# VC4.26

Let $\phi(x,y,z)=x^2+y^2=9\,$ --(1)

Therefore $\nabla\phi=i\frac{\partial \phi}{\partial x}+j\frac{\partial \phi}{\partial y}+k\frac{\partial \phi}{\partial z}=2xi+2yj\,$

n=unit normal vector at any point (x,y,z) of S =$\frac{2xi+2yj}{\sqrt{4x^2+4y^2}}=\frac{xi+yj}{3}\,$

Therefore,$F\cdot n=(zi+xj-yzk)\cdot\frac{xi+yj}{3}=\frac{zx+xy}{3}\,$ -(2)

While solving a problem connected with a cylindrical surface we should use cylindrical coordinates r,theta,z for better solution.on the surface of the cylinder(1) we have $x=3\cos\theta,y=3\sin\theta,z=z\,$ --(3)

dS=elementaty area on the cylindrical surface $x^2+y^2=9,dS=3d\theta dz\,$ --(4)

$\iint_S F\cdot n dS=\iint_S \frac{1}{3}x(z+y)3d\theta dz\,$ using (2),(4)

=$\int_{z=0}^{4}\int_{\theta=0}^{\frac{\pi}{2}}(3\cos\theta)(z+3\sin\theta)d\theta dz\,$ using (3)

=$3\int_{\theta=0}^{\frac{\pi}{2}}\cos\theta[\frac{1}{2}z^2+3z\sin\theta]_{z=0}^{4}d\theta\,$

=$3\int_{0}^{\frac{\pi}{2}} \cos\theta[8+12\sin\theta]d\theta\,$

=$3\int_{0}^{\frac{\pi}{2}}[8\cos\theta+6\sin 2\theta]d\theta\,$

=$3[8\sin\theta-3\cos 2\theta]_{0}^{\frac{\pi}{2}}=3[8+3+3]=42\,$

##### Toolbox

 Get A Wifi Network Switcher Widget for Android