$ \int_{S_{c}} \underset{\bar{}}{F}.d \underset{\bar{}}{S}\;$ where $S_{c}$ is the curved part of the truncated cone $x^2 + y^2 = (a-z)^2 , 0≤z≤(1/2)a$
Where $\underset{\bar{}}{F}$ = $(2x + y^2)\underset{\bar{}}{i} + (y+xz)\underset{\bar{}}{j} + z\underset{\bar{}}{k}$. Use a normal to $S_{c}$ that points away from the z axis.
My first thought is to split the surface area $S_{c}$ of the truncated cone into 3 parts and use the divergence theorem. $S_{1}$ = the top, $S_{2}$ = the bottom, $S_{3}$= the sides(middle part). $S_{c}$ is the union $S_{1}$∪$S_{2}$∪$S_{3}$.
For the divergence theorem you need a double integral but the question gives a single integral. How can i overcome this?
First, choose a surface internal parameters: $$ P = \{ (u,v) \in ([0..2\pi],[0..(1/2)a]) \} $$
Then describe the surface based on those parameters: $$ S(a)(u,v) = \{ (x,y,z)\in ((a-v)*\cos(u),(a-v)*\sin(u),v) \}$$
Calculating surface area: $$ A = \pi r_0 s_0 - \pi r_1 s_1 $$ where $r_0$ and $s_0$ are the radius and slant height of the cone, and the $r_1$ and $s_1$ are the radius and slant height of the (invisible) cone removed from the larger cone.
Then the integral is simply average of all samples picked from the surface multiplied by the surface area: $$ \int_{S_c}{F\cdot dS} = A \times\frac{\sum_{n=1}^{k}F(S(a)(u_n,v_n))}{k}, k>10000 $$, where $(u_n,v_n)$ are randomly picked surface point from $P$.
Note: edited the cone surface area formula to more correct one.