Peruvian marinera

Any constant lower than one will satisfy the relation of points to the region, and they will make a set of curves that lies below the curve $x^{1/4}+y^{4}=1$ itself. Inverting domain and image, we would get $x(y)=\left(1-y^{4} \right)^{4}, y \in [-1, 1]$ . From volumes of rotation theory we know that volume must be:

$\displaystyle V_R=\int_{a}^{b} \pi [f(t)]^{2} dt$ ,

where $I=[a,b]$ is the integration interval, and $f(t)$ is the function to be rotated around horizontal axis. Then:

$\displaystyle V_R=\int_{-1}^{1} \pi \left[ \left(1-y^{4} \right)^{4} \right]^{2} dy$

A glance at $x(y)$ tells us that it's an even function, and we are integrating over a symmetrical interval. Hence:

$\displaystyle V_R=2 \pi \int_{0}^{1} \left(1-y^{4} \right)^{8} dy$

Expanding the binomial we get a polynomial, which can be applied to the lower and upper extremes. Thus:

$\displaystyle V_R=2 \pi \int_{0}^{1} 1-8y^{4}+28y^{8}-56y^{12}+70y^{16}-56y^{20}+28y^{24}-8y^{28}+y^{32}dy$

$\displaystyle V_R=2 \pi \left(y-\frac{8}{5}y^{5}+\frac{28}{9}y^{9}-\frac{56}{13}y^{13}+\frac{70}{17}y^{17}-\frac{56}{21}y^{21}+\frac{28}{25}y^{25}-\frac{8}{29}y^{29}+\frac{1}{33}y^{33} \right) \biggr \rvert _0^1$

Therefore:

$\boxed{\displaystyle V_R=2 \pi \left(1-\frac{8}{5}+\frac{28}{9}-\frac{56}{13}+\frac{70}{17}-\frac{56}{21}+\frac{28}{25}-\frac{8}{29}+\frac{1}{33} \right)}$

My calculus teacher would accept this answer, but on Brilliant you need to give a numerical answer. A calculator easily computes it as $\boxed{V_R=3.323}$