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The inertia is that where a body offers a
resistance to move linearly and is directly proportional to the mass. If
this reluctance is seen in a rotating body about an axis we use the
term moment of inertia. It gives the distribution of mass about the axis of rotation. It is given as Moment of inertia = mass $\times$ radius^{2} (kgm^{2}) I_{A} = m $\times$ r^{2} Moment of inertia = m_{1}r_{1}^{2} + m_{2}r_{2}^{2} + m_{3}r_{3}^{2} + ..... + m_{n}r_{n}^{2} Where n is the number of samples taken I_{A} = $\sum$ m_{n} r_{n}^{2} There are two factors that determine the moment of inertia of rotating body
Since mass is expressed in kilogram and radius in meters.The unit of Moment of inertia is kilogram meter square (kgm^{2}). 
It is defined as the sum of the products of the mass and the square of the perpendicular distance to the axis of rotation of each particle in a body rotating about an axis. It is given as
I = Mr^{2}Where M = mass of the body,
r = radius of body from axis of rotation.
The Moment of inertia of rectangle varies in x and y axes as shown in figureThe moment of inertia I_{x} = $\int_A$ y^{2} dA
= $\int_0^h$ y^{2} (b dy)
= $\frac{hy^3}{3}$ _{}_{0}^{h}
= $\frac{bh^3}{3}$
I_{x} = I_{x'}= $\int_A$ y^{2} dA
= 4 $\int_0^h$ y2 ($\frac{b}{2}$ dy)
= 4 $\frac{b}{2}$ $\frac{y^3}{3}$_{0}^{h/2}
= $\frac{bh^3}{12}$
The moment of inertia I_{y }= $\int_A$ x^{2} dA
= $\int_0^b$ x^{2} (h dx)
= $\frac{hx^3}{3}$_{}_{0}^{b}
= $\frac{hb^3}{3}$
Iy = Iy'= $\int_A$ x^{2} dA
= 4 $\int_0^h$ x^{2} ($\frac{h}{2}$ dx)
= 4 $\frac{h}{2}$ $\frac{x^3}{3}$_{}_{0}^{b/2}
= $\frac{hb^3}{12}$
I_{x} = $\bar{I_x}$ + Ad^{2}
= $\frac{bh^3}{12}$ + (bh) $\frac{h}{2}^2$
= $\frac{bh^3}{12}$ + $\frac{bh^3}{4}$
= $\frac{bh^3}{3}$.
The moment of inertia of a solid sphere with the axis of rotation at center is
I = $\frac{2}{5}$ MR^{2}^{}Here M is the mass and R is the radius of axis of rotation.
The moment of inertia of the triangle is
I_{x} = $\frac{1}{12}$ y^{3} dx
I_{x} = $\int$ y^{2} dM = $\int$ $\int$ y^{2} $\rho$ dy dx
The moment of inertia of triangle $\bar{I_x}$ = $\frac{1}{36}$ bh^{3} and I_{x} = $\frac{1}{12}$ bh^{3}.
The Moments of Inertia of cylinder is
 Solid cylinder with axis of rotation at the center = $\frac{1}{2}$ M R^{2}
 Solid cylinder with axis of rotation as surface = $\frac{3}{2}$ M R^{2}
 Hallow cylinder with axis of rotation as center = M R^{2}
 Hallow cylinder with axis of rotation as surface = 2 MR^{2}
The moment of inertia of circle is
$\bar{I_x}$ = I_{y} = $\frac{1}{4}$ $\pi$ r^{4} and J_{o} = $\frac{1}{2}$ $\pi$ r^{4}
I = $\int_0^M$ r^{2} dm = $\int_{L/2}^{L/2}$ r^{2}$\frac{M}{L}$ dr = $\frac{1}{12}$ ML^{2}^{}
The moment of inertia along radial axis is given by I = $\frac{1}{2}$ MR^{2}
The moment of inertia along the central axis is given by I = $\frac{1}{4}$ MR^{2}
The Moment of inertia of beam varies based on the inclination as shown in figure:
The first moment of inertia tells that the moment arm will be raised to a power of one. The first moment of any area calculated about xaxis would be given by,
I_{x} = $\int$ y dA Similarly, the first moment of inertia about yaxis would be given by,
I_{y} = $\int$ x dA
The second moment of inertia is that the moment arm is raised to a power of 2. So the second moment about xaxis is given by,I_{x} = $\int y^{2} dA $
Similarly, the second moment of inertia about yaxis is given asI_{y} = $\int x^{2} dA $
In the similar way you can calculate the third moment of inertia.
The polar moment of inertia is given by
J = $\sum$ x r^{2}where r = the radius of small area, da from the perpendicular axis  for a plane area the perpendicular axis is a point
The polar moment of inertia is the sum of any two moments of inertia about axes which are at right angles to each other
J = I_{xx} + I_{yy}where I_{xx} gives the moment of Inertia in xaxis
I_{yy} gives the moment of inertia in yaxis
or
The polar moment of inertia J of an element about an axis perpendicular to its plane is nothing but the product of the area of the element and square of its distance from the axis.
The polar moment of inertia is
dJ = $\rho$^{2 }dA
= (x^{2}+y^{2})dA
= dI_{y} + dI_{x}
= (x^{2}+y^{2})dA
= dI_{y} + dI_{x}
Lets see that in the figure the moment of area by summing together l^{2}dA for all the given elements of area dA in the above region.
Hence Area moment of inertiaI = $\int$ A l^{2} dA For a rectangular region the area moment of inertia,
I_{x} = $\int$ A y^{2} dA
I_{y} = $\int$ A x^{2} dA
The above equations tells gives the area moment of inertia.I_{y} = $\int$ A x^{2} dA
The Mass Moment of Inertia of a solid tells the solid's ability to resist changes in rotational speed about a specific axis. The larger the Mass Moment of Inertia the smaller the angular acceleration would be about an axis for a given torque. The mass moment of inertia is the measurement of the distribution of the mass of an object or body relative to a given axis. It is represented by the symbol I.
For a particle of mass m the moment of inertia is
I_{o} = MR^{2}
Lets say that the body is made of sum of particles each having mass dm. The formula is
I = $\int \sum$ m r^{2} dm Similarly
I = $\int \sum$ r^{2} p $\rho$ V
where $\rho$ = density and for area dA the moment of inertia isI = $\int \sum$ r^{2} p dA
Here is the table for moment of inertiaCLICK TO EDIT PARAGRAPH CONTENT