Consider the region bounded by the three curves:
$$\begin{aligned} &y=\frac{2x}{5}\\ &y=1\\ & x=x_0 \end{aligned}$$
Assuming a homogenous distribution of mass, the formula for the coordinates of the center of mass that I was given are: $$\begin{aligned}& x_s=\frac{1}{A}\int_A x \space dA \\ & y_s=\frac{1}{A} \int _A y \space dA\end{aligned}$$
- Calculate the center of mass for $x=-0.5$
- For what $x_0$ is the center of mass on the $x$-axis. What is $x_s$ in this case?
My attempt:
I tried to sketch the curves and the bounded region first:
So I am trying to find the coordinates of the center of mass of the triangle. But how do I use those formulas? Are $x$ and $y$ in the integral functions?
$\endgroup$ 52 Answers
$\begingroup$Hint.
The answer of Matthew Leingang is general. In your case the region is a triangle so you can find the centroid (that coincides with the center of mass since the density is constant) using the fact that the coordinates of the centroid are the means of the coordinates of the vertices.
$\endgroup$ 1 $\begingroup$Let $D$ be the region described. Since the density is uniform, can assume it's constantly one. The “mass” of the region $D$ is the same as its area, $$ m = \iint_D 1\,dA = \int_{x_0}^{5/2} \int_{2x/5}^1 1\,dy\,dx $$ The $x$-coordinate of the center of mass is $$ \bar x =\frac{1}{m} \iint_D x\,dA = \frac{1}{m}\int_{x_0}^{5/2} \int_{2x/5}^1 x\,dy\,dx $$ and the $y$-coordinate is $$ \bar y =\frac{1}{m} \iint_D y\,dA = \frac{1}{m}\int_{x_0}^{5/2} \int_{2x/5}^1 y\,dy\,dx $$
To answer one of your questions: yes, the $x$ and $y$ in the integrals are functions. You can antidifferentiate a function of two variables with respect to one of them just like you can differentiate a function of two variables with respect to one of them. That is, you treat the “other” variable as a constant.
Can you take it from there?
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