Use Stokes' Theorem to evaluate

C
F · dr
where C is oriented counterclockwise as viewed from above.
F(x, y, z) = yzi + 9xzj + exyk,

C is the circle
x2 + y2 = 1, z = 3.

To evaluate the line integral using Stokes' Theorem, we need to follow a few steps.

Step 1: Find the curl of the vector field F.
The curl of a vector field F = (P, Q, R) is given by the following formula:
curl F = (∂R/∂y - ∂Q/∂z)i + (∂P/∂z - ∂R/∂x)j + (∂Q/∂x - ∂P/∂y)k.

For F(x, y, z) = yzi + 9xzj + exyk, let's find the partial derivatives:
∂P/∂z = 0
∂Q/∂x = 9z
∂R/∂y = 0
∂R/∂x = 0
∂P/∂y = 0
∂Q/∂z = 1

Therefore, the curl of F is:
curl F = (1 - 0)i + (0 - 0)j + (0 - 9z)k = i - 9zk.

Step 2: Find a surface S that is bounded by the circle C and has a counterclockwise orientation when viewed from above.
The circle C is defined by the equation x^2 + y^2 = 1 and z = 3. This represents a circle in the x-y plane centered at the origin with a radius of 1. To find a surface bounded by this circle, we can use the equation z = 3 as the upper boundary and the x-y plane as the lower boundary. So, the surface S can be defined as {(x, y, z) | x^2 + y^2 = 1, z = 3}.

Step 3: Calculate the surface integral of the curl of F over surface S.
The surface integral can be calculated using the formula:
∬S (curl F) · dS
where dS represents the vector normal to the surface S.

Since S is a flat surface in the x-y plane, the normal vector dS points in the positive z-direction. Therefore, dS = k.

Now, we can calculate the surface integral:
∬S (curl F) · dS = ∬S (i - 9zk) · k dS.

Since dS = k, the dot product (i - 9zk) · k simplifies to -9z.

Step 4: Integrate the resulting expression over the region of the circle C.
To integrate over the region of the circle C, we need to represent it in terms of parameters. A common parameterization of a circle in the x-y plane is:
x = cos(t)
y = sin(t)
z = 3

Thus, z is constant and can be taken out of the integral:
∫C F · dr = ∫C (i - 9zk) · k dS
= ∫C -9z dS

Now we can substitute in the parameterization:
∫C -9z dS = ∫C -9(3) k · k dS
= -27 ∫C dS

Since dS represents the element of arc length, integrating over the circle C with radius 1 gives us the circumference of the circle, which is 2π.

Therefore, the final result is:
∫C F · dr = -27 ∫C dS = -27(2π) = -54π.