Multivariable Calculus Divergence Theorem

Question

I have a cylinder where the top is given by $ z=-2$ the bottom is $z=2$ and the "lateral surface" is $x^2+y^2=1$. I have to find$\iint_{s}(4x+3y+z^2) \ dS$. I know this involves using the divergence theorem and then plugging in the answer into the volume of a cylinder but I have no idea how to set up the limits on the triple integral once I have found the partials and split up the integral for the divergence theorem. Thanks.

Answer 1

EDIT: The OP's desire to apply the divergence theorem is ambitious, but can be achieved. To apply the Divergence Theorem, we must have the surface integral in the form of $\iint_S \vec F\cdot\vec n dS$. So this problem gets sneaky. To this end:

Consider the vector field $\vec F_1=(4,3,(4x+3y)z/2)$. Since $\vec n = (x,y,0)$ on the lateral surface and $\vec n = (0,0,z/2)$ on the two disks, this accounts for $\iint_S (4x+3y)dS$ and gives us flux $\iiint_V \text{div}\vec F_1 dV =\iiint_V \frac12(4x+3y)dV = 0$, by symmetry. Now consider $\vec F_2 = z^2(x,y,z/2)$. By similar reasoning, its flux across $S$ is indeed $\iint_S z^2 dS$. So the flux of $\vec F_2$ is $$\iiint_V \text{div} F_2 dV = \iiint_V \frac72z^2\,dV =\frac72\int_0^{2\pi}\int_0^1\int_{-2}^2 z^2r dzdrd\theta = \frac{56}3\pi\,.$$
(Note that we do indeed get the same answer by the much simpler direct calculation.)

Answer 2

There is nothing wrong with this approach! The OP said "I know it involves using the divergence theorem.." but this does not mean that he is not interested in other approach.

Since the surface $S$ consists of three parts, then the integral can be evaluated as

$$ \iint_{S} = \iint_{S_1} + \iint_{S_2} + \iint_{S_3} \longrightarrow (1) $$

First, we parametrize the lateral surface as

$$ r(u,v)= \langle \cos u, \sin u, v \rangle, \quad -2 \leq v \leq 2,\, 0\leq u \leq 2\pi. $$

Then, we have

$$\iint_{S} (4x+3y+z^2) dS = \int_{-2}^{2}\int_{0}^{2\pi} (4\cos u + 3\sin u +v^2)|\textbf{r}_u \times \textbf{r}_v|dudv .$$

For the second and the third integrals in $(1)$, you do not need to parametrize, just use the technique. Now, you should be able to finish the problem.

Note: We used the following identity

$$ \iint_{S} f \,dS = \iint_{T} f(\mathbf{x}(u, v)) \left\|{\partial \mathbf{x} \over \partial u}\times {\partial \mathbf{x} \over \partial v}\right\| du\, dv $$