Examples of Chi-Rho in the following topics:
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- Areas in the chi-square table always refer to the right tail.
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- Define the Chi Square distribution in terms of squared normal deviates
- A Chi Square calculator can be used to find that the probability of a Chi Square (with 2 df) being six or higher is 0.050.
- The mean of a Chi Square distribution is its degrees of freedom.
- The Chi Square distribution is very important because many test statistics are approximately distributed as Chi Square.
- Chi Square distributions with 2, 4, and 6 degrees of freedom
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- $\displaystyle {\bf q} = \left [ \frac{1}{2} V^2 + w \right ] \rho {\bf V}$
- where $w=(\epsilon + P)/\rho$ is the heat function (enthalpy) per unit mass of the fluid.
- $\displaystyle \frac{\partial \rho}{\partial t} + \nabla \cdot \left ( \rho {\bf V} \right ) = \frac {\partial \rho}{\partial t} + \left ( {\bf V} \cdot \nabla \right ) \rho + \rho \nabla \cdot {\bf V} = \frac{\partial \rho}{\partial t} + \rho \nabla \cdot {\bf V} = 0$
- $\displaystyle \frac{(\partial \rho s)}{\partial t} + \nabla \cdot \left ( \rho s {\bf v} \right ) = 0.$
- $\displaystyle \rho \frac{\partial s}{\partial t} + {\bf v} \cdot \nabla (\rho s) = 0
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- Given the function $f(x,y) = x + y$ and applying the transformation, one obtains $f(\rho, \phi) = \rho \cos \phi + \rho \sin \phi = \rho(\cos \phi + \sin \phi )$.
- $\displaystyle{\frac{\partial (x,y)}{\partial (\rho, \phi)}} = \begin{vmatrix} \cos \phi & - \rho \sin \phi \\ \sin \phi & \rho \cos \phi \end{vmatrix} = \rho$
- which has been obtained by inserting the partial derivatives of $x = \rho \cos(\varphi)$, $y = \rho \sin(\varphi)$ in the first column with respect to $\rho$ and in the second column with respect to $\varphi$, so the $dx \, dy$ differentials in this transformation become $\rho \,d \rho \,d\varphi$.
- $\iint_D f(x,y)dx \, dy = \iint_T f(\rho \cos \varphi, \rho \sin \varphi)\rho$
- $\begin{aligned} \iint_D x \, dx\, dy &= \iint_T \rho \cos \phi \rho \, d\rho\, d\phi \\ &= \int_0^\pi \int_2^3 \rho^2 \cos \phi \, d \rho \, d \phi \\ &= \int_0^\pi \cos \phi \ d \phi \left[ \frac{\rho^3}{3} \right]_2^3 \\ &= \left[ \sin \phi \right]_0^\pi \ \left(9 - \frac{8}{3} \right) = 0 \end{aligned}$
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- $\displaystyle
{\iint_D f(x,y) \ dx\, dy = \iint_T f(\rho \cos \phi, \rho \sin \phi) \rho \, d \rho\, d \phi}$
- $\displaystyle
{\iiint_D f(x,y,z) \, dx\, dy\, dz = \iiint_T f(\rho \cos \phi, \rho \sin \phi, z) \rho \, d\rho\, d\phi\, dz}$
- $f(x,y,z) \longrightarrow f(\rho \cos \theta \sin \phi, \rho \sin \theta \sin \phi, \rho \cos \phi)$
- $\displaystyle
{\iiint_D f(x,y,z) \, dx\, dy\, dz \\
= \iiint_T f(\rho \sin \phi \cos \theta, \rho \sin \phi \sin \theta, \rho \cos \phi) \rho^2 \sin \phi \, d\rho\, d\theta\, d\phi}$
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- $\displaystyle \frac{\partial \rho}{\partial t} + \nabla \cdot (\rho {\bf V}) = \frac{\partial \rho'}{\partial t} + \rho_0 \nabla \cdot {\bf V}' = 0 $
- We can write $P' = (\partial P/\partial \rho)_s \rho'$ and rewrite the continuity equation to get
- $\displaystyle \frac{\partial P'}{\partial t} + \rho_0 \left ( \frac{\partial P}{\partial \rho} \right )_s \nabla \cdot {\bf V'} = 0 $
- We can write $P' = c_s^2 \rho'$.
- $\displaystyle -\omega^2 + c_s^2 k^2 - 4\pi G \rho_0 = 0$
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- $\displaystyle -\rho \frac{\partial \Phi}{\partial t} + \rho \frac{v^2}{2} + g z \rho + p = B(t), \\ \displaystyle -\rho \frac{\partial \Phi'}{\partial t} + \rho' \frac{v'^2}{2} + g \rho' z + p = B'(t).$
- $\displaystyle \rho \left ( \omega - k U \right )^2 \coth kh + \rho' \left ( \omega - k U' \right )^2 \coth kh' = k g \left (\rho' - \rho \right ),$
- $\displaystyle \frac{\omega}{k} = \frac{\rho U \coth kh + \rho' U'\coth k h'}{\rho \coth kh + \rho' \coth k h'} \pm \nonumber \\ \displaystyle ~~~ \left [ \frac{g}{k} \frac{\rho' - \rho}{\rho \coth kh + \rho' \coth k h'} - \frac{\rho \rho' \coth kh \coth kh' \left ( U - U' \right )^2}{\left (\rho \coth kh + \rho' \coth k h' \right)^2 } \right ]^{1/2}.$
- $\displaystyle \frac{\omega^2}{k^2} = \frac{g}{k} \frac{\rho' - \rho}{\rho \coth k h + \rho' \coth k h'} $
- $\displaystyle \omega^2 = k^2 \frac{g (\rho' - \rho) h h'}{\rho h'+\rho' h}$
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- $\displaystyle \frac{\partial \rho}{\partial t} + \nabla \cdot (\rho {\bf V}) = \frac{\partial \rho'}{\partial t} + \rho_0 \nabla \cdot {\bf V}' = 0 $
- We can write $\displaystyle P' = (\partial P/\partial \rho)_s \rho'$ and rewrite the continuity equation to get
- $\displaystyle \frac{\partial P'}{\partial t} + \rho_0 \left ( \frac{\partial P}{\partial \rho} \right )_s \nabla \cdot {\bf V'} = 0$
- $\displaystyle \frac{\partial ^2 P'}{\partial t^2} + \rho_0 \left ( \frac{\partial P}{\partial \rho} \right )_s \nabla \cdot \frac{\partial \bf V'}{\partial t} = 0.$
- $\displaystyle - \omega {\bf v}' + \frac{p'}{\rho_0} {\bf k} = 0$
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- $f(x,y,z) \longrightarrow f(\rho \cos \theta \sin \varphi, \rho \sin \theta \sin \varphi, \rho \cos \varphi)$
- $\displaystyle{\frac{\partial (x,y,z)}{\partial (\rho, \theta, \varphi)}} = \begin{vmatrix} \cos \theta \sin \varphi & - \rho \sin \theta \sin \varphi & \rho \cos \theta \cos \varphi \\ \sin \theta \sin \varphi & \rho \cos \theta \sin \varphi & \rho \sin \theta \cos \varphi \\ \cos \varphi & 0 & - \rho \sin \varphi \end{vmatrix} = \rho^2 \sin \varphi$
- The $dx\, dy\, dz$ differentials therefore are transformed to $\rho^2 \sin \varphi \, d\rho \,d\varphi \,dz$.
- $f(\rho \sin \varphi \cos \theta, \rho \sin \varphi \sin \theta, \rho \cos \varphi) = \rho^2$
- $\begin{aligned} \iiint_D (x^2 + y^2 +z^2) \, dx\, dy\, dz &= \iiint_T \rho^2 \ \rho^2 \sin \theta \, d\rho\, d\theta\, d\phi, \\ &= \int_0^{\pi} \sin \phi \,d\phi \int_0^4 \rho^4 d \rho \int_0^{2 \pi} d\theta \\ &= 2 \pi \int_0^{\pi} \sin \phi \left[ \frac{\rho^5}{5} \right]_0^4 \, d \phi \\ &= 2 \pi \left[ \frac{\rho^5}{5} \right]_0^4 \left[- \cos \phi \right]_0^{\pi}= \frac{4096 \pi}{5} \end{aligned}$