Divanshu's question via email about cylindrical shells

[Prove It]

View attachment 5638

Here is a sketch of the region to be rotated around the y axis.

View attachment 5639

You first need to visualise this entire region being rotated around the y axis, to get a mental picture of what the solid looks like. Then you need to imagine that the solid is made up of very thin vertically-oriented hollow cylinders. You can then approximate the volume of the solid by adding up the volumes of all the cylinders.

The curved surface of each cylinder is a rectangle. The width of each rectangle is the y value of the function. The length of the rectangle is the circumference of the cylinder, so $\displaystyle \begin{align*} 2\,\pi\,r \end{align*}$, and the radius of the cylinder is the x value of the function. So the area of each cylinder is $\displaystyle \begin{align*} 2\,\pi\,x\,y \end{align*}$, and thus the volume of each cylinder is $\displaystyle \begin{align*} 2\,\pi\,x\,y\,\Delta x \end{align*}$, where $\displaystyle \begin{align*} \Delta x \end{align*}$ is some small change in x.

So the total volume of the solid can be approximated by $\displaystyle \begin{align*} \sum{ 2\,\pi\,x\,y\,\Delta x } \end{align*}$.

If we increase the number of cylinders in the region and make each cylinder thinner, we get a better approximation of the total volume, so letting $\displaystyle \begin{align*} n \to \infty \end{align*}$ and $\displaystyle \begin{align*} \Delta x \to 0 \end{align*}$, the approximation becomes exact and the sum becomes an integral.

So the total volume of the solid is exactly

$\displaystyle \begin{align*} V &= \int_1^3{ 2\,\pi\,x\,y\,\mathrm{d}x } \\ &= 2\,\pi\int_1^3{ x\,\left( 2 + \frac{1}{5\,x} \right) \,\mathrm{d}x } \\ &= 2\,\pi \int_1^3{ \left( 2\,x + \frac{1}{5} \right) \,\mathrm{d}x } \\ &= 2\,\pi \,\left[ x^2 + \frac{x}{5} \right] _1^3 \\ &= 2\,\pi\,\left[ \left( 3^2 + \frac{3}{5} \right) - \left( 1^2 + \frac{1}{5} \right) \right] \\ &= 2\,\pi \,\left( 8 + \frac{2}{5} \right) \\ &= 2\,\pi\,\left( \frac{42}{5} \right) \\ &= \frac{84\,\pi}{5}\,\textrm{units}^3 \end{align*}$

 

[jvicens]

Thank you for sharing your calculation for the volume of the solid. Your approach of visualizing the solid as a collection of thin cylinders and then using integration to find the exact volume is a common and effective method in mathematics and physics. It is important to note that this method assumes the solid is continuous and has a smooth surface, which may not always be the case in real-world scenarios.

Additionally, as a scientist, it is important to consider the units of measurement in your calculation. In this case, the units of the volume should be in $\textrm{units}^3$, as you have correctly included in your final answer. It is also important to check the units of each term in your integral to ensure they are consistent.

Overall, your explanation and calculation are clear and well-reasoned. Keep up the good work!