Q 44 - Calculus

Question

Find each mass described in Exercises 41–48.

The mass of a cylindrical rod with a radius of 3 centimeters and a length of 20 centimeters, made of a combination of copper and aluminum in such a way that

the density of the rod x centimeters from the left end is $\rho \left(x\right) = 8.93 - 0.015x$ grams per cubic centimeter

Step-by-Step Solution

Answer

 The mass of the rod is 5,134.62 grams 

Step 1: Given Information

The mass of a cylindrical rod with a radius of 3 centimeters and a length of 20 centimeters . the density of the rod x centimeters is  $\rho \left(x\right) = 8.93 - 0.015x$

Step 2: Formula

Volume of cylinder is ${\mathrm{\pi r}}^2\mathrm{h}$

where 'r' is the radius and h is the height of the cylinder 

Each slice is of radius 3  and width  $∆x$

So, the volume becomes 

$$\begin{gathered} {\mathrm{\pi r}}^2\mathrm{h} \\ \mathrm{\pi }{\left(3\right)}^2∆\mathrm{x} \\ 9\mathrm{\pi }∆\mathrm{x} \end{gathered}$$

Step 3: Density

Density of entire slice is   

$$\begin{gathered} \rho \left(x_i\right) = 8.93 - 0.015x_i \\ \rho =\frac{M}{V} \\ 8.93 - 0.015x_i=\frac{M}{9\mathrm{\pi }∆\mathrm{x}} \\ M=\left(8.93 - 0.015x_i\right)9\mathrm{\pi }∆\mathrm{x} \end{gathered}$$

Step 4: Sum

approximate mass is given by sum 

$$\begin{gathered} \underset{i}{\sum \left(8.93 - 0.015x_i\right)9\mathrm{\pi }∆\mathrm{x}} \\ \mathrm{To} \mathrm{get} \mathrm{total} \mathrm{mass}, \mathrm{accumulate} \mathrm{the} \mathrm{mass} \mathrm{from} \mathrm{x}=0 \mathrm{to} 20 \\ \mathrm{Mass}={\int }_0^{20}\left(8.93 - 0.015\mathrm{x}\right)9\mathrm{\pi dx} \\ =9\mathrm{\pi }{\int }_0^{20}\left(8.93 - 0.015\mathrm{x}\right)\mathrm{dx} \\ =9\mathrm{\pi }{\left[8.93\mathrm{x}-\frac{0.015{\mathrm{x}}^2}{2}\right]}_0^{20} \\ =9\mathrm{\pi }\left[8.93\left(20\right)-\frac{0.015{\left(20\right)}^2}{2}\right] \\ =5134.62 \mathrm{grams} \end{gathered}$$