• Nov 15th 2009, 04:10 PM
Skittles8768
Find the mass and center of mass of the lamina for density = k
where R = a triangle with vertices (0,0), (b/2,h), (b,0)

(use double integrals to solve for both the mass and center of mass!)

ive tried this problem a million times and i cant get the right answer :-( (and yeah i checked calcchat but i need a more detailed explanation....walk me through it please? :-) )
• Nov 15th 2009, 05:04 PM
Chris L T521
Quote:

Originally Posted by Skittles8768
Find the mass and center of mass of the lamina for density = k
where R = a triangle with vertices (0,0), (b/2,h), (b,0)

(use double integrals to solve for both the mass and center of mass!)

ive tried this problem a million times and i cant get the right answer :-( (and yeah i checked calcchat but i need a more detailed explanation....walk me through it please? :-) )

First off, if $\displaystyle \delta\!\left(x,y\right)$ is the density of the lamina, then $\displaystyle M=\iint\limits_{R}\delta\!\left(x,y\right)\,dA$.

In our case, $\displaystyle R$ is a triangle with vertices $\displaystyle (0,0)$, $\displaystyle (b,0)$ and $\displaystyle (b/2,h)$. From this, we can come up with the equation of the line that connects the points $\displaystyle (b,0)$ and $\displaystyle (b/2,h)$ and the points $\displaystyle (0,0)$ and $\displaystyle (b/2,h)$. I leave it for you to verify that the equations of the lines are $\displaystyle y=-\frac{2h}{b}x+2h$ and $\displaystyle y=\frac{2h}{b}x$ respectively.

Thus, it follows that if $\displaystyle \delta\!\left(x,y\right)=k$, then

$\displaystyle M=\iint\limits_{R}k\,dA=k\cdot\text{Area of$R$}=k\cdot\left(\tfrac{1}{2}bh\right)=\tfrac12 kbh$

Now, to find center of mass, we need to evaluate $\displaystyle M_x=\iint\limits_{R}x\delta\!\left(x,y\right)\,dA$ and $\displaystyle M_y=\iint\limits_{R}y\delta\!\left(x,y\right)\,dA$.

From there, it follows that

$\displaystyle \bar{x}=\frac{M_x}{M}=\frac{\displaystyle\iint\lim its_{R}x\delta\!\left(x,y\right)\,dA}{\displaystyl e\iint\limits_{R}\delta\!\left(x,y\right)\,dA}$ and $\displaystyle \bar{y}=\frac{M_y}{M}=\frac{\displaystyle\iint\lim its_{R}y\delta\!\left(x,y\right)\,dA}{\displaystyl e\iint\limits_{R}\delta\!\left(x,y\right)\,dA}$

Note that $\displaystyle M_x=k\int_0^h\int_{\frac{b}{2h}y}^{-\frac{b}{2h}y+b}x\,dx\,dy$ and $\displaystyle M_y=k\int_0^h\int_{\frac{b}{2h}y}^{-\frac{b}{2h}y+b}y\,dx\,dy$.

I leave it for you to simplify the two integrals and figure out the coordinates for the center of mass.

Can you take it from here?