Calculating the mass of a beam of non uniform density.

[Craptola]

Been doing some physics problems from my mechanics class. This is the first time I've attempted a problem like this and am not so confident with solving it, I could be correct and just being paranoid but I just have a feeling I've made a mistake somewhere, would appreciate if someone who is comfortable with this kind of thing would tell me if/where I'm going wrong.

Homework Statement

Consider a thin beam of length L=3 m, and height and width equal to w= 2 cm and h=2 cm respectively. Assume that the composure of the beam is from a mixture of materials, that have given it a non-uniform mass density, which varies continuously along x [where x represents the coordinate for the direction along the length of the beam, measured from the left edge of the beam]. Assume that such mass density as a function of x is given by:
$$\rho = \rho _{0}e^{\alpha x}$$
where ρo= 9.10^3 kg/m3 and α=1/L
Find the total mass for the beam.

Homework Equations

$$M=\int_{0}^{3}dm$$

$$dm=\rho dv$$

The Attempt at a Solution

$$dv=0.0004dx$$
$$\therefore dm=0.0004\rho dx = 3.6e^{-\frac{1}{3}x}dx$$
$$M=\int_{0}^{3}dm = 3.6\int_{0}^{3}e^{-\frac{1}{3}x}dx$$
$$= -10.8\left ( e^{-1} - 1 \right ) = 6.83kg$$

 

[haruspex]

$$dv=0.0004dx$$

It is a good idea to work with the algebra until you have the final formula and only then substitute in the values of the known constants. You're less likely to make a mistake, it's easier to spot where a mistake has been made (e.g. by considering dimensionality) and it's easier for others to follow.

$$\therefore dm=0.0004\rho dx = 3.6e^{-\frac{1}{3}x}dx$$

Where did the minus sign come from? Did you omit that in the formula for density?

Otherwise, looks fine.

 

[Craptola]

Where did the minus sign come from? Did you omit that in the formula for density?

Yeah, that's a typo, thanks for the help.