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

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

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

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

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

From there, it follows that

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

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

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?