Using Pappus' Theorem to Find Volumes of Solids of Revolution

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SUMMARY

This discussion focuses on applying Pappus' Theorem to calculate the volume of a solid of revolution formed by rotating the region under the curve y = x^3 from x=0 to x=2 about the y-axis. The area of the region "R" is determined using the integral from 0 to 2 of x^3 dx. The volume is calculated using the method of cylindrical shells, leading to the formula V = integral(2*pi*x*height*dx). The discussion also emphasizes the relationship between the centroid's distance traveled and the volume, confirming Pappus' Theorem.

PREREQUISITES
  • Understanding of integral calculus, specifically definite integrals.
  • Familiarity with the method of cylindrical shells for volume calculation.
  • Knowledge of Pappus' Theorem and its application in geometry.
  • Ability to compute centroids of two-dimensional shapes.
NEXT STEPS
  • Study the method of cylindrical shells in detail, focusing on volume calculations.
  • Explore Pappus' Theorem further, including its proofs and applications in different contexts.
  • Practice finding centroids of various shapes to solidify understanding.
  • Investigate other methods for calculating volumes of solids of revolution, such as the disk and washer methods.
USEFUL FOR

Students and educators in calculus, mathematicians interested in geometric applications, and anyone studying the properties of solids of revolution and their volumes.

Pirata
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The Region "R" under the graph of y = x^3 from x=0 to x=2 is rotated about the y-axis to form a solid.

a. Find the area of R.
b. Find the volume of the solid using vertical slices.
c. Find the first moment of area of R with respect to the y-axis. What do you notice about the integral?
d. Find the x coordinate of the centroid of R.
e. A theorem of Pappus states that the volume of a solid of revolution equals the area of the region being rotated times the distance the centroid of the region travels. Show that this problem confirms this theorem.

The Attempt at a Solution


I was able to do part "a" as the integral from 0 to 2 of x^3 dx. Also I believe part "b" is pi*[3y^(5/3)/5] evaluated from 0 to 2
 
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I don't think "b" is right. How did you get that?
 
Well I posted what I got. Was hoping you would show me where I went wrong...
 
It's best if you show how you got what you got. Otherwise we just have to guess how you got what you got. That's not fun. Use the method of shells to find the volume. V=integral(2*pi*x*height*dx).
 

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