Solid Shape Volume by Disk Method Problem

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SUMMARY

The problem involves calculating the volume of a solid with an equilateral triangle base of side length 'a' and square cross-sections perpendicular to the x-axis. The user attempted to derive the volume using the functions f(x) = (sqrt(3)x)/3 and g(x) = (-sqrt(3)x)/3, resulting in an area A = (2sqrt(3)x)/3. However, the user incorrectly set the limits of integration, leading to an erroneous volume calculation of (4a^3)/9 instead of the correct volume of (sqrt(3)a^3)/6. Feedback emphasized the importance of accurately determining the limits of integration through visual representation.

PREREQUISITES
  • Understanding of calculus, specifically integration techniques
  • Familiarity with geometric properties of equilateral triangles
  • Knowledge of volume calculation methods for solids of revolution
  • Ability to interpret and sketch geometric shapes based on mathematical descriptions
NEXT STEPS
  • Review the method of calculating volumes using the disk method in calculus
  • Study the properties and equations related to equilateral triangles
  • Practice setting up and solving integrals with varying limits of integration
  • Explore visual aids and software tools for graphing functions and solids
USEFUL FOR

Students studying calculus, particularly those focusing on volume calculations, geometry enthusiasts, and educators seeking to enhance their teaching of solid geometry concepts.

harrietstowe
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Homework Statement


The base of a certain solid is an equilateral triangle of side a, with one vertex at the origin and an altitude along the x-axis. Each plane perpendicular to the x-axis intersects the solid in a square cross section with one side in the base of the solid. Find the volume.


Homework Equations





The Attempt at a Solution



I am pretty unsure about this problem but what I did was:
Let a be one side of this equilateral triangle
Let f(x) = (c*x)/(a*sqrt(3))
Let g(x) = (-c*x)/(a*sqrt(3))
so a*sqrt(3) is the altitude and c is a constant
Solve for c:
tan(60 deg)= (a*sqrt(3))/c
c = a
Sub that into f(x) and g(x)
f(x) = (sqrt(3)*x)/3
g(x) = (-sqrt(3)*x)/3
The next thing I did was to subtract the two functions to get the area
A=f(x)-g(x)= (2*sqrt(3)*x)/3
the lower limit of integration will be 0
the upper limit of integration will be a
So I took the integral (with respect to x) from 0 to a of the Area squared and I got (4a^3)/9
The correct answer though is (sqrt(3)*a^3)/6
I hope my set up wasn't completely off but I guess you guys can give me so feedback.
Thanks
 
Physics news on Phys.org
The integration limit is not correct. If you made a drawing and saw the picture of the body you would have found out the correct limits of integration.

ehild
 

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