# Thread: Mass of enclosed area

1. ## Mass of enclosed area

I really do not understand this and of course I need to by a few hours from now. If any one can help I would really appreciate it!

I have an area enclosed by y = 3x and y = x^2 with x greater than or equal to 1 and less than or equal to 2

I need to find the mass of the region and then find the moment about the x axis and the moment about the y axis.

Thanks for all the help I have received this term!!!!!

2. Originally Posted by Frostking
I really do not understand this and of course I need to by a few hours from now. If any one can help I would really appreciate it!

I have an area enclosed by y = 3x and y = x^2 with x greater than or equal to 1 and less than or equal to 2

I need to find the mass of the region and then find the moment about the x axis and the moment about the y axis.

Thanks for all the help I have received this term!!!!!
The mass is just the integral (the area) if density is constant. the line is above the parabola for 1 < x < 2

so we get

$m=\int_{1}^{2}(3x-x^2)dx=\frac{3}{2}x^2-\frac{1}{3}x^3 \bigg|_{1}^{2}=6-\frac{8}{3}-(\frac{3}{2}-\frac{1}{3})=\frac{13}{6}$

The moment about the y axis is

$M_y=\rho\int_{a}^{b}x[f(x)-g(x)]dx$ where rho is the density

$M_y=\int_{1}^{2}x(3x-x^2)dx=\frac{13}{4}$

The moment about the x axis is

$M_x=\rho\int_{a}^{b}\frac{1}{2}\{ [f(x)]^2-[g(x)]^2\}dx$

$M_x=\frac{1}{2}\int_{1}^{2} \{ [3x]^2-[x^2]^2 \} dx =\frac{74}{5}$

Note that the center of mass is at the point $\left(\frac{M_x}{m},\frac{M_y}{m}\right)$

Good luck.