I am studying about hydraulics, there are integrals that I don't know how to evaluate. Here is the problem:
Equation of velocity is: $$U_r=K(R^2-r^2)$$ Where $U_r$ is velocity at radius $r$ and $R$ is radius of the pipe and $K$ is constant. Find energy and momentum coefficients ($\alpha$ and $\beta$).
We have: $\alpha=\cfrac{\int v^3dA}{V^3.A}$ and $\beta=\cfrac{\int v^2.dA}{V^2A}$ (where $v$ is velocity and $V$ is average velocity).
I draw this:
In this figure $(R^2-r^2)$ lies on the shaded area. so $A=\pi(R^2-r^2)$. but I don't know how to calculate these integrals to find $\alpha$ and $\beta$
EDIT:
For $\alpha$ I tried it:
$V=\frac{Q}{A}$ where $Q$ is flow rate:
$$\alpha=\cfrac{\int v^3dA}{V^3.A}=\cfrac{\int v^3dA}{\dfrac{Q^3}{A^2}}=\frac{A^2}{Q^3}\int v^3 dA=\frac{A^2}{Q^3}\int(K(R^2-r^2))^3dA=\cfrac{K^3 (\pi R^2)^2}{Q^3}\times\int (R^2-r^2)^3dA $$ Using polar system: $dA=rdrd\theta$ But I am not sure what can I write for $(R^2-r^2)^3$ and what is the integral bound? I don't know how to change it to polar system integral.
$\displaystyle \int v^3dA = 2\pi K^3 \int_r^{R} (R^2-\rho^2)^3 \rho \, d\rho$
$\rho(R^2 - \rho^2)^3 = R^6 \rho - \rho^7 - 3\rho^3R^4 + 3 \rho^5R^2$
So your integral becomes
$ \displaystyle 2\pi K^3 \int_r^{R} (R^6 \rho - \rho^7 - 3\rho^3R^4 + 3 \rho^5R^2) d\rho$
Can you take it from here?