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

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.

• Parametrize the cylinder using (guess what) cylindrical coordinates. Then, calculate the tangent vectors of such parametrization. Their cross product is a normal vector to the surface. – user512346 Nov 11 '18 at 15:36

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?

• ok so when a cylinder is closed it has both a top and bottom disk and when it is open it is just the mantle? thanks for your help btw – THN Nov 11 '18 at 19:56
• You are welcome, @TCarr. When it has both a top and bottom disk then it is closed, just like a closed tin can. When at least one of the disks is missing, it is not a closed surface anymore. – Ernie060 Nov 11 '18 at 19:58
• if i wanted this cylinder in the first quadrant, would you define a new surface over which to calculate the flux or would you just change the limits of integration for the mantle ( I assume its the former) – THN Nov 11 '18 at 20:26
• Well, if you want to integrate over a closed surface, you need to change the surface and some of the limits of integration. Note that the surface then consists of five pieces: a quarter of a mantle, a quarter of the top and bottom disc and also two rectangles (parts of the $xz$ and $yz$-plane). If you restrict the parametrisations of the mantle and of the top and bottom discs, you have the first three parts. Then you still need to parametrise the rectangles, but that's not too difficult. If you only want to integrate over e.g. the cylinder mantle, changing the limits will suffice. – Ernie060 Nov 11 '18 at 20:32
• Will the xz and yz planes have normal vectors of -i and -j respectively as their equations are x=0 and y=0 and so the flux through them both will be zero? – THN Nov 11 '18 at 21:35