= Solution
For <pressure lobes in a symmetric translating lubrication gap>, the height integrals are
$$
\int_{-1}^1\widehat h^{-2}\,d\widehat x=\int_{-1}^1\frac{d\widehat x}{1+\widehat x^2}=\frac\pi2,\qquad
\int_{-1}^1\widehat h^{-3}\,d\widehat x=\left[\frac{\widehat x}{\sqrt{1+\widehat x^2}}\right]_{-1}^1=\sqrt2.
$$
Hence
$$
\boxed{Q=\frac\pi{4\sqrt2}.}
$$
The denominator of the <pressure gradient> is positive. Its numerator changes sign when $\sqrt{1+\widehat x^2}=\pi/(2\sqrt2)$, that is at $\widehat x=\pm x_*$ with $x_*=(\pi^2/8-1)^{1/2}$. The <gradient> is positive for $|\widehat x|>x_*$, zero at those two points, and negative for $|\widehat x|<x_*$. Integration, with either endpoint condition, gives
$$
\boxed{\widehat p(\widehat x)=6\arctan\widehat x-\frac{3\pi}{\sqrt2}\frac{\widehat x}{\sqrt{1+\widehat x^2}}.}
$$
This <function> is odd and vanishes at $-1,0,1$. It has a positive maximum at $-x_*$ and a negative minimum at $x_*$. Thus the sketch has a positive <pressure> lobe in the converging left half of the gap and a negative gauge-<pressure> lobe in the diverging right half.
\Image[/past-exam-of-the-theoretical-mechanics-course-of-the-university-of-oxford/2011-lubrication-pressure.png]
{title=The <lubrication pressure> in a symmetric translating gap and its two stationary points}
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