# Effie's question via email about a volume by revolution

#### Prove It

##### Well-known member
MHB Math Helper
What is the volume of the solid formed by the region bounded between the functions \displaystyle \begin{align*} y 2\,x^2 \end{align*} and \displaystyle \begin{align*} y = x + 1 \end{align*} being rotated around the line \displaystyle \begin{align*} y = 3 \end{align*}?
To start with, we should find the points of intersection of the two functions, as these will be the terminals of our regions of integration.

\displaystyle \begin{align*} 2\,x^2 &= x + 1 \\ 2\,x^2 - x - 1 &= 0 \\ 2\,x^2 - 2\,x + x - 1 &= 0 \\ 2\,x\,\left( x - 1 \right) + 1 \,\left( x - 1 \right) &= 0 \\ \left( x - 1 \right) \left( 2\,x + 1 \right) &= 0 \\ x - 1 &= 0 \textrm{ or } 2\,x + 1 = 0 \\ x &= 1 \textrm{ or } x = -\frac{1}{2} \end{align*}

As we are rotating around the line \displaystyle \begin{align*} y = 3 \end{align*}, it would make sense to move everything down by 3 units. The areas and volumes will still be exactly the same, but we will be rotating around the x axis.

The furthest function from the x axis will be \displaystyle \begin{align*} y = 2\,x^2 - 3 \end{align*}, so we should evaluate the volume formed by rotating the area between that function and the x axis between \displaystyle \begin{align*} x = -\frac{1}{2} \end{align*} and x = 1 around the x axis, and then subtract the volume formed by rotating the area between the function \displaystyle \begin{align*} y = x - 2 \end{align*} and the x axis between \displaystyle \begin{align*} x = -\frac{1}{2} \end{align*} and x = 1 around the x axis.

To evaluate each volume, picture each region being broken up into a number of rectangles. Each rectangle will then be rotated around the x axis to form a cylinder. The volume of a cylinder is \displaystyle \begin{align*} \pi\,r^2\,h \end{align*}. Each cylinder will have a radius that is the same as the y value, and a height that is equal to \displaystyle \begin{align*} \Delta x \end{align*}, a small change in x.

So the outer volume we can approximate by \displaystyle \begin{align*} \sum{ \pi\,\left( 2\,x^2 - 3 \right) ^2 \,\Delta x } \end{align*}, and if we increase the number of cylinders (making each cylinder thinner) we get a better approximation. As \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 outer volume is exactly \displaystyle \begin{align*} \int_{-\frac{1}{2}}^1{ \pi\,\left( 2\,x^2 - 3 \right) ^2\,\mathrm{d}x } \end{align*}. By identical arguments, the inner volume to be subtracted is exactly equal to \displaystyle \begin{align*} \int_{-\frac{1}{2}}^1{ \pi\,\left( x - 2\right) ^2 \,\mathrm{d}x } \end{align*}. Thus the total volume we want to evaluate is

\displaystyle \begin{align*} v &= \int_{-\frac{1}{2}}^1{ \pi\,\left( 2\,x^2 - 3 \right) ^2\,\mathrm{d}x } - \int_{-\frac{1}{2}}^1{ \pi\,\left( x - 2 \right) ^2\,\mathrm{d}x } \\ &= \pi\int_{-\frac{1}{2}}^1{ \left[ \left( 2\,x^2 - 3 \right) ^2 - \left( x - 2 \right) ^2 \right] \,\mathrm{d}x } \\ &= \pi \int_{-\frac{1}{2}}^1{ \left[ 4\,x^4 - 12\,x^2 + 9 - \left( x^2 - 4\,x + 4 \right) \right] \,\mathrm{d}x } \\ &= \pi\int_{-\frac{1}{2}}^1{ \left( 4\,x^4 - 13\,x^2 + 4\,x + 5 \right) \,\mathrm{d}x } \\ &= \pi\,\left[ \frac{4\,x^5}{5} - \frac{13\,x^3}{3} + 2\,x^2 + 5\,x \right] _{-\frac{1}{2}}^1 \\ &= \pi\,\left\{ \left[ \frac{4\,\left( 1 \right) ^5}{5} - \frac{13\,\left( 1 \right) ^3}{3} + 2\,\left( 1 \right) ^2 + 5\,\left( 1 \right) \right] - \left[ \frac{4\,\left( -\frac{1}{2} \right) ^5}{5} - \frac{13\,\left( -\frac{1}{2} \right) ^3}{3} + 2\,\left( -\frac{1}{2} \right) ^2 + 5\,\left( -\frac{1}{2} \right) \right] \right\} \\ &= \pi \,\left\{ \left( \frac{4}{5} - \frac{13}{3} + 2 + 5 \right) - \left[ \frac{4\,\left( -\frac{1}{32} \right)}{5} - \frac{13\,\left( -\frac{1}{8} \right) }{3} + 2\,\left( \frac{1}{4} \right) - \frac{5}{2} \right] \right\} \\ &= \pi\,\left[ \left( \frac{12}{15} - \frac{65}{15} + \frac{105}{15} \right) - \left( -\frac{1}{40} + \frac{13}{24} + \frac{1}{2} - \frac{5}{2} \right) \right] \\ &= \pi\,\left[ \frac{52}{15} - \left( -\frac{3}{120} + \frac{65}{120} - \frac{240}{120} \right) \right] \\ &= \pi\,\left[ \frac{52}{15} - \left( -\frac{178}{120} \right) \right] \\ &= \pi\,\left( \frac{52}{15} + \frac{89}{60} \right) \\ &= \pi\,\left( \frac{208}{60} + \frac{89}{60} \right) \\ &= \pi\,\left( \frac{297}{60} \right) \\ &= \frac{99\,\pi}{20}\,\textrm{units}^3 \end{align*}

#### Prove It

##### Well-known member
MHB Math Helper
Using the same region, what will be the volume of the solid if that region is rotated around the line \displaystyle \begin{align*} x = -1 \end{align*}?
To do this, picture what it would look like if the region was rotated about the vertical line x = -1. This solid can be thought of as a bunch of vertically oriented thin hollow cylinders being stuck together. If we move everything right by 1 unit, the volumes will be the same, but we will rotate around the y axis. So we really want the area between \displaystyle \begin{align*} y = x \end{align*} and \displaystyle \begin{align*} y = 2\,\left( x - 1 \right) ^2 \end{align*} with \displaystyle \begin{align*} \frac{1}{2} \leq x \leq 2 \end{align*}.

The curved surface of each cylinder is a rectangle. The width of the rectangle is equal to the vertical distance between the two functions, so \displaystyle \begin{align*} y_2 - y_1 = x - 2\,\left( x - 1 \right)^2 = x - 2\,\left( x^2 - 2\,x + 1 \right) = x - 2\,x^2 + 4\,x - 1 = 5\,x - 2\,x^2 - 1 \end{align*}, and the length is the same as the circumference of the cylinder, so \displaystyle \begin{align*} 2\,\pi\,r \end{align*}. Since the radius of each cylinder is the x value of the function, that means the area of the curved surface of each cylinder is \displaystyle \begin{align*} 2\,\pi\,x\,\left( 5\,x - 2\,x^2 - 1 \right) \end{align*}, and thus the volume of each cylinder is \displaystyle \begin{align*} 2\,\pi\,x\,\left( 5\,x - 2\,x^2 - 1 \right) \,\Delta x \end{align*}, where \displaystyle \begin{align*} \Delta x \end{align*} is a small change in x.

We can therefore approximate the total volume by adding up the volumes of all these cylinders, so \displaystyle \begin{align*} \sum{ 2\,\pi\,x\,\left( 5\,x - 2\,x^2 - 1 \right) \,\Delta x } \end{align*}. As we increase the number of cylinders, thereby making each one thinner, the approximation gets better. So as \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 volume of the solid is exactly

\displaystyle \begin{align*} V &= \int_{\frac{1}{2}}^2{ 2\,\pi\,x\,\left( 5\,x - 2\,x^2 - 1 \right) \,\mathrm{d}x } \\ &= 2\,\pi\int_{\frac{1}{2}}^2{ \left( 5\,x^2 - 2\,x^3 - x \right) \,\mathrm{d}x } \\ &= 2\,\pi \,\left[ \frac{5\,x^3}{3} - \frac{x^4}{2} - \frac{x^2}{2} \right] _{\frac{1}{2}}^2 \\ &= 2\,\pi\,\left\{ \left[ \frac{5\,\left( 2 \right) ^3}{3} - \frac{2^4}{2} - \frac{2^2}{2} \right] - \left[ \frac{5\,\left( \frac{1}{2} \right) ^3}{3} - \frac{\left( \frac{1}{2} \right) ^4}{2} - \frac{\left( \frac{1}{2} \right) ^2}{2} \right] \right\} \\ &= 2\,\pi\, \left[ \left( \frac{40}{3} - 8 - 2 \right) - \left( \frac{5}{24} - \frac{1}{32} - \frac{1}{8} \right) \right] \\ &= 2\,\pi \,\left[ \left( \frac{40}{3} - \frac{30}{3} \right) - \left( \frac{20}{96} - \frac{3}{96} - \frac{12}{96} \right) \right] \\ &= 2\,\pi \, \left( \frac{10}{3} - \frac{5}{96} \right) \\ &= 2\,\pi\,\left( \frac{320}{96} - \frac{5}{96} \right) \\ &= 2\,\pi\,\left( \frac{315}{96} \right) \\ &= 2\,\pi\,\left( \frac{105}{32} \right) \\ &= \frac{105\,\pi}{16}\,\textrm{units}^3 \end{align*}