Instead of just getting 4/Binomial[2/p,1/p] (or equivalent but messy 𝝘 functions) from Assuming[p > 0, 2 Integrate[(1 - Abs[x^p])^(1/p), {x, -1, 1}]], Assuming[p > 0, Area[ImplicitRegion[Abs[x^p] + Abs[y^p] < 1, {x, y}]]] does weird things that don't always work: 1/3 Area[ImplicitRegion[Abs[x]^(1/3)+Abs[y]^(1/3)<1,{x,y}]] 1/2 2/3 2/3 Area[ImplicitRegion[Abs[x]^(2/3)+Abs[y]^(2/3)<1,{x,y}]] 1 2 3/2 (2 2^(2/3) √π Gamma[2/3])/Gamma[1/6]-(2^(2/3) √(3 π) Gamma[5/6])/Gamma[-(2/3)] 2 π 5/2 - 5 √(2/(5+√5)) π Gamma[7/5])/(Gamma[-2/5] Gamma[4/5]))+(2^(1/5) √π Gamma[7/5])/Gamma[9/10] 3 3 3^(1/3) AppellF1[4/3,-(1/3),-(1/3),7/3,-(1/(-1+(-1)^(2/3))),1/(1+(- 1)^(1/3))] 2 √3 (2^(1-1/Sqrt[3]) Sqrt[\[Pi]/3] Gamma[1/(2 Sqrt[3])])/Gamma[1/6 (3+Sqrt[3])] 4 (8 Gamma[5/4]^2)/√π 9/2 (2^(5/9) √π Gamma[11/9])/Gamma[13/18]+(π Csc[2 π/9] Gamma[11/9])/(Gamma[4/9] Gamma[7/9]) 5 (2 2^(3/5) √π Gamma[6/5])/Gamma[7/10] Notice the AppellF1 for p=3, which must equal 4/Binomial[2/3,1/3], unbeknownst to FullSimplify. Is it beknownst to anybody? --rwg