`y=x^(2/3), y=4` 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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gsarora17 eNotes educator| Certified Educator

Consider 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 ;




Refer to attached image. The plot of `y=x^(2/3)` is red in color.

The graphs intersect at the coordinates `(-8,4),(8,4)`

Let's evaluate the area of the bounded region,












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

Since the graph is symmetrical about the y-axis,

So, `M_y=0` and `barx=0` ,

`M_x=rhoint_(-8)^8 1/2([4^2]-[x^(2/3)]^2)dx`













The coordinates of the center of mass are `(0,20/7)`


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