Setting up an integral for surface area

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SUMMARY

The discussion focuses on setting up an integral to calculate the surface area generated by rotating the region bounded by the equation x + y² = 1 and the y-axis around the y-axis. The correct formula for surface area is SA = 2π ∫ x ds, where ds represents the differential arc length. A proposed solution includes the integral 4π ∫ (-y² + 1)√(1 + 4y²) dy with limits from 0 to 1, indicating a solid understanding of the relationship between the surface area and the given bounds.

PREREQUISITES
  • Understanding of surface area calculations in calculus
  • Familiarity with integral calculus and differential arc length (ds)
  • Knowledge of the equation of a curve and its geometric interpretation
  • Experience with rotation of solids about an axis
NEXT STEPS
  • Study the derivation of the surface area formula SA = 2π ∫ x ds
  • Learn about the method of cylindrical shells for volume and surface area
  • Explore the concept of parametric equations in surface area calculations
  • Practice setting up integrals for different shapes and their rotations
USEFUL FOR

Students in calculus, particularly those studying surface area and volume of solids of revolution, as well as educators looking for examples of integral applications in geometry.

Abyssnight
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Homework Statement


Set up an integral that represents the area of the surface generated when the region is bounded by x + y2 = 1 and the y-axis, then rotated about the y-axis. All in one-variable.


Homework Equations


SA = 2pi \int xds
possible x2 + y2 = r2

The Attempt at a Solution


I tried to set up an integral as a circle and somehow ended up getting 2pi\int(r/2)(2pi r) dr

When I picture it, it would be a sphere with a radius of 1. Still don't truly understand on how to make a integral to show it
 
Physics news on Phys.org
I looked at it again
would one possible answer be: 4pi\int(-y^2 + 1)\sqrt{1 + 4y^2}dy
with limits of integration (0,1)
 

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