Question

Φ1 = ∬S1 (-2u^6cos(v) - 2u^3sin(v)) du dv

= ∫(0->1) ∫(0->2π) (-2u^6cos(v) - 2u^3sin(v)) dv du

Answer 1

`\begin{align}\sf\:\Phi_1 &= \iint_(S_1) (-2u^6\cos(v) - 2u^3\sin(v)) \, du \, dv \\ &= \int_(0)^(1) \int_(0)^(2\pi) (-2u^6\cos(v) - 2u^3\sin(v)) \, dv \, du \end{align} \\`