= Solution
{c}
Let the constant in the integrated <Reynolds equation> be $C$:
$$
\frac{H^3}{12}\frac{dp'}{dx'}+\frac H2=C,
\qquad \frac{dp'}{dx'}=12C H^{-3}-6H^{-2},
\qquad H=1+\frac{x'^2}{2}.
$$
Both endpoint <pressures> are zero, so the <integral> of $dp'/dx'$ on the real line vanishes. Define
$$
I_m=\int_{-\infty}^{\infty}\frac{dx'}{(1+x'^2/2)^m}.
$$
The substitution $x'=\sqrt2\tan\theta$ gives
$$
I_m=\sqrt2\int_{-\pi/2}^{\pi/2}\cos^{2m-2}\theta\,d\theta,
\qquad I_2=\frac{\pi}{\sqrt2},\qquad
I_3=\frac{3\pi\sqrt2}{8}.
$$
The second value follows by integrating the supplied identity for $\cos^4\theta$; the nonconstant cosine terms have zero <integral> over this interval. Hence $12CI_3-6I_2=0$, giving
$$
\boxed{C=\frac{I_2}{2I_3}=\frac23,\qquad q=-C=-\frac23.}
$$
The <pressure gradient> is now
$$
\frac{dp'}{dx'}=\frac{8-6H}{H^3}
=\frac{2-3x'^2}{(1+x'^2/2)^3}.
$$
Direct differentiation shows
$$
\frac d{dx'}\left(\frac{2x'}{H^2}\right)
=\frac2{H^2}-\frac{4x'^2}{H^3}
=\frac{2-3x'^2}{H^3}.
$$
The candidate tends to zero at both ends, so the additive constant is zero and
$$
\boxed{p'(x')=\frac{2x'}{(1+x'^2/2)^2}.}
$$
Together with the local <Couette-Poiseuille flow> profile, this solves the <sliding-plate flow through a parabolic gap>.
To restore dimensions, integrate $u=Uu'$ across $dz=h_0\,dz'$. The signed dimensional <volume flux per unit width>, positive in the $+x$ direction, is
$$
\boxed{Q=Uh_0q=-\frac23Uh_0.}
$$
Thus the magnitude drawn in the direction of the moving plate is $2Uh_0/3$. For a plate width $B$ large enough to neglect side-wall effects, the total <volumetric flow rate> is $-2BUh_0/3$; without a specified width the answer is per unit width.
The <Couette flow> and <plane Poiseuille flow> contributions to the dimensional <volume flux per unit width> are respectively
$$
Q_C=-\frac{Uh(x)}2<0,\qquad
Q_P=-\frac{h(x)^3}{12\mu}\frac{dp}{dx}
=Uh_0\left(\frac H2-\frac23\right)
=Uh_0\left(\frac{x'^2}{4}-\frac16\right).
$$
They point in the same direction precisely when $Q_P<0$, giving
$$
\boxed{-\sqrt{\frac{2Rh_0}{3}}<x<\sqrt{\frac{2Rh_0}{3}}.}
$$
At the endpoints the <plane Poiseuille flow> contribution is zero. Outside that interval it opposes the <Couette flow> contribution, while their sum remains the constant negative <volume flux per unit width> $-2Uh_0/3$. Thus \b[pressure assists the moving plate near the narrowest gap] and drives a compensating return contribution farther away.
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