Respuesta :
Answer:
mass, m=0.4525
moment, M = 0.4222
x-coordinate = 0.9330
Explanation:
lets write out the useful formulas.[tex]\\[/tex]
the [tex]mass, m = \int\limits \,\int\limits {p} \, dA[/tex] [tex]\\[/tex]
where [tex dA=dxdy[/tex] [tex]\\[/tex]
hence we can express the mass inters of x and y axes [tex]\\[/tex]
[tex]mass,m=\int\limits \int\limits {p} \,dx \, dy[/tex] [tex]\\[/tex]
Next we find expression for the moment [tex]\\[/tex]
[tex]moment ,M_{y}=mx = \int\limits \int\limits x{p} \,dx \,dy[/tex]
[tex]\\[/tex]
lastly, the x-coordinate
[tex]\\[/tex]
[tex]x=\frac{moment,M_{y} }{mass,m}[/tex]
[tex]\\[/tex]
Now from the question we were given the expression for y as
[tex]\\[/tex]
[tex]y= \frac{1}{\sqrt{x^{2}+6x+13 }}[/tex]
[tex]\\[/tex]
and the points of x which are
[tex]\\[/tex]
x=2 from 0. [tex]\\[/tex]
also the densit, p=1
from the expression for the mass [tex]\\[/tex]
[tex]mass,m=\int\limits \int\limits {p} \,dx \,dy[/tex] [tex]\\[/tex]
[tex]mass, m =\int\limits^2_0 \int\limits^y_0 {1} \,dx \,dy \\[/tex]
[tex][tex]mass.m=\int\limits^2_0 \, dx[y]^{y} _{0} \\[/tex] [\int\limits^y_0 \,dy ]\\[/tex]
[tex]mass.m=\int\limits^2_0 \,ydx[/tex] [tex]\\[/tex]
substitute the value of y [tex]\\[/tex]
[tex]mass,m=\int\limits^2_0 \,\frac{1}{\sqrt{x^{2}+6x+13}} dx\\[/tex]
we can manipulate the expression under the square-root to something familiar under integration [tex]\\[/tex]
[tex](x^{2}+6x)+9+4 \\(x+3)^{2}+4\\[/tex]
Now we can easily integrate the expression using the integral of the inverse of an hyperbolic function. the identity is shown below [tex]\\[/tex]
[tex]\int\limits^a_b {\frac{1}{\sqrt{(u+a)^{2}+c^{2} } } } \, dx = sinh^{-1} \frac{u+a}{c} +k\\k = constant[/tex]
Hence
[tex]mass,m=[sinh^{-1} (\frac{x+3}{2} )]^{2} _{0} \\[/tex]
[tex]mass,m=[sinh^{-1} (\frac{2+3}{2} )]-[sinh^{-1} (\frac{0+3}{2} )]\\[/tex]
Note: [tex]sinh^{-1}x \neq (sinhx)^{-1}=\frac{1}{sinhx} \\[/tex]
Rather we use the expansion [tex]sinh^{-1}x=ln(x+\sqrt{1+x^{2} })\\[/tex]
[tex]mass,m=1.6473-1.19483\\mass,m=0.4525\\[/tex]
Next we solve for the moment about the y-axis
from the expression for moment ,we have
[tex]moment ,M_{y} =mx = \int\limits \int\limits x{p} \, dx \, dy[/tex]
[tex]\\[/tex]
By following the same method of integral for the expansion process like we did above, we arrive at
[tex]moment ,M_{y} =\int\limits^2_0 {\frac{x}{\sqrt{(x+3)^{2} +4 } } } \, dx\\[/tex]
integrating at the definite points,we have
[tex]moment ,M_{y}=[\sqrt{(2+3)^{2}+4} -3sinh^{-1} (\frac{2+3}{2} )] - [\sqrt{(0+3)^{2}+4} -sinh^{-1} (\frac{0+3}{2} )]\\[/tex]
[tex]moment ,M_{y} =0.4433-0.02106\\moment ,M_{y} =0.4222\\[/tex]
Finally we solve for the x-coordinate, using the ratio of the moment to the mass we obtained earlier
[tex]x=\frac{moment}{mass}\\ x=\frac{0.4222}{0.4525} \\x=0.9330\\[/tex]
