Recall the Laplace-Young equation governing the shape $z=h(x)$ of a meniscus near a vertical planar all $$ \ell_c^2\frac{h_{xx}}{(1 + h_x^2)^{3/2}} = h $$ Integrating the full Laplace-Young equation once (without making an assumption of shallow slopes), show that the rise height of the meniscus on the all, $h_0$ is given in terms of the contact angle of the liquid $\theta$ by $$ h_0 = \pm\ell_c[2(1 - \sin\theta)]^{1/2} $$ and discuss when each of the $\pm$ branches of the result are appropriate.
Show that the total area displaced by the meniscus is given by $$ A = \int_{0}^{\infty}h~{\rm d}x = \ell_c^2\cos\theta $$ and comment on this result in the light of the generalized Archimedes' principle discussed in lecture 3
I've done all of the question except the integral at the end. I have $h$ in terms of $h_x$ and so I have no idea how to go about integrating this
If the contact angle between the tangent line of the meniscus and the $y-$axis (wall) is $\theta$ and $\alpha$ is the angle formed by the tangent line and the $x-$axis , we have
$$h_x(0) = - \frac{\sin \alpha} { \cos \alpha} = - \frac{\sin (\pi/2 -\theta)}{ \cos (\pi/2 -\theta )}= -\frac{\cos \theta}{\sin \theta} = -\cot \theta .$$
We also have the far-field condition $h_x(\infty)= 0$.
Integrating both sides of the Laplace-Young equation we get
$$\int_0^\infty h(x) \, dx = l_c^2 \int_0^ \infty \frac{h_{xx}}{(1 + h_x^2)^{3/2}} \, dx.$$
Changing variables as $z = h_x$ we have $dz = h_{xx} dx$ and
$$\begin{align}\int_0^\infty h(x) \, dx &= l_c^2 \int_{h_x(0)}^{h_x(\infty)} \frac{dz}{(1 + z^2)^{3/2}} \\&= l_c^2 \int_{-\cot \theta}^{0} \frac{dz}{(1 + z^2)^{3/2}} \\ &= -\left.l_c^2 \frac{z}{\sqrt{1 +z^2}}\right|_0^{-\cot \theta} \\ &= l_c^2\frac{\cot \theta}{\sqrt{1 + \cot ^2 \theta}} \\ &= l_c^2 \cos \theta \end{align} $$