Amejka – Rolston
Senior IDS
Using SolidWorks in Calculus
Calculus is involved in a variety of design software programs that many
engineers and designers use. One of these design software programs is
SolidWorks. SolidWorks is a computer aided design program (CAD) that can be
used to create three-dimensional parts. SolidWorks can then calculate and
display values for that three-dimensional solid such as volume, mass, and
surface area. In this case, the equation f(x) = cos(x) + 5 was rotated 360° around
the x-axis. The solid that resulted is shown in Figure 1.
Figure 1. Solid formed when f(x) was rotated around the x-axis
To create the three-dimensional part, a sketch was drawn on the software
and the equation inputted with the boundaries from x = 0 to x = 4. Next, lines
were drawn from the x-axis to the equation line at both x = 0 and x = 4 to close
off the drawing completely. Figure 2 shows the sketch of the drawing with
dimensional lines included.
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Amejka – Rolston
Senior IDS
Figure 2. Closed-off sketch of f(x) from x = 0 to x = 4
Once the sketch was fully closed off, the part was rotated around the xaxis to create a three-dimensional solid. Figure 3 shows the drawing of the part
with the Top and Isometric views on the top row and the Front and Right views
on the bottom row.
Figure 3. Drawing of the part with four different views
The end goal was to verify that the SolidWorks method of volume
calculation and the calculus method of volume calculation produced the same
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Amejka – Rolston
Senior IDS
results. The volume was calculated using SolidWorks after the threedimensional part was created. SolidWorks determined the volume to be 297.46
cubic inches, which is highlighted in a box in Figure 4.
Figure 4. Screen shot of the mass properties window
Calculus was used to confirm the value that SolidWorks calculated for the
volume. The calculus topic is known as solids of revolution. The method used
was the disk method, because when the solid is cut into small cross-sections, the
small cross-sections are disks. Figures 5, 6, and 7 show the solid cut into
various disk cross-sections at various spots.
Figure 5. Large
cross-section
Figure 6. Small
cross-section
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Figure 7. Medium
cross-section
Amejka – Rolston
Senior IDS
The disk method uses the equation πr2, where r is the radius (in this case
this is the y-value). This and the antiderivative are shown in figure 8. The
antiderivative was found using the even function rules for integration.
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1. ∫0 𝜋𝑟 2 𝑑𝑟 =∫0 𝜋𝑦 2 𝑑𝑦
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∫0 𝜋(cos(𝑥) + 5)2 𝑑𝑥
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2. 𝜋(cos(𝑥)2 + 10 cos(𝑥) + 25)=𝜋(4 sin(2𝑥) + 10 sin(𝑥) + 25.5𝑥)
Figure 8. Integration of f(x)
The integrand was integrated from 0 to 4, since those were the boundaries
used to sketch the solid. The Fundamental Theorem of Calculus states that
𝑏
∫𝑎 𝑓(𝑥) = 𝐹(𝑏) − 𝐹(𝑎) where F is the antiderivative of f(x). For this case, we’d
use 4 as b and 0 as a. The steps of integration and the calculations are shown in
figure 9.
When 4 is plugged in for x:
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𝜋( sin(2(4)) + 10 sin(4) + 25.5(4))=297.444
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When 0 is plugged in for x:
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𝜋(4 sin(2(0)) + 10 sin(0) + 25.5(0))=0
So, 297.444 – 0 = 297.444
Figure 9. Steps of integration and the final volume
The volume that was calculated using SolidWorks was 297.46 cubic
inches, and the volume that was calculated by hand using the disk method and
integration was 297.444 cubic inches. There is virtually no variation between the
two values and virtually no percent error.
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