Differences in equation of moments.

Question

Context: http://archives.math.utk.edu/visual.calculus/5/work.2/

The formula in question:

$$\mathbf{M}_{x}=\int_{a}^{b}\frac{\rho}{2} \left \{ \left [ f\left ( x \right ) \right ]^{2} - \left [ g\left ( x \right ) \right]^{2} \right \}dx$$

Questions:

• Why $\rho$ is divided by 2 and $f(x)$ and $g(x)$ are squared?
• Why is $\mathbf{M}_{x}$ different from $\mathbf{M}_{y}$?

Thanks in advance

Answer 1

The moments are actually computed from double integrals, which can be integrated first on $y$:

$$\int_{x=a}^b\int_{y=f(x)}^{g(x)}\rho x\,dy\,dx=\int_{x=a}^b\rho xy\Big|_{y=f(x)}^{g(x)}\,dx=\int_{x=a}^b\rho x(g(x)-f(x))\,dx$$and $$\int_{x=a}^b\int_{y=f(x)}^{g(x)}\rho y\,dy\,dx=\int_{x=a}^b\rho\frac{y^2}2\Big|_{y=f(x)}^{g(x)}\,dx=\int_{x=a}^b\rho\frac{g(x)^2-f(x)^2}2\,dx.$$

The asymmetry comes from the fact that the domain is delimited by two explicit curves $y=f(x),y=g(x)$.

Similarly, the expression of the mass is obtained from $$\int_{x=a}^b\int_{y=f(x)}^{g(x)}\rho \,dy\,dx=\int_{x=a}^b\rho y\Big|_{y=f(x)}^{g(x)}\,dx=\int_{x=a}^b\rho (g(x)-f(x))\,dx.$$

Answer 2

It seems that $$ M_y = \int_a^b \rho y\left(f(x) -g(x)\right)dx $$ Now we have

$$ y = f(x)\\ y = g(x). $$ So mean value gives $$ y = \frac{f(x) +g(x)}{2} $$ Subbing in for $y$ in the integral yeilds the desired result.