Verified answer

First of all, you haven't completely described the circle C. Did you mean to say that it lies in the xz-plane? Also, if I've misunderstood your description of the orientation, then my answer will be off by a factor of -1.

Lucky for you, you've still given enough information to solve the problem. Let D denote the disk bounded by C. Note that

curl(-zi + yj + 6xk) = -7j,

so by Stokes' Theorem, the given integral is equal to

∫D -7j . dA.

Since -j is the unit normal to the disk D (if I understood what you were trying to say about the orientation), this integral is simply 7 times the area of D, which is:

7 * (π 4^2) = 112π.

Alternatively, we can compute the integral without Stokes' Theorem. First recall that

∫ from 0 to 2π of sin^2 t dt = ∫ from 0 to 2π of cos^2 t dt = π.

This is easy to see. The integrals must sum to 2π, since their sum is simply the integral of the constant function 1. The integrals must be equal, by symmetry, so it follows that they are both equal to π.

Now parametrize C by

p(t) = (4 cos t, y0, 4 sin t).

Then p'(t) = (-4 sin t, 0, 4 cos t), so

∫c (-zi + yj + 6xk)

= ∫ from 0 to 2π of (-4sin t i + y0 j + 6*4 cos t k) . (-4 sin t, 0, 4 cos t) dt

= ∫ from 0 to 2π of 16 (sin^2 t + 6 cos^2 t) dt

= 16 * (π + 6π)

= 112π.