Q 43 - Calculus

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.

Step-by-Step Solution

Answer

The mass of the rod is 139.32 grams

Step 1: Given Information

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

Step 2: Find Volume

Volume of cuboid = length time width times height

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

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

Step 3: Density function

Let xi represents slice of the rod at some point 

The density throughout enter slice is 

$$\begin{gathered} \rho \left(x_i\right) = 4.2 + 0.4x_i - 0.03{x_i}^2 \\ Now we use the density formula \\ \rho =\frac{M}{V} \\ M is the mass and V is the volume \\ 4.2 + 0.4x_i - 0.03{x_i}^2=\frac{M}{{(1.5)}^2∆x} \\ M=\left( 4.2 + 0.4x_i - 0.03{x_i}^2\right){(1.5)}^2∆x \end{gathered}$$

Step 4: Integration

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

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