In a problem from my multivariable integration class, i've reached this problem. I will thank any comment with advice or answer. The problem asks me to calculate the integral $$\iint_S(x)dy\wedge dz+(x+y)dz\wedge dx+(x^2+2z)dx\wedge dy$$ Being $S$ the surface of the solid $V$ limited by: $$S_1=\{(x,y,z)\in\mathbb{R}:2x^2+y^2=4z\},$$ $$S_2=\{(x,y,z)\in\mathbb{R}:x^2+2z=2\}.$$ I'm told to solve twice it using two different methods: direct integration and Gauss' Theorem (divergence). I've started trying with Gauss' Theorem, but i don't really get if i'm getting it correctly. The theorem tells that (under certaing region and surface conditions that this problem verifies) given $V$ a solid limited by a closed surface $S$, $N$ the normal vector, and $F=(P,Q,R)$ a vectorial field of class $C^1$, $$\iint_{\partial V}F=\iint_S(F\cdot N)d\sigma = \iiint_V\text{div}(F)dxdydz.$$ Being $\text{div}(F)=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}$.
I've started trying to find $P,Q,R$ in my example, but I'm not sure if that's possible, at least with my actual knowledge. What can I do to find my $\text{div}(F)$?.
For the direct integration part, I have no idea about how to do it. I specially struggle finding the integration limits.
Bounty edit: I need a step-by-step solution for both methods: direct integration and Gauss' Theroem (Divergence Theorem).
i) $ D = \{ (x,y,0)| 4x^2+y^2\leq 4\} $ Here $S_i$ is a graph surface over $D$. And assume that they enclosed a solid body $T$
ii) Consider a vector field $ X$ so that we have 1-form $ w(V) = \langle X,V \rangle $. Hence we have a formula $$ d(\ast w) = {\rm div}\ X\cdot \nu $$ where $ \ast$ is a Hodge star and $\nu$ is a volume form (cf 12 page in the book Differential forms and application - do Carmo).
When we define $\ast w:= x dy \wedge dz +(x+y) dz\wedge dx + (x^2+2z) dx\wedge dy $, then by definition of Hodge star, $$ w= xdx + (x+y)dy + (x^2 +2z)dz $$
That is, we can find $X =(x,x+y,x^2+2z)$ so that ${\rm div}\ X =4$
iii) (cf 63 page)
\begin{align*}\int_{\partial T}\ \ast w&=_{{\rm Stokes\ Theorem}}\ \int_T \ d(\ast w) \\&=\int_T\ {\rm div}\ X\cdot \nu\\&=4\int_{D}\ \bigg(1-x^2-\frac{y^2}{4}\bigg) \cdot d{\rm Area}_D\\ &=4\int_{-2}^2\ \frac{4}{3} \sqrt{1-\frac{y^2}{4}}^3 \ dy \\&=4\int^{\pi/2}_{-\pi/2} \ \frac{8}{3}\cos^4 t\ dt = 4\pi \end{align*} [Another solution] By considering $S_1\bigcap S_2 $, we define $ D\ :\ x^2+ \frac{y^2}{4} \leq 1$
Hence the solid $V$ has bottom face $B$ : $(x,y, \frac{2x^2+y^2}{4}),\ (x,y)\in D$ and top face $T$ : $(x,y,1-\frac{x^2}{2}),\ (x,y)\in D$.
Hence when $w= xdydz + (x+y)dz dx+(x^2+2z)dxdy $, then $$ \int_S \ w=\int_T\ w- \int_B\ w$$
Here \begin{align*} \int_T \ w &= \int_D\ x dy d(1-\frac{x^2}{2} ) \\&+ (x+y) d(1-\frac{x^2}{2}) dx +(x^2+2( 1-\frac{x^2}{2} ) )dxdy \\&=\int_D\ (x^2+2)dxdy \\ \int_B\ w &=\int\ xdyd ( \frac{2x^2+y^2}{4} ) +(x+y)d(\frac{2x^2+y^2}{4}) dx + (x^2+2\frac{2x^2+y^2}{4} ) dxdy\\&=\int_D\ x^2dxdy \\ \int_S\ w&=\int_D\ 2dxdy =4\pi \end{align*}