Cubic!

The polar equation of the curve is $1 = r(\cos^3\theta + \sin^3\theta)$ for $0 \le \theta \le \tfrac12\pi$ . Thus the desired area is $\begin{aligned} A & = \; \tfrac12\int_0^{\frac12\pi} \frac{d\theta}{(\sin^3\theta + \cos^3\theta)^2} \; = \; \tfrac12\int_0^{\frac12\pi} \frac{\sec^6\theta}{(1 + \tan^3\theta)^2}\,d\theta \\ & = \; \tfrac12\int_0^\infty \frac{(1 + x^2)^2}{(1 + x^3)^2}\,dx \; = \; \tfrac1{18}\int_0^\infty\left(\frac{2}{1+x} + \frac{1+x}{1 - x + x^2}\right)^2\,dx \\ & = \; \tfrac{1}{18}\int_0^\infty \left(\frac{4}{(1+x)^2} + \frac{4}{1 - x + x^2} + \frac{(1+x)^2}{(1 - x + x^2)^2}\right)\,dx \; = \; \tfrac1{18}\int_0^\infty \left(\frac{4}{(1+ x)^2} + \frac{5}{1 - x + x^2} + \frac{3x}{(1 - x + x^2)^2}\right)\,dx \\ & = \; \tfrac{1}{18}\left[-\frac{4}{1+x} + \frac{10}{\sqrt{3}}\tan^{-1}\big(\tfrac{2x-1}{\sqrt{3}}\big) - \frac{3}{2(x^2-x+1)}\right]_0^\infty + \tfrac{1}{12}\int_0^\infty \frac{dx}{(x^2 - x + 1)^2} \; = \; \tfrac{11}{36} + \tfrac{10}{27\sqrt{3}}\pi + \tfrac{1}{12}\int_0^\infty \frac{dx}{(x^2 - x + 1)^2} \end{aligned}$ The substitution $x = \tfrac12(1 + \sqrt{3}\tan\phi)$ now gives $\begin{aligned} A & = \; \tfrac{11}{36} + \tfrac{10}{27\sqrt{3}}\pi + \tfrac{2}{9\sqrt{3}}\int_{-\frac16\pi}^{\frac12\pi} \cos^2\phi\,d\phi \; = \; \tfrac{11}{36} + \tfrac{10}{27\sqrt{3}}\pi + \tfrac{1}{9\sqrt{3}}\int_{-\frac16\pi}^{\frac12\pi} (\cos2\phi + 1)\,d\phi \\ & = \; \tfrac{11}{36} + \tfrac{10}{27\sqrt{3}}\pi + \tfrac{1}{9\sqrt{3}}\Big[\tfrac12\sin2\phi + \phi\Big]_{-\frac16\pi}^{\frac12\pi} \; = \; \tfrac13 + \tfrac{4}{9\sqrt{3}}\pi \end{aligned}$ so that $p = \tfrac13$ and $q = \tfrac{16}{243}$ . Thus $4\sqrt{\tfrac{p}{q}} = \boxed{9}$ .