We can determine the volume of a solid with similar cross-sections using definite integrals. In this post, let's explore the disc method in finding the volume of a solid of revolution.
We move on to a particular case of finding volume \(V\) using integration. For these cases, we will be dealing with a solid of revolution. There are three general variations of finding \(V\).
This post discusses the first of these: the disc method.
Approximation
Let's say you're finding the volume \(V\) of a solid of revolution by approximating it first. The object consists of a uniform cross-section throughout its whole length. We generated the solid by revolving the area along its axis of rotation, which must lie along the area boundary.
We must divide the solid into multiple parts to estimate the total volume. We can imagine these parts as a series of similar plane sections stacked on each other to make up the 3D object.
For a solid of revolution, we can divide it into a disc or a shell. For this post, we will be using small discs - thin prisms with a circular base.
From this, we begin the process of approximating:
First, divide the 3D object into multiple regular discs.
Then, we solve the volume \(V\) of each divided disc. From Cavalieri's theorem, \(V=Ah\) where \(A\) is the uniform cross-sectional area and \(h\) is the length of the solid. Since it is a circular disc, \(A\) is equal to \(\pi r^2\) where \(r\) is its radius.
Finally, we add all individual \(V\) to get the total approximate volume.
Since this is an approximation, how can we have a more accurate answer? We increase the number of discs \(n\).
When we increase the number of discs, it will be closer to the actual value.
As the number of discs increases, the smaller its volume \(dV\) and thickness \(h\) become.
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From the approximate solution, we know that if we want an exact answer, the volume and height of the discs must be tiny. Building from this, we can solve the precise volume. We will only need to modify two things in the approximate solution:
We need to set up a solid disc that is so small we can express its volume as the differential \(dV\)
We need an operation that will sum up all of these tiny divisions
We can combine all \(dV\) through integration; hence, the general equation to solve for the volume \(V_T\) is:
\(V_T=\int_{a}^{b}{dV}\)
\(V_T\) is the total volume
\(dV\) is the differential volume of the division.
\(a\) is the lower limit position (endpoint with the lowest value)
\(b\) is the upper limit position (endpoint with the highest value)
This equation is similar to solving the volume of solids using integration with uniform cross-sections. What makes this unique is the differential partition which is always a disc; hence, the name: the disc method.
One essential detail for differential discs is that they must be perpendicular to the axis of rotation.
Differential Disc
Let's discuss the differential disc with volume \(dV\). Using Cavalieri's theorem, the differential volume is \(dV=A \cdot dh\).
\(A\) is the cross-sectional area of the solid
\(dh\) is the differential thickness
The cross-sectional area is a circle. We can further expand this as: \(A=\pi r^2\); hence, \(dV\) becomes:
\(dV=\pi r^2 dh\)
Let's expound more on these variables:
The variable \(r\) is the radius function. We need to express it in terms of the differential variable. The radius function will depend on the orientation of the solid if it is along the x, y, or z-axis.
The differential thickness \(dh\) is the distance between two points; however, because it is so minuscule, we represent it with a position variable \(x\), \(y\), or \(z\). Again, the thickness will also depend on the orientation.
These two variables \(r\) and \(dh\) must be consistent with the differential variable.
Going back to \(V_T\), we can summarize the disc method with the following general equation:
\(V_T=\pi \int_{a}^{b} r^2 dh\)
Summary
Solving the volume by integration expands on the approximation solution.
The logic is to subdivide the volume needed with multiple solids, solve for the volume of each part, and add all of these volumes.
It is necessary to keep the volume small to increase the accuracy of the solution; hence, we must represent the volume as differential \(dV\).
We can combine all \(dV\) through integration; hence, the general equation to solve for the volume is: \(V_T=\int_{a}^{b}{dV}\).\(V_T\) is the total volume, \(dV\) is the differential volume of the division, \(a\) is the lower limit position (endpoint with the lowest value), and \(b\) is the upper limit position (endpoint with the highest value)
We can expound the differential solid with volume \(dV\) with Cavalieri's theorem: \(dV=A \cdot dh\). \(A\) is the cross-sectional area of the solid, and \(dh\) is the differential thickness.
The cross-sectional area is always a circle since we deal with a disc. We can further expand the area as: \(A=\pi r^2\); hence, \(dV\) becomes: \(dV=\pi r^2 dh\)
One essential detail for differential discs is that they must be perpendicular to the axis of rotation.
The variable \(r\) is a radius function. The radius function will depend on the orientation of the solid if it is along the x, y, or z-axis.
The differential thickness \(dh\) is the distance between two points; however, because it is so minuscule, we represent it with a position variable \(x\), \(y\), or \(z\) which will depend on the orientation.
These two variables \(r\) and \(dh\) must be consistent with the differential variable.
We can summarize the disc method with the following general equation: \(V_T=\pi \int_{a}^{b} r^2 dh\)
