I got this question in an online assessment where you weren't allowed to use a programming language or online resources.
The question defines a squircle by $y^4 + x^4 = R^4$. It's area would be $4\int^R_0 (R^4-x^4)^\frac{1}{4}dx$.
I know the answer uses Gamma functions, so how could you even be expected to work this out? What about if you assume you have a means of calculating Gamma functions, how would you get to this point?
We can compute the special case $R = 1$ without loss of generality and then multiply the result by $R^2$ to get the general case.
This means we want to compute
$$I = \int_{x=0}^1 (1-x^4)^{1/4} \, dx.$$
The choice $$t = x^4, \quad x = t^{1/4}, \quad dx = \frac{1}{4} t^{-3/4} \, dt,$$ yields $$I = \frac{1}{4} \int_{t=0}^1 (1-t)^{1/4} t^{-3/4} \, dt.$$ Then using the beta function identity $$B(a,b) = \int_{t=0}^1 t^{a-1} (1-t)^{b-1} \, dt = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)},$$ we obtain $$I = \frac{1}{4} \frac{\Gamma(1/4)\Gamma(5/4)}{\Gamma(3/2)} = \frac{2}{\sqrt{\pi}} \Gamma(5/4)^2.$$
So the general area will be $$4R^2 I = \frac{8}{\sqrt{\pi}} \Gamma(5/4)^2 R^2.$$ The value of $\Gamma(5/4)$ has no further simplification.
As an exercise, what is the enclosed area for the family of such curves given by $$x^{2n} + y^{2n} = 1$$ for positive integers $n$?
