`y=x^(2/3), y=0, x=8` Find the x and y moments of inertia and center of mass for the laminas of uniform density `p` bounded by the graphs of the equations.

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For an irregularly shaped planar lamina of uniform density `rho` , bounded by graphs `y=f(x)` ,`y=g(x)` and `a<=x<=b` , the mass (m) of this region is given by:


where A is the area of the region.

The moments about the x and y-axes are given by the formula:



The center of mass `(barx,bary)` is given by `barx=(M_y)/m`  and `bary=M_x/m` ,



Now we given `y=x^(2/3),y=0,x=8`

The plot of the functions is attached as image and the bounds of the limits can be found from the same.

Area of the region A =`int_0^8x^(2/3)dx`

Use the power rule,








Now let's evaluate the moments about the x and y-axes,



`=rhoint_0^8 1/2x^(4/3)dx`

Take the constant out and apply the power rule,







` ` `=rho(3/7)(2)^6`













Now let's find the center of mass,


Plug in the value of `M_y` and  `A` ,




Plug in the values of `M_x` and `A` ,




The coordinates of the center of mass are,`(5,10/7)`

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