Calculate the surface integral $\iint_{S}F\cdot dS$

Question

Calculate the surface integral $$\iint_{S}F\cdot dS$$ where $F\left(x,y,z\right) = xy\mathbf i + yz\mathbf j+zx\mathbf k$ and S is the surface of the cylinder $x^2+y^2 \le 1, 0 \le z\le 1$ oriented by the outwards normal.

Answer 1

the straightforward approach is to paramerize the surface

$S = (\cos t, \sin t, z)$

$dS = \frac {\partial S}{\partial t} \times \frac{\partial S}{\partial z} = (\cos t, \sin t, 0)$

$\iint (\cos t \sin t, \sin t \,z, \cos t\,z)\cdot(\cos t, \sin t, 0) dz dt$

$\int_0^{2\pi}\int_0^{1} \cos^2 t \sin t + \sin^2 t \,z \,dz dt$

$\int_0^{2\pi}\cos^2 t \sin t + (1/2) \sin^2 t dt$

$(1/3)cos^3 t + (1/4) (t + \sin t \cos t) |_0^{2\pi}$

$\pi/2$

Alternatively, you can cap the top and the bottom and apply the divergence theorem. The surface integrals across the caps = 0.

$\iiint x+y+z\; dz\, dy\, dx$

Convert to cylindrical

$\iiint r cos t + r sin t + rz \; dz\, dr\, dt$

$\iint r cos t + r sin t + (1/2) r \; , dr\, dt$

$\int (1/2) cos t + (1/2) sin t + (1/4) \; dt$

$ \pi/2$