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A worksheet prepared by: Prof. Maria Girardi

Math 550A -- Spring 1995

CAS: Maple V Release 4
Title: V4-95finalp5.mws

URL: http://www.math.sc.edu/~girardi -- then link to Math550

Objective:

Solution to 1995 Final Exam problem 5

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1995 Final Exam problem 5

> restart : with(linalg):

Warning, new definition for norm

Warning, new definition for trace

> F_1 := -3*y^2;

> F_2 := 4*z;

> F_3 := 6*x;

> F := [ F_1, F_2, F_3 ];

> curl_F := curl( F , [x,y,z] );

The curve c(t) traces out, in counterclockwise direction, the triangle with vertices:

> A := vector([2,0,0]) ;

> B := vector([0,2,1]) ;

> C := vector([0,0,0]);

Stokes's Theorem gives that

int_c F dot ds = int int_S curl(F) dot dS

where S is the triangluar surface bounded by c, with upward pointing normal.

So to parameterize S, we note that :

> CA := matadd( A , -1*C) ;
CB := matadd( B , -1*C) ;
n := crossprod( CA, CB ) ;

So the triangluar surface S sits in the plane:

> dotprod(n , matadd(vector([x,y,z]) , -1*C)) = 0 ;

and so S has a parameterization:

> x:= x :
y:= y :
z:= y/2:
Phi(x,y) := vector([ x,y,z]);

where, by projecting S into the xy-plane to get a triangle with vertices: (2,0) , (0,2), and (0,0),

we see that the limits for Phi are: (e.g.)

0 <= y <= 2-x and 0 <= x <= 2 .

We will need:

> T_x := [ diff(x,x), diff(y,x) , diff(z,x) ] ;

> T_y := [ diff(x,y), diff(y,y) , diff(z,y) ] ;

> cross_of_T_x_and_T_y := crossprod( T_x , T_y ) ;
length_of_cross_of_T_x_and_T_y := norm( cross_of_T_x_and_T_y , 2 ) ;
n := normalize( cross_of_T_x_and_T_y);

Way # 1 :

> integrand1 := dotprod(curl_F,cross_of_T_x_and_T_y);

> Int(Int( integrand1, y=0..2-x) , x = 0 .. 2) =
int(int( integrand1, y=0..2-x) , x= 0 .. 2);

Way # 2 :

> integrand2 := dotprod(curl_F , n) * length_of_cross_of_T_x_and_T_y ;

> Int(Int( integrand2, y=0..2-x) , x = 0 .. 2) =
int(int( integrand2, y=0..2-x) , x= 0 .. 2);

>

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