# Thread: Problem with moments and center of mass

1. ## Problem with moments and center of mass

I have a real problem here. It's from Stewart's Early Transcendental's 5th Edition. Section 8.3 problem 33. I'm studying from this book as a hobby. I'm just trying to solve for M sub x. The formula is density times the integral from -1 to +1 of the function 1/2(2-2x)^2 dx. The solutions manual acts like it's an even symmetrical function and converts it to
2*density integral from 0 to +1 of the same function. I cant see how that function is an even function. But even if I try to solve it without calling it an even function I cant match the book's answer. I cant find this problem anywhere on the web and I dont know how to draw it here. I'm hoping that someone out there has this book and understands this problem. Thanks

2. ## Re: Problem with moments and center of mass

$\displaystyle M_x = \rho \int_a^b \frac{1}{2}[f(x)]^2 \, dx$

$\displaystyle \rho = 1$

$\displaystyle f(x) = 2 - |2x|$

$\displaystyle [f(x)]^2 = 4x^2 - 8|x| + 4$

$\displaystyle M_x = \int_{-1}^1 2x^2 - 4|x| + 2 \, dx$

since the integrand is an even function, i.e. $\displaystyle f(-x) = f(x)$ ...

$\displaystyle M_x = 2 \int_0^1 2x^2 - 4x + 2 \, dx = 2\left[\frac{2x^3}{3} - 2x^2 + 2x\right]_0^1 = \frac{4}{3}$

3. ## Re: Problem with moments and center of mass

Thanks Skeeter so it's the absolute value sign that causes that function to become even? I dont think I've ever seen an absolute value sign in the integrand. The student guide never mentions the abolute value sign but I remember from the discussion on cusps that you need them. Thanks a lot I feel better now!