Introduction
When you were a kid first learning how to calculate volume, the problems you solved involved simple shapes: a cube, triangular prism, sphere, or cylinder. The volume of these solids can all be calculated with a straightforward equation- however, more uniquely shaped solids rarely- if ever- have pre-determined formulas for their volume. Fortunately, we still have methods to calculate the volume of more adventurous solids using a bit of calculus. Let’s look at an example.
Imagine you are a manufacturer working with an artist who needs your help to produce a sculpture of a very specific shape. You need to know the volume of the shape before making it so you can buy the right amount of supplies. The artist wants the cross section of the sculpture to be equilateral triangles. The sides of the triangle in these cross sections, however, will vary in length. You have determined that the side lengths of the triangles can be modeled by the distance between the equations \( f(x)= \frac{1}{4}\ x^3-x^2+3 \) and \( g(x)=-x^2+2x \) on the interval \( 0 \le\ x \le\ 2 \). Difficult to visualize? Take a look at the graph below of \( f(x) \) and \( g(x) \). This will be the shape of the sculpture if you are looking down on it from above. Imagine stacking infinitely many equilateral triangles vertically between the two lines – this will give us our solid.
But how can we get the volume of such a crazy shape? Before we jump into the method let’s leave out the calculus so we can get an idea of what our method is trying to accomplish.
An Approximation of our Method
Remember when we said that our solid would be formed by placing infinitely many equilateral triangles between our two equations? That is exactly what our method will do to determine the volume. But since an infinite number of triangles sounds overwhelming, let’s see what this looks like if we use just 10 triangles.
First we will determine what the area of our first triangle will be. Since we are placing it at \( x=0 \), the length of one side will be \( f(0)-g(0) \): \[ \Big( \frac{1}{4}\ 0^3-0^2+3 \Big) - \Big( -0^2+2(0) \Big) =3 \]
Next, the area of an equilateral triangle is \( \frac{ \sqrt{3}\ }{4}s^2 \), where \( s \) is the length of the sides. Thus, the area will be: \[ \frac{ \sqrt{3}\ }{4}3^2=3.897 \]
Remember, we are splitting our solid up into 10 triangles. Since the length of our interval is 2, this means that each of our triangles will become triangular prisms of width \( 2/10=0.2 \). Multiplying this by our area, we find that the volume of our first triangle is: \[ 3.897*0.2=0.779 \textrm{ units}^2 \]
We’ve finished the first triangle- let’s do the same with the remaining 9 triangles:
All of these cross sections give us the following approximation of our sculpture:
Lastly, we add together the volumes of each triangular prism. This gives us an approximate volume of 2.708 units squared.
The Exact Method
Now that we’ve found an approximate volume for our sculpture, we have a better idea of how cutting a solid into cross sections and adding those together helps us find our volume. Our last approximation wasn’t perfect because we only used 10 cross sections. This is obvious from looking at the image of our model: the edges of the solid aren’t smooth like our equations. However, if we were to divide the solid into an infinite number of cross sections, our triangles would be so similar that the edge of our model would be smooth like we want.
In order to build out solid from an infinite number of cross sections, all we have to do is use an integral. So, if we change the equation, we used for each cross section above \[ V= \frac{ \sqrt{3}\ }{4} s^2 \cdot h= \frac{ \sqrt{3}\ }{4} (f(x)-g(x))^2 \cdot h \] so that it becomes \[ \int_{a}^{b} \frac{ \sqrt{3}\ }{4} s^2 \,dx = \frac{ \sqrt{3}\ }{4} \int_{a}^{b} (f(x)-g(x))^2 \,dx \] where \( a \) and \( b \) represent the beginning and end respectively of the interval we are interested in.
Let’s use this equation to calculate the exact volume of our example: \[ \frac{ \sqrt{3}\ }{4} \int_{0}^{2} (f(x)-g(x))^2 \,dx \] \[ = \frac{ \sqrt{3}\ }{4} \int_{0}^{2} \Big( \frac{1}{4}\ x^3-x^2+3-(-x^2+2x) \Big)^2 \,dx \] \[ = \frac{ \sqrt{3}\ }{4} \int_{0}^{2} \Big( \frac{1}{4}\ x^3-x^2+3+x^2-2x \Big) ^2 \,dx = \frac{ \sqrt{3}\ }{4} \int_{0}^{2} \Big( \frac{1}{4}\ x^3-2x+3 \Big) ^2 \,dx \] \[ = \frac{ \sqrt{3}\ }{4} \int_{0}^{2} \Big( \frac{1}{16}\ x^6- \frac{1}{2}\ x^4+ \frac{3}{4}\ x^3- \frac{1}{2}\ x^4+4x^2-6x+ \frac{3}{4}\ x^3-6x+9 \Big) \,dx \] \[ = \frac{ \sqrt{3}\ }{4} \int_{0}^{2} \Big( \frac{1}{16}\ x^6-x^4+ \frac{3}{2}\ x^3+4x^2-12x-9 \,dx \] \[ = \frac{ \sqrt{3}\ }{4} \Big[ \frac{1}{112}\ x^7- \frac{1}{5}\ x^5+ \frac{3}{8}\ x^4+ \frac{4}{3}\ x^3-6x^2+9x \Big] \] \[ = \frac{ \sqrt{3}\ }{4} \Big[ \Big( \frac{2^7}{112}\ - \frac{2^5}{5}\ + \frac{3 \cdot 2^4}{8}\ + \frac{4 \cdot 2^3}{3}\ -6 \cdot (2)^2+9(2) \Big) - \Big( \frac{0^7}{112}\ - \frac{0^5}{5}\ + \frac{3 \cdot 0^4}{8}\ + \frac{4 \cdot 0^3}{3}\ -6 \cdot (0)^2+9(0) \Big) \Big] \] \[ = \frac{ \sqrt{3}\ }{4} \Big( \frac{128}{112}\ - \frac{32}{5}\ + \frac{48}{8}\ + \frac{32}{3}\ -24+18 \Big) \] \[ = 2.342 \textrm{ units}^2 \]
Now that we’ve computed the integral, we know the exact volume of this solid. This method works for more than this shape too- even if we use different equations or change our cross sections to a different shape, the same process works to find the volume. We can generalize this to the equation: \[ \int_{a}^{b} A(x) \,dx \] where \( A(x) \) is the equation for the area of the cross section. Instead of equilateral triangles, they could be rectangles, semi-circles, ellipses, or even octagons. With this method, there is no limit on the different types of shapes we can calculate the volume of.
Why this example?
The equations were also chosen for two main reasons:
- The area between these lines produces a strange but easily distinguishable shape. This should help the learner identify what is happening in the model. They should easily be able to identify where these curves can be found on the model, which will aid their understanding of how the equations translate into the resulting solid. Since visualizing these solids is the most challenging part of this method, it is extremely important that the student can easily understand how the word problem resulted in the 3-D object.
- These equations are basic polynomials that the student should be comfortable working with. As a result, they should be able to follow the math easily (without being distracted by a new function or difficult integration technique) and better understand the method.
The cross section for this example was chosen to be equilateral triangles for two main reasons:
- The area of an equilateral triangle is fairly easy to calculate compared to, say, the area of a pentagon or octagon. This allows the learner to focus on the method rather than the numbers in the area equation. Once they understand this more basic example, they can apply the method to these more involved examples.
- Again, the hardest part of this method is visualizing the solids that are created. The triangular cross sections result in a very distinct shape. This is intended to help the reader interpret the final model and understand the visualization, since the triangular points are easy to find in the model compared to, say, an ellipsis (which to a new learner may seem similar to a polynomial curve). Additionally, choosing a shape with straight sides helps the student distinguish this method from that of the washer method.
Author: Sarah Bombrys
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