= Solution
Orient $C$ counterclockwise around the body. For an arc element $ds$, the body-outward <unit normal vector> obeys $\mathbf n\,ds=(dy,-dx)$. The <pressure> acts normally into the body, so its <force per unit length> in the suppressed third direction is
$$
d\mathbf F=-p\mathbf n\,ds=(-p\,dy,p\,dx).
$$
Integrating gives
$$
\boxed{F_x=-\oint_Cp\,dy,\qquad F_y=\oint_Cp\,dx.}
$$
Taking positive <torque> to mean counterclockwise about the origin, the infinitesimal moment is $dM=x\,dF_y-y\,dF_x=p(x\,dx+y\,dy)$. Hence
$$
\boxed{M=\oint_Cp(x\,dx+y\,dy).}
$$
All three expressions are per unit span. They are the normal-<pressure> contributions assumed by the inviscid model; a viscous <traction> would require extra terms. A constant <pressure> contributes zero net <force> and <torque> on a closed body, since $\oint dx=\oint dy=\oint d(x^2+y^2)=0$.
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