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From "Advanced Calculus: Demystified: A Self-teaching Guide", 2007, David Bachman, Example 10-3, page 141:

Let vector field:

$$W = (y,x,z)$$

Let S be the portion of a cylinder parameterized by:

$$psi(theta, z) = (cos theta, sin theta, z)$$
$$0 le theta le frac{pi}{4}$$
$$0 le z le 1$$

The vector $(-1, 0, 0)$ defines an orientation on S at point (1,0,0).

Integral W over the surface S with orientation vector (-1, 0, 0) at point (1,0,0).

For the integral below, why does the parameterization of $psi$ disagree with the specified orientation? and therefore the final result need to be negated?

$$
int_{S} textbf{W} cdot dtextbf{s} = int^{1}_0 int^{frac{pi}{4}}_{0} textbf{W}(psi(theta,z)) cdot left(begin{vmatrix}
frac{partial{psi}}{partial{theta}} times
frac{partial{psi}}{partial{z}}
end{vmatrix} right) dtheta dz$$

I can kind of see that if you pick point (1,0,0) and evaluate W at this point, then:

W(1,0,0) = (0,1,0)

Which is basically the unit $vec{j}$ vector.

then, if you take the the cross product of the two tangents vectors of the surface S at the point (1,0,0) then it gives you a normal vector to the surface:

$$begin{vmatrix}
frac{partial{psi}}{partial{theta}} times
frac{partial{psi}}{partial{z}}
end{vmatrix} = (1,0,0)
$$

Which is basically the unit $vec{i}$ vector at point (1,0,0).

But, they say the orientation of the integral setup is $-vec{i} = (-1, 0, 0)$.
Why?

Continuation of solution from book:

$$int_{S} textbf{W} cdot dtextbf{s} = int^{1}_0 int^{frac{pi}{4}}_{0} textbf{W}(psi(theta,z)) cdot left(begin{vmatrix}
frac{partial{psi}}{partial{theta}} times
frac{partial{psi}}{partial{z}}
end{vmatrix} right) dtheta dz$$

$$=int_{1}^{0} int_{0}^{frac{pi}{4}} (sin theta, cos theta, z) cdot (cos theta, sin theta, 0) dtheta dz$$

$$=int_{1}^{0} int_{0}^{frac{pi}{4}} 2 sin theta cos theta dtheta dz$$

$$=int_{1}^{0} int_{0}^{frac{pi}{4}} sin 2theta cos theta dtheta dz$$

$$=int_{1}^{0} frac{1}{2} dz $$

$$=frac{1}{2} $$

At this point, they say, orientation of $psi$ disagrees with the specified orientation therefore the correct answer is $-frac{1}{2}$.

how are they settings up the integral such that its orientation disagrees?

For a surface integral, the orientation is chosen by the person setting up the problem to represents the direction in which the flux of the vector field needs to pass through the surface for the resulting surface integral to be positive in value.

For a given surface, there are generally, two choice for the orientation vectors:

(1) normal vectors on "top" of the surface, and

(2) normal vectors on the "bottom" of the surface.

For some surfaces, the orientation could also be interpreted as a choice between:

(1) normal vectors on the "inner surface"

(2) normal vectors on the "outer surface"

If the cross product in the surface integral disagree with this chosen orientation for a given surface, then it is necessary to negate the surface integral to match the orientation. (This is just the rule.)

If the surface integral problem says, assume the induced orientation for the surface integral, then, this just means that we will choose the orientation vector to be the same direction as the cross product in the surface integral. (thus, no sign flipping required in this case, we just need to obtain the orientation vector from the cross product in the surface integral to make sense of the sign of the resultant integral value.)