Find the volume of a described solid

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SUMMARY

The volume of the solid described in the discussion is calculated using the formula for the volume of isosceles right triangles as cross sections. The base of the triangles is determined by the equation \( \sqrt{9 - x^2} \), leading to the integral \( \pi \int_{-3}^{3} (4.5 - 0.5x^2) \, dx \). The final computed volume is confirmed to be \( 18\pi \). The approach and calculations presented are accurate and valid.

PREREQUISITES
  • Understanding of calculus, specifically integration techniques
  • Familiarity with the properties of isosceles right triangles
  • Knowledge of the equation of a circle in Cartesian coordinates
  • Ability to visualize geometric shapes and their cross sections
NEXT STEPS
  • Study the method of calculating volumes of solids with known cross sections
  • Learn about the applications of double integrals in finding volumes
  • Explore graphing techniques for visualizing regions and cross sections
  • Review integration techniques for polynomial functions
USEFUL FOR

Students in calculus courses, educators teaching solid geometry, and anyone interested in mastering volume calculations of solids with varying cross sections.

Physicsisfun2005
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I don't have the answers in the back of my book and I really want to know if i did this correctly since its graded. the Problem is: Let the region bounded by [tex]x^2+y^2=9[/tex] be the base of a solid. Find the Volume if cross sections taken perpendicular to the base are isosceles right triangles.
i know a triangle is [tex].5bh[/tex] and the base of it will be [tex]\sqrt{9-x^2}[/tex] so for volume the final answer is [tex]\pi\int 4.5-.5x^2 dx[/tex] with the limits from -3 to 3 and i get [tex]18\pi[/tex] for volume.....is this right?
 
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Your calculation looks correct to me! To double check, you can graph the region and the cross sections to make sure they match up with your calculations. Also, it never hurts to double check your integration and make sure you didn't make any mistakes. But overall, it looks like you have the right approach and answer. Great job!
 

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