# Surface integrals where normal changes?

I am having problems getting my head around this problem:

Evaluate the surface integral $$\int_S \vec F\bullet d\vec s$$ where $\vec F=x \vec i-y \vec j +z \vec k$ and where the surface S is the cylinder defined by $x^2+y^2\le 4$ and $0\le z \le 1$. Verify your answer using the Divergence Theorem.

The thing I am confused about is do we integrate over the ends of the cylinder and if so how? I think that we should since the divergence theorem requires a closed surface, but if I am right how do we do the surface integral since the normal changes direction (with a non-continuous derivative) between the top and bottom and the curved edges and therefore I cannot see how split the surface integral up into the three sections (top, bottom and curved side), how do we do this?

The divergence theorem

$$\int_{\Omega} \nabla \cdot \mathbf{F} \, d\mathbf{x}=\int_{\partial \Omega}\mathbf{F} \cdot \mathbf{n} \, dS$$

is applicable when the closed surface $\partial \Omega$ is piecewise smooth.

The discontinuity of the normal vector $\mathbf{n}$ is restricted to a set of measure $0$ and does not affect the integration. Just integrate over the top and bottom disks with $\mathbf{n} = \pm \mathbf{e_z}$ and over the side using $\mathbf{n} = \mathbf{e_r} = \cos \theta \mathbf{e_x} + \sin \theta \mathbf{e_y}.$

• So do the following? $$\int_{S_1}\mathbf{F} \cdot \mathbf{e_z} \, dS_1+\int_{S_2}\mathbf{F} \cdot \mathbf{(-e_z)} \, dS_2+\int_{S_3}\mathbf{F} \cdot \mathbf{e_r} \, dS_3$$ where $S_1$ is the top, $S_2$ is the bottom and $S_3$ is the side? If so please can you tell me why this works? – user135842 Dec 4 '14 at 8:19
• Yes that is correct. You can split the integral for the same reason that with $a < c< b$, $\int_a^b f = \int_a^cf + \int_c^bf$ is true in 1-dimension -- and you can arbitrarily change the value of $f$ at $c$ without changing the values of the integrals. – RRL Dec 4 '14 at 8:24
• The normal vector is continuous on the interior of $S_i$ and has a well defined limit as the boundary of $S_i$ is approached from the interior. – RRL Dec 4 '14 at 8:59
• Sorry just to check that in this case $d\vec S_1 \ne d\vec S_2$ but $dS_1 =dS_2$ – user135842 Dec 4 '14 at 10:21
• Yes -- the normal vectors to $S_1$ and $S_2$ point in opposite directions. But $dS = rdrd\theta$ on both surfaces. The only non-zero contribution is $\int_{S_1}\mathbf{F}\cdot\mathbf{e_z}dS = 4\pi$ -- which matches the volume integral of the divergence. – RRL Dec 4 '14 at 17:29