Question

Find each mass described in Exercises 41–48.

The mass of a 12-inch rod with a square cross section of side length 1.5 inches, with density x inches from the left end given by $\rho \left(x\right)=4.2+0.4x-0.03x^2$grams per

cubic inch.

Short answer

The mass of the rod is139.32 grams

Step-by-step solution

Given Information 

The mass of a 12-inch rod with a square cross section of side length 1.5 inches, with density $\rho \left(x\right)=4.2+0.4x-0.03x^2$

Find Volume 

Volume of cuboid = length time width times height

The length and height of cuboid is 1.5 inches and width is $∆x$

$$\begin{gathered} Volume=(1.5)(1.5)∆x \\ V={(1.5)}^2∆x \end{gathered}$$

Density function 

Let xi represents slice of the rod at some point

The density throughout enter slice is

$$\begin{gathered} \rho \left(x_i\right)=4.2+0.4x_i-0.03{x_i}^2 \\ Nowweusethedensityformula \\ \rho =\frac{M}{V} \\ MisthemassandVisthevolume \\ 4.2+0.4x_i-0.03{x_i}^2=\frac{M}{{(1.5)}^2∆x} \\ M=\left(4.2+0.4x_i-0.03{x_i}^2\right){(1.5)}^2∆x \end{gathered}$$

Integration 

The approximate of the entire rod is the sum from x=0 to 12

$$\begin{gathered} M=\sum _{i=0}^{12}\left(4.2+0.4x_i-0.03{x_i}^2\right){(1.5)}^2∆x \\ M={\int }_0^{12}\left(4.2+0.4x-0.03{xi}^2\right){(1.5)}^{2}dx \\ M=2.25{\int }_0^{12}\left(4.2+0.4x-0.03{xi}^2\right)dx \\ M=2.25{\left(4.2x+\frac{0.4x^2}{2}-\frac{0.03x^3}{3}\right)}_0^{12} \\ =2.25\left(4.2\left(12\right)+\frac{0.4{\left(12\right)}^2}{2}-\frac{0.03{\left(12\right)}^3}{3}\right)-\left(2.25\left(4.2\left(0\right)+\frac{0.4{\left(0\right)}^2}{2}-\frac{0.03{\left(0\right)}^3}{3}\right)\right) \\ =139.32 \end{gathered}$$