = \frac{4ab}{15}\cdot\frac{a^2b^2}{16c^4} = \frac{a^3b^3}{60c^4}. $$

Additional Notes

• A note on the source computation: the source's derivation reached the correct relation $4c^2u^3=ab(u^2-v^2)$, but then divided both sides by $u^2$ incorrectly, writing "$4c^2u=ab(u^2-v^2)$" (which is not what dividing $4c^2u^3=ab(u^2-v^2)$ by $u^2$ actually gives). This error led to a self-detected sign contradiction mid-derivation ("this gives a negative value under the square root... I made an error") and an attempted recovery that integrated over the wrong branch of $u$ (negative $u$, which is unbounded, not a closed loop, since $u=\frac xa+\frac yb<0$ is not where the true first-quadrant loop lies), ultimately giving the structurally different and incorrect answer $\dfrac{\pi c^3}{\sqrt{ab}}$. The correct relation $v^2=u^2\left(1-\frac{4c^2}{ab}u\right)$, worked through above, gives a genuine closed loop over $u\in\left[0,\frac{ab}{4c^2}\right]$ and a purely rational closed form.
• Verified independently two ways for $a=1.3,\,b=0.9,\,c=0.7$: (1) direct quadrature of $ab\int_0^{1/k}u\sqrt{1-ku}\,du$ gives $A\approx0.111177$; (2) the closed form $\dfrac{a^3b^3}{60c^4}\approx0.111177$. Both match to $9$ significant figures. (For comparison, the source's formula $\frac{\pi c^3}{\sqrt{ab}}$ would give $\approx0.996$ for these parameters — off by roughly a factor of $9$.)