Question

Use Stokes' Theorem to evaluate C F · dr where C is oriented counterclockwise as viewed from above. F(x, y, z) = (x + y2)i + (y + z2)j + (z + x2)k, C is the triangle with vertices (5, 0, 0), (0, 5, 0), and (0, 0, 5).

Answer 1

Answer:

$\mathbf{\int_C F*dr= -125}$

Step-by-step explanation:

Given that:

$F(x,y,z) = ( x+ y^2) i + (y +z ^2) j+(z + x^2)k$   , 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 : $\int_c F .dr = \iint _s \ curl \ F. dS$

To estimate curl F , we need to find the partial derivatives:

So;

$P = x+y^2$

partial derivative is:

$\dfrac{\partial P }{\partial y }= 2y$

$\dfrac{\partial P }{\partial z }= 0$

$Q = y + z^2$

partial derivative is:

$\dfrac{\partial Q }{\partial x }= 0$

$\dfrac{\partial Q }{\partial z }= 2z$

$R = z +x^2$

partial derivative is:

$\dfrac{\partial R }{\partial x }= 2x$

$\dfrac{\partial R }{\partial y }= 0$

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,

$a \times b = \begin {vmatrix} \begin{array} {ccc} i &j&k \\-5&5&0 \\-5&0&5 \\ \end {array} \end {vmatrix}$

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 $x_0 = (5,0,0)$

Then

$n*(x-x_0) =0$

$(1,1,1)*(x-5,y-0,z-0) =0$

$x-5+y+z =0$

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

$\iint_S \ curl \ F .dS= \iint _S (-2z,-2y,-2x) *(1,1,1) \ dS$

$\iint_S \ curl \ F .dS= \iint _S -2z-2y-2x \ dS$

$\iint_S \ curl \ F .dS= \iint _S -2(5-x-y)-2y-2x \ dS$

$\iint_S \ curl \ F .dS= \iint _S -(10) \ dS$

$\int_C F*dr= \int ^5_0 \ \int ^{5-x}_0 -10 \ dy \ dx$

$\int_C F*dr= -10 \int^5_0 (5-x) \ dx$

$\int_C F*dr= -10 \begin {bmatrix} 5x - \dfrac{x^2}{2} \end {bmatrix}^5_0$

$\int_C F*dr= -10 \begin {bmatrix} 25 - \dfrac{25}{2} \end {bmatrix}$

$\int_C F*dr= -10 \begin {bmatrix} \dfrac{25}{2} \end {bmatrix}$

$\mathbf{\int_C F*dr= -125}$