Volume of Solid using cross sections

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SUMMARY

The discussion focuses on calculating the volume of a solid with a base defined by the curves y=x+1 and y=x^2-1, utilizing square cross-sections perpendicular to the x-axis. The user correctly identifies the need to find the area of the square cross-section as the square of the difference between the two curves. The integral setup is debated, with two potential forms presented: the integral from -1 to 2 of ((x+1)-(x^2-1))^2 dx and the integral of ((x+1)^2)-((x^2-1)^2) dx. The user seeks clarification on the correct expression for the side length of the square at a given x-value.

PREREQUISITES
  • Understanding of definite integrals in calculus
  • Familiarity with the concept of cross-sections in solid geometry
  • Knowledge of polynomial functions and their graphs
  • Ability to compute areas and volumes using integration
NEXT STEPS
  • Study the method of calculating volumes using cross-sections in calculus
  • Learn about the properties of definite integrals and their applications
  • Explore the differences between various types of cross-sectional shapes
  • Practice problems involving volume calculations of solids with different bases and cross-sections
USEFUL FOR

Students studying calculus, particularly those focusing on volume calculations of solids, educators teaching integration techniques, and anyone interested in geometric applications of calculus.

BigJon
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Homework Statement


Find the volume of a solid whose base bounded by y=x+1 and y=x^2-1, with cross section of a square perpendicular to x-axis


The Attempt at a Solution


So i set up the problem like this delta volume= y^2*delta x, that being the area of the square where i get loss is that I am supposed to subtract the curves from one another but don't know if its setup like this:

integral from -1 to 2 ((x+1)-(x^2-1))^2 dx or ((x+1)^2)-((x^2-1)^2) dx, from here i know how to do the rest
 
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What do you think the length of the side of the square is for a given value of x? Then what would be the area of that square?
 
Haha thanx, now that you put it that way :)
 

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