Hardcore rotation problem in multivariable calculus

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SUMMARY

The discussion centers on deriving the Cartesian equation for a torus formed by rotating the circle defined by (x-2)² + y² = 1 around the y-axis. The solution involves substituting x with r = √(x² + z²) in the original equation, leading to the equation (√(x² + z²) - 2)² + y² = 1. This substitution is justified as any point (x, y, z) on the torus must satisfy the original circle's equation, confirming the relationship between the radius of the torus and the x-coordinate of the circle.

PREREQUISITES
  • Understanding of multivariable calculus concepts, particularly rotation of shapes.
  • Familiarity with Cartesian equations and their transformations.
  • Knowledge of the geometric properties of a torus.
  • Proficiency in algebraic manipulation and simplification of equations.
NEXT STEPS
  • Study the geometric properties of a torus and its equations.
  • Learn about the implications of rotating shapes in multivariable calculus.
  • Explore examples of Cartesian equations derived from rotating different shapes.
  • Investigate the use of parametric equations in describing surfaces like the torus.
USEFUL FOR

Students and educators in multivariable calculus, mathematicians interested in geometric transformations, and anyone seeking to understand the relationship between 2D shapes and their 3D counterparts.

Nikitin
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Homework Statement


Let C be the circle defined by (x-2)2+y2 = 1. If this circle is rotated along the y-axis, a torus will form. What is the Cartesian equation for the torus?

The Attempt at a Solution



The solution manual says you just switch the x in (x-2)2+y2 = 1 with r=√(x2+z2) and simplify.Why would that work? Why does the x in the first equation equal the radius of the torus? I don't get it.
 
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help?
 
Hi Nikitin! :smile:

if (x,y,z) is on the torus, then so must be (√(x2+z2),y,0)

but that must be on the original circle, so (√(x2+z2) -2)2+y2 = 1 :wink:
 

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