Finding the volume of a cone with a elliptic base

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SUMMARY

The volume of a cone with an elliptic base can be computed using the formula V = (1/3)πab, where 'a' and 'b' are the lengths of the semimajor and semiminor axes of the ellipse. In this discussion, the specific values are a = 4 and b = 6, leading to a height 'h' of 20. The integration approach attempted by the user was unnecessary, as the direct formula provides the correct volume without complications. The key takeaway is that both the formula and integration yield the same result for this geometric problem.

PREREQUISITES
  • Understanding of geometric formulas, specifically for volume calculation
  • Familiarity with the properties of ellipses, including semimajor and semiminor axes
  • Basic knowledge of calculus, particularly integration techniques
  • Proficiency in using π (pi) in mathematical equations
NEXT STEPS
  • Study the derivation of the volume formula for cones with elliptical bases
  • Learn about the properties and equations of ellipses in greater detail
  • Explore integration techniques relevant to volume calculations in calculus
  • Investigate applications of geometric volume calculations in real-world scenarios
USEFUL FOR

Students studying geometry, mathematics educators, and anyone involved in calculus or engineering applications requiring volume calculations of conical shapes.

Ahlahn
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Finding the volume of a cone with a elliptic base!

Homework Statement



The area of an ellipse is (pi)ab, where a and b are the lengths of the semimajor and semiminor axes. Compute the volume of a cone of height h = 20 whose base is an ellipse with semimajor and semiminor axes a = 4 and b = 6.

Homework Equations


The Attempt at a Solution



I tried to use the law of similar triangles to obtain the area of the elliptic at the cross section.

4/20 = a/20-y and 6/20 = b/20-y
b=6(20-y)/20 and a = 4(20-y)/20

Since the area of an elliptic is (pi)a*b, I tried to integrate by plugging in the above equations for a and b on the interval [0,20]. I failed.

HELP!
 
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You've set up the problem exactly correctly. The volume is (1/3)*pi*a*b where a=4 and b=6. The integral also gives you (1/3)*pi*a*b where a=4 and b=6. How exactly did you 'fail'?
 

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