Is the surface of a vase an easy integral

2 Pi Integrate[(2+Sin[x])(Sqrt[1+((Cos[x])^2)]),{x,0,2 Pi}]

No, it is not.

Problem 17 in Section 19.3

{(x^2),(y^2),(z^2)}

{a+d Cos[θ]Sin[ϕ],b+d Sin[θ]Sin[ϕ],c+d Cos[ϕ]}

Feed sphere into the vector field

FullSimplify[{(x^2),(y^2),(z^2)}/.{x-> a+d Cos[θ]Sin[ϕ],y-> b+d Sin[θ]Sin[ϕ],z-> c+d Cos[ϕ]}]

FullSimplify[Cross[D[{a+d Cos[θ]Sin[ϕ],b+d Sin[θ]Sin[ϕ],c+d Cos[ϕ]},θ],D[{a+d Cos[θ]Sin[ϕ],b+d Sin[θ]Sin[ϕ],c+d Cos[ϕ]},ϕ]]]

- (d^2) Sin[ϕ]{ Cos[θ] Sin[ϕ],Sin[θ] Sin[ϕ],Cos[ϕ]}

FullSimplify[Expand[{(((a+d Cos[θ] Sin[ϕ]))^2),(((b+d Sin[θ] Sin[ϕ]))^2),(((c+d Cos[ϕ]))^2)}.{-(d^2) Cos[θ] ((Sin[ϕ])^2),-(d^2) Sin[θ] ((Sin[ϕ])^2),-(d^2) Cos[ϕ] Sin[ϕ]}]]

-Integrate[Integrate[-(d^2) Sin[ϕ] (Cos[ϕ] (((c+d Cos[ϕ]))^2)+((a^2) Cos[θ]+(b^2) Sin[θ]) Sin[ϕ]+2 d (a ((Cos[θ])^2)+b ((Sin[θ])^2)) ((Sin[ϕ])^2)+(d^2) (((Cos[θ])^3)+((Sin[θ])^3)) ((Sin[ϕ])^3)),{θ,0,2 Pi}],{ϕ,0,Pi}]