Volume of a solid with known cross sections

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SUMMARY

The volume of the solid with known cross sections is determined by integrating the area of the square cross sections between the points of intersection of the curves defined by the equations y² = 4x and x² = 4y. The limits of integration are established as 0 to 4, corresponding to the intersection points (0, 0) and (4, 4). The area of each square cross section is calculated as s², where s represents the distance between points A and B on the curves. The final volume can be computed using the integral formula ∫ from 0 to 4 of A(x) dx, where A(x) is the area function derived from the side length of the square.

PREREQUISITES
  • Understanding of calculus, specifically integration techniques.
  • Familiarity with the equations of curves and their intersections.
  • Knowledge of geometric properties of squares and area calculations.
  • Ability to graph functions and interpret graphical data.
NEXT STEPS
  • Study the method of finding volumes of solids of revolution using integration.
  • Learn how to derive area functions from geometric shapes, particularly squares.
  • Explore the use of definite integrals in calculating volumes of solids with known cross sections.
  • Investigate the graphical interpretation of curve intersections and their significance in calculus.
USEFUL FOR

Students studying calculus, particularly those focusing on volume calculations and integration techniques. This discussion is also beneficial for educators teaching geometric applications of calculus.

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Homework Statement



Any cross sectional slice of a certain solid in a plane perpendicular to the x-axis is a square with side AB, with A lying on the curve [tex]y^2 = 4x[/tex] and B on the curve [tex]x^2 = 4y[/tex]. Find the volume of the solid lying between the points of intersection of these two curves.

Homework Equations


[tex]\int ^{b}_{a} A(x)dx[/tex]

The Attempt at a Solution


I'm not sure if I'm going in the right direction, but so far I've put the curves in terms of y, leaving me with [tex]y = 2\sqrt{x}[/tex] and [tex]y = \frac{x^2}{4}[/tex]. After graphing, I also know that the limits of integration will be from 0 to 4 since the points of intersection are at (0, 0) and (4, 4). From here on, I'm completely lost.

Thanks :)
 
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The area of a square is s^2 where s is the length of one side. So, what is the length of one side? The distance from A to B, so find that from your graph (at an arbitrary x value and the expression should be in terms of y.)
 

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