**Calculus**

Manmohan Dash<g6pontiac@gmail.com>

Question.

Calculate *Curl F* and then use Stokes’ theorem to compute the flux of

*Curl*through the given surface as a line integral.

**F****= <**

*F**y*,

*x*,

*x*

^{2}+

*y*

^{2}> , the upper hemisphere;

*x*

^{2}+

*y*

^{2}+

*z*

^{2}= 1,

*z*≥ 0.

Answer; *curl F* =

i |
j |
k |

∂_{x} |
∂_{y} |
∂_{z} |

F_{x} |
F_{y} |
F_{z} |

Now expanding the terms, the first component of curl of ** F** is, (

*curl*)

**F***= ∂*

_{x}*− ∂*

_{y}F_{z}*and the 2nd term here has to be zero, as there is no z dependence. Thus the first term is 2*

_{z}F_{y}*y*.

Similarly the *y* component of curl of ** F**, is: (

*curl*)

**F***= ∂*

_{y}*− ∂*

_{x}F_{z}*= 2*

_{z}F_{x}*x*.

But the 3rd term of the curl, (*curl F*)

*= ∂*

_{z}*− ∂*

_{x}F_{y}*is zero, since x-partial of 2nd component, ∂*

_{y}F_{x}*, and y-partial of the 1st component of*

_{x}F_{y}**, ∂**

*F**, are both 1.*

_{y}F_{x}So, *Curl F* = < 2

*y*, 2

*x*, 0 > .

Now Stoke’s Theorem is the following; ∫* _{S }curlF*.

**= ∮**

*dS***.**

*F***.**

*dl*The line integral of ** F** is over the contour of the circle on the

*z*= 0 plane, as S on the surface integral on LHS is the upper hemisphere.

So, the dot product of ** F** and

**is this: (**

*dl**y*+

**i***x*).(

**j***dx*+

**i***dy*) =

**j***ydx*+

*xdy*.

Consequently the integral is ∮*ydx* + *xdy* where the limits of integration vary from (0, 1).

So RHS = ∫_{0}^{1}*dx*√1 - x^{2} + ∫_{0}^{1}*dy*√1 − *y*^{2} = 2∫_{0}^{1}*dx*√1 − *x*^{2}.

We replace *x* = *sinθ* so that *dx* = − *cosθdθ*.

We then have RHS = − 2∫_{0}* ^{π/}*

^{2}

*dθcos*

^{2}

*θ*= − 2[1/2(

*θ*+

*sinθcosθ*)]

_{0}

^{π/}^{2}= −

*π/*2.

So the flux of the *Curl F* through the given surface is −

*π/*2.

Document generated by eLyXer 1.2.5 (2013-03-10) on 2015-05-30T22:42:34.204000

Categories: Calculus, Mathematics, Stoke's Theorem, Vectors

## Leave a Reply