Answer:
[tex]M=168k[/tex]
[tex](\bar{x},\bar{y})=(5,\frac{85}{28})[/tex]
Step-by-step explanation:
Let's begin with the mass definition in terms of density.
[tex]M=\int\int \rho dA[/tex]
Now, we know the limits of the integrals of x and y, and also know that ρ = ky², so we will have:
[tex]M=\int^{9}_{1}\int^{4}_{1}ky^{2} dydx[/tex]
Let's solve this integral:
[tex]M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx[/tex]
[tex]M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx[/tex]
[tex]M=k\int^{9}_{1}21dx[/tex]
[tex]M=21k\int^{9}_{1}dx=21k*x|^{9}_{1}[/tex]
So the mass will be:
[tex]M=21k*8=168k[/tex]
Now we need to find the x-coordinate of the center of mass.
[tex]\bar{x}=\frac{1}{M}\int\int x*\rho dydx[/tex]
[tex]\bar{x}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}x*ky^{2} dydx[/tex]
[tex]\bar{x}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}x*y^{2} dydx[/tex]
[tex]\bar{x}=\frac{1}{168}\int^{9}_{1}x*\frac{y^{3}}{3}|^{4}_{1}dx[/tex]
[tex]\bar{x}=\frac{1}{168}\int^{9}_{1}x*21 dx[/tex]
[tex]\bar{x}=\frac{21}{168}\frac{x^{2}}{2}|^{9}_{1}[/tex]
[tex]\bar{x}=\frac{21}{168}*40=5[/tex]
Now we need to find the y-coordinate of the center of mass.
[tex]\bar{y}=\frac{1}{M}\int\int y*\rho dydx[/tex]
[tex]\bar{y}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}y*ky^{2} dydx[/tex]
[tex]\bar{y}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}y^{3} dydx[/tex]
[tex]\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{y^{4}}{4}|^{4}_{1}dx[/tex]
[tex]\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{255}{4}dx[/tex]
[tex]\bar{y}=\frac{255}{672}\int^{9}_{1}dx[/tex]
[tex]\bar{y}=\frac{255}{672}8=\frac{2040}{672}[/tex]
[tex]\bar{y}=\frac{85}{28}[/tex]
Therefore the center of mass is:
[tex](\bar{x},\bar{y})=(5,\frac{85}{28})[/tex]
I hope it helps you!
