`y=sqrt(x) +1 , y=1/3x + 1` 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:


`m=rhoA` , where A is the area of the region.

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

`M_x=rhoint_a^b 1/2([f(x)]^2-[g(x)]^2)dx`


The center of mass `(barx,bary)` is given by:



We are given:`y=sqrt(x)+1,y=1/3x+1`

Refer to the attached image, plot of `y=sqrt(x)+1` is red in color and plot of `y=1/3x+1` is blue in color.The curves intersect at `(0,1)` and `(9,4)` .

Now let's evaluate the area of the region bounded by the graphs of the given equations,




Evaluate using power rule,








Now let's find the moments about the x- and y-axes using the formulas stated above,

`M_x=rhoint_0^9 1/2([sqrt(x)+1]^2-[1/3x+1]^2)dx`

`M_x=rhoint_0^9 1/2([x+2sqrt(x)+1]-[(1/3x)^2+2(1/3x)(1)+1^2])dx`

`M_x=rhoint_0^9 1/2(x+2sqrt(x)+1-x^2/9-2/3x-1)dx`

Take the constant out,



Apply the basic integration rules i.e sum and power rules,
























Now let's evaluate the coordinates of the center of mass by plugging in the values of the moments and area,









The center of mass is `(18/5,5/2)`


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