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Divanshu's question via email about cylindrical shells

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
cylindrical shells question.JPG

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

cylindrical shells question graph.JPG

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

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