Math Problem Statement

Seja S ⊂ ℝ³ uma superfície de revolução que resulta da rotação, em torno do eixo dos zz, duma curva Γ de Jordan, ou seja, simples e fechada, definida no plano radial ρOz, em que ρ > 0 é a coordenada radial. (a) Mostre que a área da superfície S é dada pela expressão A(S) = 2π∫_Γ ρ ds. (b) Use o resultado anterior para mostrar que a área da superfície de um toro de raios 0 < r < R é 4π²Rr.

Solution

This problem concerns a surface of revolution in R3\mathbb{R}^3 that results from rotating a Jordan curve Γ\Gamma around the zzzz-axis. Let me break down each part:

Part (a)

The goal is to show that the surface area A(S)A(S) of the surface SS is given by the integral: A(S)=2πΓρds,A(S) = 2\pi \int_{\Gamma} \rho \, ds, where ρ\rho is the radial coordinate, and dsds is the differential arc length element along the curve Γ\Gamma.

Explanation:

For a surface of revolution, the formula for the surface area is obtained by rotating a curve around an axis (here, the zz-axis). The general formula for the surface area of revolution is: A=2πρ(s)ds,A = 2\pi \int \rho(s) \, ds, where ρ(s)\rho(s) is the distance from the curve to the axis of rotation (in this case, the zz-axis). The curve Γ\Gamma is parameterized in terms of arc length ss, and dsds represents an infinitesimal element of this curve.

Thus, the formula A(S)=2πΓρdsA(S) = 2\pi \int_{\Gamma} \rho \, ds is a general result for the surface area of revolution, obtained by integrating over the curve Γ\Gamma.

Part (b)

Now, we are tasked with using the result from part (a) to show that the surface area of a torus with radii 0<r<R0 < r < R is 4π2Rr4\pi^2 Rr.

Explanation:

A torus can be described as the surface obtained by rotating a circle of radius rr (lying in the plane perpendicular to the zz-axis) around a circle of radius RR (centered on the zz-axis). Using cylindrical coordinates, the circle that is rotated has a radial distance ρ=R+rcos(θ)\rho = R + r\cos(\theta), where θ\theta is the angle parameter of the small circle of radius rr.

From part (a), the surface area formula is: A(S)=2πΓρds.A(S) = 2\pi \int_{\Gamma} \rho \, ds. For the torus, the radial distance ρ=R+rcos(θ)\rho = R + r\cos(\theta), and we are integrating over the circular arc of length 2πr2\pi r. Thus, the surface area of the torus is: Atorus=2πR2πr=4π2Rr.A_{\text{torus}} = 2\pi \cdot R \cdot 2\pi r = 4\pi^2 Rr. This confirms the result.

Would you like a more detailed explanation or any clarifications on these concepts?

5 related questions:

  1. How is the arc length differential dsds computed for a given curve Γ\Gamma?
  2. How does the surface of revolution formula change for different axes of rotation?
  3. What is the difference between a solid torus and a surface torus in mathematical terms?
  4. How would you compute the volume of the torus using the same rotational principles?
  5. Can this surface area formula be generalized to other coordinate systems, such as spherical coordinates?

Tip:

When dealing with surfaces of revolution, always check the radial distance ρ\rho to the axis of rotation, as it directly impacts the surface area formula.

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Math Problem Analysis

Mathematical Concepts

Surface of revolution
Differential geometry
Integration

Formulas

Surface area of revolution: A(S) = 2π ∫_Γ ρ ds
Torus surface area: 4π²Rr

Theorems

Surface area formula for revolutions

Suitable Grade Level

Undergraduate Mathematics (Calculus and Geometry)