Find the mass m, the moment my, and the x-coordinate x of the center of mass of the uniform flat plate of density rho = 1 bounded by the curve y = 1 x2 + 6x + 13 , the line x = 2, and the x and y axes.

mass, m=0.4525

moment, M = 0.4222

x-coordinate = 0.9330

Explanation:

lets write out the useful formulas.

the

where [tex  dA=dxdy[/tex]  

hence we can express the mass inters of x and y axes  

 Next we find expression for the moment

lastly, the x-coordinate

Now from the question we were given the expression for y as

and the points of x which are

x=2 from 0.

also the densit, p=1

from the expression for the mass

[\int\limits^y_0  \,dy ]\\" alt="mass.m=\int\limits^2_0 \, dx[y]^{y} _{0} \\" /> [\int\limits^y_0  \,dy ]\\" />

substitute the value of y

we can manipulate the expression under the square-root to something familiar under integration

Now we can easily integrate the expression using the integral of the inverse of an hyperbolic function. the identity is shown below

Hence

Note:

Rather we use the expansion

Next we solve for the moment about the y-axis

from the expression for moment ,we have

By following the same method of integral for the expansion process like we did above, we arrive at

integrating at the definite points,we have

Finally we solve for the x-coordinate, using the ratio of the moment to the mass we obtained earlier

gets less

falling water hydro = water .. spectcular places these plants are