Surface integral examples

Example 1

For the cylinder of radius 3 and height 5 given by x2 + y2 = 32 and 0$ \le$z$ \le$5, let the charge density be proportional to the distance from the xy-plane. Find the total charge on the cylinder.

Solution: Since the distance from the point (x, y, z) to the xy-plane is z, let the charge density be f (x, y, z) = kz for some constant k. The total charge on the cylinder is the surface integral of f over the cylinder S: $ \iint_{{S}}^{}$f dS.

Parameterize the cylinder by

$\displaystyle \Phi$($ \theta$, t) = (3 cos$ \theta$, 3 sin$ \theta$, t)    

for 0$ \le$$ \theta$$ \le$2$ \pi$ and 0$ \le$t$ \le$5. The total charge on the cylinder is

$\displaystyle \iint_{{S}}^{}$f dS = $\displaystyle \int_{{0}}^{{5}}$$\displaystyle \int_{{0}}^{{2\pi}}$f ($\displaystyle \Phi$($ \theta$, t))$\displaystyle \left\Vert\vphantom{ \frac{\partial \mathbf{\Phi}}{\partial {\tex... ...times \frac{\partial \mathbf{\Phi}}{\partial t}({\textstyle \theta},t) }\right.$$\displaystyle {\frac{{\partial \mathbf{\Phi}}}{{\partial {\textstyle \theta}}}}$($ \theta$, t) x $\displaystyle {\frac{{\partial \mathbf{\Phi}}}{{\partial t}}}$($ \theta$, t)$\displaystyle \left.\vphantom{ \frac{\partial \mathbf{\Phi}}{\partial {\textsty... ...s \frac{\partial \mathbf{\Phi}}{\partial t}({\textstyle \theta},t) }\right\Vert$d$ \theta$ dt    

Calculate the components of the above integral as follows.

f ($\displaystyle \Phi$($ \theta$, t)) = kt    
$\displaystyle {\frac{{\partial \mathbf{\Phi}}}{{\partial {\textstyle \theta}}}}$($ \theta$, t) = (- 3 sin$ \theta$, 3 cos$ \theta$, 0)    
$\displaystyle {\frac{{\partial \mathbf{\Phi}}}{{\partial t}}}$($ \theta$, t) = (0, 0, 1)    
$\displaystyle {\frac{{\partial \mathbf{\Phi}}}{{\partial {\textstyle \theta}}}}$($ \theta$, t) x $\displaystyle {\frac{{\partial \mathbf{\Phi}}}{{\partial t}}}$($ \theta$, t) = $\displaystyle \left\vert\vphantom{ \begin{array}{ccc} \mathbf{i} & \mathbf{j} &... ...textstyle \theta}& 3\cos{\textstyle \theta}& 0 0 & 0 & 1 \end{array} }\right.$$\displaystyle \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} -3\sin{\textstyle \theta}& 3\cos{\textstyle \theta}& 0 0 & 0 & 1 \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \ma... ...style \theta}& 3\cos{\textstyle \theta}& 0 0 & 0 & 1 \end{array} }\right\vert$    
  = i3 cos$ \theta$ - j(- 3 sin$ \theta$)    
  = (3 cos$ \theta$, 3 sin$ \theta$, 0)    
$\displaystyle \left\Vert\vphantom{ \frac{\partial \mathbf{\Phi}}{\partial {\tex... ...times \frac{\partial \mathbf{\Phi}}{\partial t}({\textstyle \theta},t) }\right.$$\displaystyle {\frac{{\partial \mathbf{\Phi}}}{{\partial {\textstyle \theta}}}}$($ \theta$, t) x $\displaystyle {\frac{{\partial \mathbf{\Phi}}}{{\partial t}}}$($ \theta$, t)$\displaystyle \left.\vphantom{ \frac{\partial \mathbf{\Phi}}{\partial {\textsty... ...s \frac{\partial \mathbf{\Phi}}{\partial t}({\textstyle \theta},t) }\right\Vert$ = $\displaystyle \sqrt{{9 \cos^2{\textstyle \theta}+ 9 \sin^2{\textstyle \theta}}}$    
  = 3    

Therefore, the total charge is

$\displaystyle \iint_{{S}}^{}$f dS = $\displaystyle \int_{0}^{5}$$\displaystyle \int_{0}^{{2\pi}}$kt3 d$ \theta$ dt    
  = $\displaystyle \int_{0}^{5}$$\displaystyle \left(\vphantom{ 3t {\textstyle \theta}\Big\vert _{\theta=0}^{\theta=2\pi} }\right.$3t$ \theta$$\displaystyle \Big\vert _{{\theta=0}}^{{\theta=2\pi}}$$\displaystyle \left.\vphantom{ 3t {\textstyle \theta}\Big\vert _{\theta=0}^{\theta=2\pi} }\right)$dt    
  = $\displaystyle \int_{0}^{5}$k6$ \pi$t dt    
  = k3$ \pi$t2$\displaystyle \Big\vert _{0}^{5}$ = 3$ \pi$(25)k = 75$ \pi$k    

Example 2

Let S be the same cylinder as above (x2 + y2 = 9 for 0$ \le$z$ \le$5).

Let F be the vector field F(x, y, z) = (2x, 2y, 2z).

Find the integral of F over S. Use outward pointing normal.

\includegraphics[width=1.6in]{cylinderflux.eps}

What is the sign of integral? Since the vector field and unit normal point outward, the integral better be positive.

As above, parameterize the cylinder by

$\displaystyle \Phi$($ \theta$, t) = (3 cos$ \theta$, 3 sin$ \theta$, t)    

for 0$ \le$$ \theta$$ \le$2$ \pi$ and 0$ \le$t$ \le$5. We calculated above that

$\displaystyle {\frac{{\partial \mathbf{\Phi}}}{{\partial {\textstyle \theta}}}}$($ \theta$, t) x $\displaystyle {\frac{{\partial \mathbf{\Phi}}}{{\partial t}}}$($ \theta$, t) = (3 cos$ \theta$, 3 sin$ \theta$, 0)    

Is (3 cos$ \theta$, 3 sin$ \theta$, 0) an outward pointing normal? It is a normal at the point $ \Phi$($ \theta$, t) = (3 cos$ \theta$, 3 sin$ \theta$, t). As shown in the below figure, it is an outward pointing normal.

\includegraphics[width=5in]{cylindernormal.eps}

Calculate total flux:

$\displaystyle \iint_{{S}}^{}$F . dS = $\displaystyle \int_{{0}}^{{5}}$$\displaystyle \int_{{0}}^{{2\pi}}$F($\displaystyle \Phi$($ \theta$, t)) . $\displaystyle \left(\vphantom{ \frac{\partial \mathbf{\Phi}}{\partial {\textsty... ...times \frac{\partial \mathbf{\Phi}}{\partial t}({\textstyle \theta},t) }\right.$$\displaystyle {\frac{{\partial \mathbf{\Phi}}}{{\partial {\textstyle \theta}}}}$($ \theta$, t) x $\displaystyle {\frac{{\partial \mathbf{\Phi}}}{{\partial t}}}$($ \theta$, t)$\displaystyle \left.\vphantom{ \frac{\partial \mathbf{\Phi}}{\partial {\textsty... ...times \frac{\partial \mathbf{\Phi}}{\partial t}({\textstyle \theta},t) }\right)$d$ \theta$ dt    
  = $\displaystyle \int_{0}^{5}$$\displaystyle \int_{0}^{{2\pi}}$F(3 cos$ \theta$, 3 sin$ \theta$, t) . (3 cos$ \theta$, 3 sin$ \theta$, 0)d$ \theta$ dt    
  = $\displaystyle \int_{0}^{5}$$\displaystyle \int_{0}^{{2\pi}}$(6 cos$ \theta$, 6 sin$ \theta$, 2t) . (3 cos$ \theta$, 3 sin$ \theta$, 0)d$ \theta$ dt    
  = $\displaystyle \int_{0}^{5}$$\displaystyle \int_{0}^{{2\pi}}$18 cos2$ \theta$ +18 sin2$ \theta$ d$ \theta$ dt                
  = $\displaystyle \int_{0}^{5}$$\displaystyle \int_{0}^{{2\pi}}$18 d$ \theta$ dt    
  = $\displaystyle \int_{0}^{5}$18(2$ \pi$)dt    
  = 18(2$ \pi$)(5) = 180$ \pi$    

There are two orientations for the cylinder. If we had instead chose the orientation given by the inward point normal, what must $ \iint_{{S}}^{}$F . dS be? -180$ \pi$


Duane Nykamp
nykamp@math.umn.edu
2005-11-10