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).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).

[tex]\left[\begin{array}{ccc}\vec i&\vec j&\vec k \\\frac{\partial}{\partial x} &\frac{\partial}{\partial y} &\frac{\partial}{\partial z } \\x+y^2&y+z^2&z+x^2\end{array}\right] = 2z\vec i -2x\vec j + 2y \vec z[/tex]

Since the plane is oriented counterclockwise. The equation of the plane is

[tex]x+y+z=5[/tex]

Now, let's use Stokes' theorem and get the surface integral set up

[tex]\int\limits_C {\vec F\bullet d\vec r} =\iint_S{curl \vec F\bullet d\vec S}[/tex]

[tex]=\iint_S {(2z\vec i-2x\vec j+2y\vec z)\bullet d\vec S}[/tex]

You finish the work.