Question

Suppose that a thin metal bar of length 10 cm and cross-sectional area 1 cm has density given by p(x)=x³ g/cm³ , where x=0 represents the left end of the bar. (a) Compute the mass of the bar.

Answer 1

To calculate the mass of a metal bar with a varying density of p(x)=x³ g/cm³, integrate the density function across the bar's length, yielding a mass of 2500 grams.

Step-by-step explanation:

To compute the mass of the bar with the given density distribution p(x)=x³ g/cm³, we integrate the density over the length of the bar. The mass m can be found using the integral:

m = ∫ p(x) dV

Since the cross-sectional area A is 1 cm², the differential volume element dV is A dx = dx. The limits of integration for x range from 0 to 10 cm, corresponding to the length of the bar. Therefore, the integral becomes:

m = ∫_{0}^{10} x³ dx

To solve this, we find the antiderivative of x³, which is x⁴/4. Evaluating the integral from 0 to 10, we get:

m = (1/4) x⁴ |_{0}^{10} = (1/4)[10⁴ - 0⁴] = (1/4)[10000 - 0] = 2500 g

Thus, the mass of the bar is 2500 grams.