Find the volume of a solid with given cross-section

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SUMMARY

The volume of a solid with a circular base of radius 4 and isosceles triangular cross-sections can be calculated using integration. The correct approach involves determining the base of the triangle as 2*sqrt(16-x^2), leading to the area formula for the triangle being A = (1/2) * base * height = (1/2) * (2*sqrt(16-x^2)) * (2*sqrt(16-x^2)). The integration of the volume from -4 to 4 yields a final volume of 512/3. The initial misunderstanding arose from incorrectly calculating the base of the triangle, which resulted in an incorrect factor of 4 in the final answer.

PREREQUISITES
  • Understanding of calculus, specifically integration techniques.
  • Familiarity with the equations of circles, particularly y^2 + x^2 = r^2.
  • Knowledge of the area formula for isosceles triangles.
  • Ability to set up and interpret volume integrals based on cross-sectional areas.
NEXT STEPS
  • Study the derivation of volume formulas for solids with varying cross-sections.
  • Learn about the application of the Fundamental Theorem of Calculus in volume calculations.
  • Explore different types of cross-sectional shapes and their impact on volume calculations.
  • Practice solving similar problems involving integration of areas to find volumes.
USEFUL FOR

Students in calculus courses, educators teaching volume calculations, and anyone interested in solid geometry and integration techniques.

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Homework Statement


The base of a solid is a circle of radius 4. Find the volume of the solid if each cross-section perpendicular to a fixed diameter of the circle is an isosceles triangle with height equal to its base.


Homework Equations


Volume of an object with given cross-section equation



The Attempt at a Solution


I know how to solve this (and similar questions) without problem, I just want to know if there is another way to solve them than the one I am using. For this question I found the equation of the circle (which is y=sqrt(16-x^2)) and then the area formula for the isosceles triangle (which I found to be y=b^2/2). Using the cross-section equation, I set the problem up as dV=(1/2)b^2dx, and then using the equation of the circle in place of b: dV=(8-(x^2/2))dx. To finally solve the question I then integrated this formula over [-4,4] and came up with an answer of 128/3, which I then multiplied by 4 to find the correct answer of 512/3. What I want to know is the 'proper' way to solve this question rather than having to multiply by 4 at the end. I don't know if all questions will allow me to multiply by 4 (or 1/4) at the end to find the correct answer, so I'm hoping someone can show me another way to go about this question. If anyone can help it would be greatly appreciated, thanks in advance.
 
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The equation of the circle is actually y^2+x^2=16. The cross-section at x goes from -sqrt(16-x^2) to +sqrt(16-x^2). The makes the base of your triangle 2*sqrt(16-x^2), not sqrt(16-x^2). That's what causing the factor of 4 error.
 

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