$$ \int_{S_{c}} \underset{\bar{}}{F}.d \underset{\bar{}}{S}, $$
where $S_{c}$ is the curved part of the truncated cone given by $x^2 + y^2 = (a-z)^2$, $0≤z≤(1/2)a$ and $\underset{\bar{}}{F} = (2x + y^2)\underset{\bar{}}{i} + (y+xz)\underset{\bar{}}{j} + z\underset{\bar{}}{k}$.
Use a normal to $S_{c}$ that points away from the $z$ axis and give answer in the form $p\pi a^3/q$.
Hints:
a) Use the divergence theorem
b) The surface of the truncated cone can be split into 2 flat pieces and a curved part.
c) The volume of a cone is 1/3 * base * height
Despite the hints given, I'm still not sure where to start. Would really appreciate some guidance.
The divergence theorem says $\iint F \cdot dS = \iiint \nabla \cdot FdV$. We find that $\nabla F = 4$. Now find the volume $V$ of the truncated cone, and multiply by $4$. By the divergence theorem this is equivalent to the surface integral on the entire surface of the truncated cone.
If you want the surface integral for just the curved part of the cone, then calculate $\int F\cdot dS$ on $z=0$ (hint: this evaluates to zero, can you see why?) and $\int F \cdot dS$ on $z=\frac{a}{2}$, and subtract this from your $4V$.
Edit for clarity:
Your question asks you to find the flux integral of $F = (2x+y^2)\mathbf{i} + (y+xz)\mathbf{j} + z\mathbf{k}$ on the blue surface shown above. There are two ways to do this.
The first way is to compute this directly through the formula $\int \mathbf{F}\cdot d\mathbf{S}$.
This involves finding the normal vector to the blue surface, parameterising the blue surface, and taking the surface integral. Often, this is the more complicated way of doing things.
The second option is to use the divergence theorem.
In this question, the way to do this would be to first calculate the volume enclosed by the three surfaces shown. This question would require a little more work though, since you're only interested in the flux through the blue surface. The divergence theorem would give you the flux through all surfaces enclosing the volume, so you would need to subtract the flux through the green surfaces for your final answer.
Doing this by the divergence theorem, the first step would be to work out $\nabla \cdot \mathbf{F}$. This gives $2+1+1=4$. Since we see that the divergence of $\mathbf{F}$ is a constant and not dependent on $x, y$ or $z$, our volume integral becomes $$\iiint 4 dV = 4\iiint dV$$ So all you need to do is find the volume and multiply by 4 (If the divergence were dependent on $x, y$ or $z$, your integral could be slightly more complicated). Calculating the volume of the truncated cone we have $$\frac{1}{3}(\pi a^2a-\pi\frac{a^2}{4}\frac{a}{2})=\frac{7\pi a^3}{24} $$ So the divergence through the enclosing surfaces is $\frac{7\pi a^3}{6}.$
The flux integral through the bottom surface where $z=0$ will be given by $$\int \mathbf{F}\cdot \mathbf{n} dS.$$ Since $F = (2x+y^2)\mathbf{i} + (y+xz)\mathbf{j} + 0\mathbf{k}$ on $z=0$, and $n=-\mathbf{k}$, we can see that $\mathbf{F}\cdot \mathbf{n}=0.$ On the upper surface, where $z=\frac{a}{2}$, we have $F = (2x+y^2)\mathbf{i} + (y+xz)\mathbf{j} + \frac{a}{2}\mathbf{k}$. Our $\mathbf{n}$ remains the same (except for a change in sign, since the normal must point outward from the volume), so $\mathbf{F}\cdot \mathbf{n}=\frac{a}{2}.$ The integral over the surface can be calculated using polar co-ordinates as follows $$\iint \mathbf{F}\cdot \mathbf{n} dS = \frac{a}{2}\int_0^{2\pi}\int_0^\frac{a}{2}rdrd\theta= a\pi\Big[\frac{r^2}{2}\Big]_0^{\frac{a}{2}}=\frac{a^3\pi}{8}.$$
So your final answer should be $$\frac{7\pi a^3}{6} - \frac{a^3\pi}{8} = \frac{25a^3\pi}{24}.$$