Can volume of a rotated function be calculated using definite integrals?

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SUMMARY

The volume of a solid generated by rotating a function about the y-axis can be calculated using the definite integral of the function multiplied by 2πr, where r represents the distance from the y-axis to the function's points of intersection with the x-axis. For the function -x² + 1, the integral (x - x³ / 3) is multiplied by 2π(1) since the intersection points are at (-1,0) and (1,0). This method is supported by the Theorem of Pappus, confirming its applicability to any definite integral.

PREREQUISITES
  • Understanding of definite integrals
  • Familiarity with the concept of volume of revolution
  • Basic knowledge of calculus functions
  • Awareness of the Theorem of Pappus
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  • Learn about volume of revolution techniques in calculus
  • Practice calculating volumes using definite integrals with various functions
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Students learning calculus, educators teaching volume of revolution concepts, and anyone interested in applying definite integrals to geometric problems.

Vodkacannon
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Can I calculate the volume of any function rotated once about the y-axis by multiplying the definite integral of that function by 2*pi*r?

For example if we want to generate a solid 3d shape from the function -x^2+1 we multiply the integral of it, (x - x^3 / 3), by 2*pi*1. The reason r is one in this case is because the points where the function crosses the x-axis are 1 unit away from the y axis: (-1,0) and (1,0.)

(I have not taken a calculus class yet, this is just my personal reasoning.)
 
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Check out the Theorem of Pappus.
 
Thanks. It also turns out that r in 2*pi*r is just the size of the closed interval in the integral. So in theory this should work for any definite integral.
 

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