Calculate the area

We are going to use symmetry, and definite integration (using walli's) to get the answer

Area between curves $=4×$ Area between curves in first quadtant

Parametric form of curve is $(\sqrt{3}\cos^{4}\theta,\sqrt{3}\sin^{4}\theta)$

$dy=4\sqrt{3}\sin^{3}\theta \cos\theta d\theta$

$=4×$ (Area under circle-Area under curve)

$=4×(\frac{3\pi}{4}-\displaystyle\int_{0}^{\pi}{2} 12\cos^{5}\theta \sin^{3}\theta d\theta$

$=3\pi-48×\frac{(4×2)(2)}{8×6×4×2}$ $\color{#3D99F6}{\text{(Using walli's formula)}}$

$=3\pi-2$

$\lambda=3,\phi=2$

$\lambda×\phi=\boxed{6}$

This is the graph of the given system of equations:

graph where the blue colour represents the circle $x^2+y^=3$ and the red colour represents the curve $\displaystyle\,\sqrt{|x|}+\sqrt{|y|}=3^{1/4}$ .

Now the shaded region is the required area which is given by $4\times\left(\displaystyle\int_{0}^{\sqrt{3}} \sqrt{3-y^2}-(3^{1/4}-\sqrt{|y|})^2\, \mathrm{d}x\,\right)=\,\displaystyle\boxed{3\pi-2}$ . So required value is $3\times2=\boxed{6}$ .

NOTE: A better phrasing would be $\left(\lambda\times\pi+\phi\right)$ to avoid ambiguity which would yield the answer -6 since it is unstated whether $\phi>0$ or $\phi<0$ .

i think all solution involve a bit complex integrals.

The Area required = area of circle -4*area of curve in first quadrant

The latter is readily evaluated by taking curve in first quadrant

And using integral of ydx!