Answer:
[tex]\mathbf{\int_C F*dr= -125}[/tex]
Step-by-step explanation:
Given that:
[tex]F(x,y,z) = ( x+ y^2) i + (y +z ^2) j+(z + x^2)k[/tex] , where C is the triangle with vertices (5, 0, 0), (0, 5, 0), and (0, 0, 5).
The objective is to use Stokes' Theorem to evaluate CF. dr
Stokes Theorem : [tex]\int_c F .dr = \iint _s \ curl \ F. dS[/tex]
To estimate curl F , we need to find the partial derivatives:
So;
[tex]P = x+y^2[/tex]
partial derivative is:
[tex]\dfrac{\partial P }{\partial y }= 2y[/tex]
[tex]\dfrac{\partial P }{\partial z }= 0[/tex]
[tex]Q = y + z^2[/tex]
partial derivative is:
[tex]\dfrac{\partial Q }{\partial x }= 0[/tex]
[tex]\dfrac{\partial Q }{\partial z }= 2z[/tex]
[tex]R = z +x^2[/tex]
partial derivative is:
[tex]\dfrac{\partial R }{\partial x }= 2x[/tex]
[tex]\dfrac{\partial R }{\partial y }= 0[/tex]
These resulted into
curl F = (0 - 2z)i + ( 0 -2x) j + ( 0 - 2y) k
= ( -2z, -2x, -2y )
The normal vector and the equation of the plane can be expressed as follows:
If a = (0,5,0 - ( 5,0,0)
a = ( -5,5,0 )
Also ,
b = (0, 0,5) - (5,0,0)
b = (-5. 0,5)
However,
[tex]a \times b = \begin {vmatrix} \begin{array} {ccc} i &j&k \\-5&5&0 \\-5&0&5 \\ \end {array} \end {vmatrix}[/tex]
a × b = (25 - 0)i - (-25-0)j+ (0+25)k
a × b = 25i +25j +25k
∴ the normal vector can be n = (1,1,1)
If we assume x to be x = (x,y,z)
and [tex]x_0 = (5,0,0)[/tex]
Then
[tex]n*(x-x_0) =0[/tex]
[tex](1,1,1)*(x-5,y-0,z-0) =0[/tex]
[tex]x-5+y+z =0[/tex]
collecting like terms
x +y +z = 5
now, it is vivid that from the equation , the plane of the normal vector =(1,1,1)
Similarly, x+y+z = 5 is the projection of surface on the xy - plane such that the line x +y = 5
Thus; the domain D = {(x,y) | 0 ≤ x ≤ 5, 0 ≤ y ≤ 5 - x}
To evaluate the line integral using Stokes' Theorem
[tex]\iint_S \ curl \ F .dS= \iint _S (-2z,-2y,-2x) *(1,1,1) \ dS[/tex]
[tex]\iint_S \ curl \ F .dS= \iint _S -2z-2y-2x \ dS[/tex]
[tex]\iint_S \ curl \ F .dS= \iint _S -2(5-x-y)-2y-2x \ dS[/tex]
[tex]\iint_S \ curl \ F .dS= \iint _S -(10) \ dS[/tex]
[tex]\int_C F*dr= \int ^5_0 \ \int ^{5-x}_0 -10 \ dy \ dx[/tex]
[tex]\int_C F*dr= -10 \int^5_0 (5-x) \ dx[/tex]
[tex]\int_C F*dr= -10 \begin {bmatrix} 5x - \dfrac{x^2}{2} \end {bmatrix}^5_0[/tex]
[tex]\int_C F*dr= -10 \begin {bmatrix} 25 - \dfrac{25}{2} \end {bmatrix}[/tex]
[tex]\int_C F*dr= -10 \begin {bmatrix} \dfrac{25}{2} \end {bmatrix}[/tex]
[tex]\mathbf{\int_C F*dr= -125}[/tex]
