Find the center of mass for the region bounded by y = x² and y = x + 2 with density p(x,y) = x.

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To determine the bounded region, graph the given equations. (Please see attached figure.)

On the bounded region, draw a dot to represent its center of mass. To get its coordinates, we have to solve for its mass. Also, let's consider the bounded region as a thin plate.

The formula to compute its mass is:

`M = rho int_a^b f(x) - g(x)dx`

where ` rho ` represents the density, f(x) is the upper curve and g(x) is the lower curve.

The upper graph of the bounded region is y = x +2 and the lower curve is y=x^2. And bounded region starts at x=-1 and ends at x=2.

So, the mass of the plate is:

`M=x int_(-1)^2 (x+2-x^2)dx`

`M = x int_(-1)^2 (-x^2+x+2)dx`

`M=x[(-x^3/3+x^2/2+2x)|_(-1)^2]`

`M=9/2x`

Next, let's determine the moments of the plate. This is the tendency of the plate to rotate about the x and y axis.

The formula of moment of the plate about the x-axis is:

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

So, the moment of the plate about the x-axis is:

`M_x=1/2x int_(-1)^2 ((x+2)^2 - (x^2)^2) dx`

`M_x =1/2 x int_(-1)^2 (x^2+2x + 4-x^4 )dx`

`M_x = 1/2x int_(-1)^2 (-x^4 + x^2+2x+4)dx`

`M_x=1/2x[(-x^5/5+x^3/3+x^2+4x)|_(-1)^2]`

`M_x=36/5x`

The formula of moment of the plate about the y-axis is:

`M_y= rho int_a^b x[f(x) - g(x)]dx`

So, the moment of the plate about the y-axis is:

`M_y = x int_(-1)^2 x(x+2-x^2)dx`

`M_y=x int_(-1)^2 (-x^3 +x^2+2x)dx`

`M_y = x [(-x^4/4+x^3/3+x^2)|_(-1)^2]`

`M_y=9/4x`

Now that the mass and the moments of the plates are known, the coordinates of the center of mass can now be solved. The formula are:

`x_c= M_y/M`

`y_c = M_x/M`

Plugging in the mass and the moments, the coordinates will be:

`x_c= (9/4x)/(9/2x)=1/2`

`y_c=(36/5x)/(9/2x)=8/5`

Therefore, the center of mass of the bounded region is `(1/2,8/5).`

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