Volume Problems: Solving for Area and Volume with Integrals

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SUMMARY

The discussion focuses on solving volume problems using integrals, specifically regarding a solid with a base defined by the curve y=1-sinx and an oil storage tank shaped by revolving the curve y=9*x^4/625. Participants clarify the area of the isosceles right triangle cross-section as (1-sin(x))^2/4 and confirm the need to integrate to find the volume of the solid. For the oil tank, the volume is calculated using the disk method, with the radius expressed as x=((625*y)/9)^(1/4), leading to the conclusion that the depth of oil increases at a rate of approximately 0.153 ft/min when the depth is 4 feet.

PREREQUISITES
  • Understanding of integral calculus, specifically volume calculations using integrals.
  • Familiarity with the disk and shell methods for finding volumes of revolution.
  • Knowledge of trigonometric functions and their properties, particularly sine functions.
  • Ability to manipulate and solve equations involving polynomial functions.
NEXT STEPS
  • Study the disk method for calculating volumes of solids of revolution in more detail.
  • Learn about the shell method and how it differs from the disk method for volume calculations.
  • Explore the properties of trigonometric functions, focusing on sine and its applications in calculus.
  • Practice solving integrals involving polynomial functions and their applications in real-world problems.
USEFUL FOR

Students studying calculus, particularly those focusing on integral applications in volume calculations, as well as educators seeking to enhance their teaching methods for these concepts.

  • #31
Sure. Estimate with rounded numbers. 200ft^3/(8*ft^3/min)=25 minutes. 5 ft/25 min=0.2ft/min roughly on average. That makes 0.15ft/min sound pretty reasonable. Not 'too small'.
 
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  • #32
Thanks for your help
 
  • #33
nns91 said:
Thanks for your help

Very welcome!
 

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