Solving a Navier-Stokes equation I was reading a paper in solving a Navier-Stokes equation applied for 1D fluid flow in hydraulic fracturing. The below part I couldn't understand. Anyone please explain for me how do they get (4) from (2) and (3) and how to get (5) from (4)?
$$\frac{∂p}{∂x} = μ \left(\frac{∂^2 u_x}{∂y^2} + \frac{∂^2 u_x}{∂z^2}\right) ; \tag 1$$
$μ$: fluid viscosity
$u_x$: fluid velocity in x direction
$p$: fluid net pressure
The elliptic cross section shape corresponds to an ellipse with axes of width $2a$ and height $2b$ (image attached)
$z^2/a^2 +y^2/b^2 =1.  \tag 2$
The boundary of the ellipse is the locus of points $(x, y)$ such that:
$f(y,z)=1-(z^2/a^2 +y^2/b^2 )=0. \tag 3$
Note that: 
$\nabla^2 f(y,z)=-2(a^2+b^2 )/(a^2 b^2 ). \tag 4$
By inspection (2) and (3) we can find out a solution to (1) with non-slip boundary conditions ($u=0$) on the ellipse boundary:
$$u_x (y,z)= \frac{∂p}{∂x}   (a^2 b^2)/2μ(a^2+b^2 )  (z^2/a^2 +y^2/b^2 -1); \tag  5$$
Then the average fluid velocity passing through the elliptic cross section of area $\pi ab$ is:
Thank you.
 A: Probably the most challenging part is computing the average velocity as follows. 
The average velocity is obtained by integrating over the cross-section $E$,
$$\begin{align}\bar{u} &= \frac{1}{\pi ab}\int_E u_x(y,z) \, dA \\ &= \frac{1}{\pi ab}\int_{-b}^b \left(\int_{-a\sqrt{1 - y^2/b^2}}^{a\sqrt{1 - y^2/b^2}}    u_x(y,z) \, dz\right) \, dy \\ &=  \frac{4}{\pi ab}\int_{0}^b \left(\int_{0}^{a\sqrt{1 - y^2/b^2}}    u_x(y,z) \, dz\right) \, dy \\ &= \frac{1}{2\mu}\frac{\partial p}{\partial x}\frac{a^2b^2}{a^2+b^2}\frac{4}{\pi ab}\int_{0}^b \left(\int_{0}^{a\sqrt{1 - y^2/b^2}}   (z^2/a^2 + y^2/b^2 -1 ) \, dz\right) \, dy \end{align}$$ 
Changing variables with $s = z/a$ and $t = y/b$ we have
$$\begin{align}\bar{u} &= \frac{2}{\pi \mu}\frac{\partial p}{\partial x}\frac{a^2b^2}{a^2+b^2}\int_{0}^1 \left(\int_{0}^{\sqrt{1 - t^2}}   (s^2 + t^2 -1 ) \, ds\right) \, dt \end{align}$$
Finally change variables again to polar coordinates to get
$$\begin{align}\bar{u} &= \frac{2}{\pi \mu}\frac{\partial p}{\partial x}\frac{a^2b^2}{a^2+b^2}\int_{0}^{\pi/2} \left(\int_{0}^{1}   (r^2 -1 ) \, r\,dr\right) \, d\theta \end{align}$$
You should be able to complete the exercise from here.  Note that the pressure gradient must be negative to cause fluid to flow in the positive $x$-direction, and, thus, $\bar{u}$ is positive.
