Effie's question via email about a volume by revolution

In summary, to find the volume of the solid formed by the region bounded by the functions $\displaystyle \begin{align*} y = 2\,x^2 \end{align*}$ and $\displaystyle \begin{align*} y = x + 1 \end{align*}$ rotated around the line $\displaystyle \begin{align*} y = 3 \end{align*}$, we need to first find the points of intersection of the two functions. Then, we need to rotate the area between the function $\displaystyle \begin{align*} y = 2\,x^2 - 3 \end{align*}$ and the x-axis between $\displaystyle \
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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

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

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

FAQ: Effie's question via email about a volume by revolution

What is a volume by revolution?

A volume by revolution, also known as a solid of revolution, is a three-dimensional shape created by rotating a two-dimensional shape around an axis. This results in a solid shape with a curved surface.

How is the volume of a shape calculated by revolution?

The volume of a shape by revolution is calculated using integration. The formula for calculating the volume is V = π∫ab f(x)2dx, where a and b are the limits of integration and f(x) is the function representing the shape.

What is the difference between a solid of revolution and a regular solid shape?

A solid of revolution is a shape created by rotating a two-dimensional shape around an axis, while a regular solid shape is created by manipulating three-dimensional shapes such as cubes, cylinders, and spheres. In other words, a solid of revolution has a curved surface while a regular solid shape has flat surfaces.

What are some real-life applications of volumes by revolution?

Volumes by revolution have many real-life applications, such as in engineering, architecture, and design. For example, they are used to calculate the volume of pipes, tunnels, and containers with curved shapes. They are also used in creating curved surfaces in products like car bodies and furniture.

Are there any limitations to using volumes by revolution to calculate volume?

Yes, there are limitations to using volumes by revolution to calculate volume. This method is only applicable to shapes with rotational symmetry, meaning the resulting shape is the same when rotated around an axis. It also assumes the shape being rotated is continuous and has a smooth surface. Additionally, the accuracy of the calculated volume depends on the accuracy of the measurements and the integration process.

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