Applications of the Definite Integral- Volume

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SUMMARY

The discussion focuses on calculating the volume of a church steeple modeled as a frustum of a pyramid with square cross sections. The steeple has a height of 30 feet, a base side length of 3 feet (36 inches), and a top side length of 6 inches. The volume is computed using the formula for the volume of a frustum, which is derived from the general pyramid volume formula, V = (1/3)Ah, where A is the area of the base and h is the height. The integration of the area of the cross sections, which shrink quadratically, leads to the final volume calculation.

PREREQUISITES
  • Understanding of definite integrals
  • Familiarity with the volume formula for pyramids and frustums
  • Knowledge of linear transformations in geometry
  • Ability to convert units (inches to feet)
NEXT STEPS
  • Study the derivation of the volume formula for frustums of pyramids
  • Learn about integration techniques for calculating volumes of solids of revolution
  • Explore applications of definite integrals in real-world volume calculations
  • Practice problems involving linear transformations and their effects on geometric shapes
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Students and educators in calculus, geometry enthusiasts, and professionals involved in architectural design or engineering who require a solid understanding of volume calculations using definite integrals.

ashleyk
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Hi, I am having trouble visualizing this problem, if anyone can help me see it, I know how to do the integral part.

A church steeple is 30 feet tall with square cross sections. The square at the base has side 3 feet, the square at the top has side 6 inches, and the sides varies linearly in between. Compute the volume.

I assume the 3 feet and then 6 inches has to be in the same form so it would 36 inches and 6 inches. I also know that volume for a square would be length times width times height or one side raised to the 3rd power. I am just not sure to start.
 
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A general fact about "pyramid" shapes, ie, shapes formed by starting with a base of a certain (2D) shape and sweeping out a volume by translating the shape and simultaneously linearly shrinking it to a point, is that their volume is 1/3 A*h, where A is the area of the base and h is the distance from the plane of the base to the tip. To prove this, note that as the side lengths linearly shrink, the area of the cross sections shrink quadratically, and integrating this out gives the result. In this problem, just take the difference of two such shapes.
 

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