How do I compute this surface integral without using the divergence theorem.

Question

I have the vector field $$\underline F = zx\underline i +xy\underline j + yz\underline k $$ And the closed surface S composed of the cylinder $$x^2 + y^2 = R^2 $$ and with $$0\le z\le H $$ I have the integral $$\iint_S \underline F\cdot \underline n $$ where $$\underline n $$ is the outward unit normal vector. How do I evaluate this integral without using the divergence theorem, I'm unsure how to proceed given the presence of the unit normal vector. Any help would be appreciated.

Answer 1

Note that the closed surface $S$ consists out of $3$ pieces: the cylinder mantle, the bottom disc and the upper disc. So you need to calculate three integrals.

Cylinder mantle: I'll set up one integral as an example. A parametrisation is $$ \Sigma: \begin{cases} x = R \cos t, \\ y = R \sin t, \\ z = u \end{cases} \qquad \qquad \text{with $t\in[0,2\pi]$ and $u \in [0,H]$.} $$ The normal $n$ is equal to the cross product of the partial derivatives: $$ n(t,u) = \frac{\partial \Sigma}{\partial t} \times \frac{\partial \Sigma}{\partial u} = \begin{bmatrix} - R \sin t \\ R \cos t \\ 0 \end{bmatrix} \times \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} R \cos t \\ R \sin t \\ 0 \end{bmatrix}. $$ This is an outward point normal vector. Please note that $n$ is not a unit vector! Then the inner product becomes $F(t,u)\cdot n(t,u)= R^2 \cos^2 t u + R^3 \cos t \sin^2 t$. The integral over the cylinder becomes $$ \iint_{\Sigma_1} F\cdot n = \int_0^{2\pi} \int_0^H R^2 \cos^2 t u + R^3 \cos t \sin^2 t \,du dt. $$

Upper and bottom disc: For the upper/bottom disc you need to find a parametrisation and set up an integral as hereabove. You need to make sure that the normal vector points upward on the upper disc, i.e., the $z$-coordinate needs to be positive. (For the bottom disc the $z$-coordinate of the normal vector needs to be negative.)

Can you take it from here?