Imagine a loaf of bread cut into 10 slices of equal thickness. It may seem obvious that the volume of the entire loaf of bread is equal to the sum of the volume of each individual slice of bread. Though each slice of bread possibly has the same thickness, each slice will likely have a different volume as the end of the loaf is a different shape than the middle of the loaf.
You've seen a similar idea applied when finding the area under the curve using a Reimann Sum or Numerical Integration methods such as the Trapezoidal Rule or Simpson's Rule. The curve is split into subintervals and the area of each subinterval is summed to approximate the total area under the curve. The same idea applies when finding the volume of a solid of revolution using the disk method!
Definition of the Disk Method
The disk method is an integration method that slices a solid of revolution up into a series of three-dimensional disks and sums the volume of each disk to find the total volume of the solid.
The disk method is a method for calculating the volume of a solid of revolution that is used when integrating along an axis that is parallel to the axis of integration. The method involves splitting the solid into infinitely many disks and summing the volume of each disk.
The disk method slices the solid of revolution into a series of flattened disks. Each disk is contained in a plane that is perpendicular to the axis of rotation.
To find the volume of the entire solid, the volume of each disk is added together.
Graph of the Disk Method
To better understand the idea behind the disk method, take a look at the image below.
Figure 1. Volume of a disc perpendicular to the \(x-\)axis obtained by slicing the solid of revolution
The cross-section of a disk is a circle with an area of \(\pi r^{2}\), so you can find the volume of each disk by multiplying its area by its thickness, so
\[V_{\text{disk}}=\pi r^2 \Delta x,\]
where \( \Delta x \) is a small subinterval of the integration interval.
To find the volume of the solid rotated around the \(x\)-axis, you slice the solid such that the slices are contained in planes perpendicular to the \(x-\)axis. The volume of each slice, or disk, is added up, giving an estimation of the volume of the solid. You obtain the exact volume of the solid by slicing the solid into infinitely many disks and integrating instead.
If the solid is obtained by a rotation around the \(y-\)axis, then the disks have to be in a plane perpendicular to the \(y-\)axis instead. In general, the slices are contained in planes that are perpendicular to the axis of rotation.
Equation for the Disk Method
The cross-section of a disk is a circle with an area of \(\pi r^{2}\), so you can find the volume of each disk by multiplying its area by its thickness, so
\[V_{\text{disk}}=\pi r^2 \Delta x,\]
where \( \Delta x \) is the thickness of the disk, and is the length of a small subinterval of the integration interval.
In order to obtain the volume of the complete solid, you need take some other things into consideration.
The radius of each disk is now given by the function, so \(r\) becomes \(f(x)\).
The disk becomes very thin, so \(\Delta x\) becomes \(\mathrm{d}x \).
Instead of adding all the disks, you integrate, which means summation, \(\sum \), becomes integration, \(\int\).
The volume \( V \) of a solid generated by revolving the bounded region \(y=f(x)\) and the \(x\)-axis on the interval \([a, b]\) around the \(x\)-axis is then given by
\[V=\int_a^b \pi \left[f(x) \right]^2 \, \mathrm{d}x.\]
Note that in the above formula the disks are perpendicular to the \(x\)-axis because the revolution was around the \(x-\)axis.
If the revolution is done around the \(y-\)axis, then the formula is adapted as
\[V=\int_a^b \pi \left[ f(y) \right] ^2\, \mathrm{d}y, \]
where the disks are perpendicular to the \(y\)-axis.
Examples of the Disk Method
Let's practice finding volumes of solids using the disk method by looking at some examples!
For the function
\[f(x)=x^2-4x+4,\]
consider the region bounded by the curve \(y=f(x)\), \(x=0\), \(x=4\), and the \(x\)-axis.
Figure 2. Region bound between \(x=0,\) \(f(x),\) \(x=4,\) and the \(x-\)axis
Find the volume of the solid obtained from rotating the above region around the \(x\)-axis.
Answer:
The solid of revolution for this case can be shown in the following picture.
Figure 3. Solid obtained by revolving the above region around the \(x-\)axis
The slices of the region are perpendicular to the \(x\)-axis, so you need to use the formula
\[V=\int_a^b \pi [f(x)]^{2}\,\mathrm{d}x.\]
Since you need to square the function, you will need to do a bit of algebra, that is
\[ \begin{align} \left( f(x) \right)^2 &= (x^2-4x+5)^2 \\ &= \left( x^2+(-4x+5) \right)^2 \\&= x^4+2x^2(-4x+5)+(-4x+5)^2 \\ &= x^4-8x^3+10x^2+16x^2-40x+25 \\ &= x^4 -8x^3+26x^2-40x+25. \end{align}\]
The above expression might look intimidating, but its integral can be found by just using the Power Rule. Plugging in \(a=0\), \(b=4\), and \(f(x)=x^{2}-4x+5\), you get
\[ \begin{align} V &= \int_0^4 \pi \left( x^2-4x+5 \right)^2 \,\mathrm{d}x \\ &= \pi \int_0^4 (x^4-8x^3+26x^2-40x+25)\,\mathrm{d}x. \end{align}\]
Start by finding the indefinite integral with the help of the Power Rule, that is
\[ \int ( x^4-8x^3+26x^2-40x+25 )\,\mathrm{d}x = \frac{1}{5}x^5-2x^4+\frac{26}{3}x^3-20x^2+25x,\]
then use the Fundamental Theorem of Calculus to evaluate the definite integral, obtaining
\[ \begin{align} \int_0^4 \left( x^2-4x+5 \right)^2\,\mathrm{d}x &= \left[\frac{4^{5}}{5} - 2(4)^{4} + \frac{26}{3} (4)^{3} - 20(4)^{2} +25(4)\right] \\ &- \left[\frac{0^{5}}{5} - 2(0)^{4} + \frac{26}{3}(0)^{3} - 20(0)^{2} + 25(0) \right]. \end{align}\]
You can do the arithmetic with the help of a calculator and find
\[\int_0^4 \left( x^2-4x+5 \right)^2\,\mathrm{d}x = \frac{412}{15}.\]
Finally, multiply the value of the definite integral by \(\pi\) to get the volume of the solid of revolution, that is
\[ V = \frac{412}{15}\pi.\]
How about a solid that was obtained by rotating a region around the \(y-\)axis?
For the function
\[g(x)=x^2,\]
consider the region bounded by the curve \(y=g(x)\), the \(y\)-axis, and the horizontal line \(y=1\).
Figure 4. Region bound between \(y=0,\) \(g(x),\) \(y=1,\) and the \(y-\)axis
Find the volume of the solid obtained by rotating the above region around the \(y-\)axis.
Answer:
This time the solid of revolution is pictured in the following image.
Example 5. Solid obtained by revolving the above region around the \(y-\)axis
In this example, the slices of the region are perpendicular to the \(y\)-axis, so you will need to use the formula
\[V=\int_a^b \pi \left[ x(y) \right]^2 \,\mathrm{d}y.\]
However, you are given \( y \) as a function of \(x,\) so you need to rewrite the opposite way, that is, you need to write \(x\) as a function of \(y.\) This can be done with the help of the square root function, so
\[ y=x^2 \rightarrow x=\sqrt{y}.\]
For the bounds of integration \(a\) and \(b\) note that the area to be revolved is bound between the origin and the horizontal line \(y=1,\) so \( a=0 \) and \( b=1.\) This gives you
\[ \begin{align} V &= \int_0^1 \pi [\sqrt{y}]^{2}\, \mathrm{d}y \\ &= \pi \int_0^1 y\,\mathrm{d}y \\ &= \frac{\pi}{2} \left( 1^2-0^2\right) \\&= \frac{\pi}{2}. \end{align}\]
The Shell Method vs the Disk Method
The shell method, also known as the method of cylindrical shells, is another method used to calculate the volume of a solid of revolution. The difference between the shell method and the disk method is the shape of the solid of partition.
With the disk method, you split the solid into infinitely many disks. However, the shell method slices the solid into infinitely many hollow cylinders instead. This method will not be addressed in this article.
The Disk Method - Key takeaways
- The Disk Method is a method for calculating the volume of a solid of revolution that is used when integrating along an axis that is parallel to the axis of revolution.
- The Disk Method involves splitting the solid into infinitely many disks and adding the volume of each disk by means of integration.
- The equation for finding the area of a solid of revolution using the disk method depends on which axis the solid is revolved around.
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