Problem 31 - Thomas Calculus

Question

a. A torus of revolution (doughnut) is obtained by rotating a circle \(C\) in the \(x z\) -plane about the \(z\) -axis in space. (See the accompanying figure.) If \(C\) has radius \(r>0\) and center \((R, 0,0),\) show that a parametrization of the torus is $$\begin{aligned} \mathbf{r}(u, v)=&((R+r \cos u) \cos v) \mathbf{i} \\\ &+((R+r \cos u) \sin v) \mathbf{j}+(r \sin u) \mathbf{k} \end{aligned}$$ where \(0 \leq u \leq 2 \pi\) and \(0 \leq v \leq 2 \pi\) are the angles in the figure. b. Show that the surface area of the torus is \(A=4 \pi^{2} R r\)

Step-by-Step Solution

Answer

The surface area of the torus is \(A = 4\pi^2 Rr\).

Step 1: Understanding the Problem

We are given a circle in the xz-plane with radius \(r\) and center \((R, 0, 0)\). This circle is rotated about the z-axis to form a torus. We need to confirm the given parametrization and then use it to find the surface area of the torus.

Step 2: Parametrization of the Circle

The circle \(C\) has radius \(r\) and center \((R, 0, 0)\) in the \(xz\)-plane. The equation of circle can be expressed as \((R + r \cos u, 0, r \sin u)\), where \(u\) is the parameter angle varying from 0 to \(2\pi\). This term traces the points along the circle.

Step 3: Rotation About the z-axis

The parameterization reflects the rotation of the circle around the z-axis. By letting \(v\) be the angle of rotation, we obtain the rotated x and y coordinates: \(x = (R + r \cos u) \cos v\) and \(y = (R + r \cos u) \sin v\). The z-coordinate remains \(z = r \sin u\) after rotation. This gives the parameterization: \(\mathbf{r}(u, v) = ((R+r \cos u) \cos v) \mathbf{i} + ((R+r \cos u) \sin v) \mathbf{j} + (r \sin u) \mathbf{k}\).

Step 4: Calculating the Differential Area Element

Finding a surface area involves integrating over a surface. Calculate the partial derivatives \(\mathbf{r}_u\) and \(\mathbf{r}_v\), then compute the cross product \(\mathbf{r}_u \times \mathbf{r}_v\). The magnitude of this cross product gives the differential area element \(dS = ||\mathbf{r}_u \times \mathbf{r}_v|| \, dudv\).

Step 5: Surface Area Integration

The magnitude of the cross product \(||\mathbf{r}_u \times \mathbf{r}_v||\) simplifies to \(r(R + r \cos u)\). The surface area \(A\) is calculated by integrating over the ranges of \(u\) and \(v\): \[A = \int_0^{2\pi} \int_0^{2\pi} r(R + r \cos u) \, dudv = r \int_0^{2\pi} (\int_0^{2\pi} (R + r \cos u) \, du) \, dv.\]

Step 6: Calculating the Surface Area

Integrate first with respect to \(u\), subject to the limits 0 to \(2\pi\).\[\int_0^{2\pi} (R + r \cos u) \, du = \left(R(2\pi) + \int_0^{2\pi} r \cos u \, du \right) = 2\pi R\] since \(\int_0^{2\pi} \cos u \, du = 0\). Then integrate with respect to \(v\):\[A = \int_0^{2\pi} (r \cdot 2\pi R) \, dv = 4\pi^2 Rr\].

Key Concepts

Torus Surface Area

A torus is a three-dimensional shape that looks like a doughnut. It is formed when a circle is rotated around a line external to itself, like spinning a hula hoop around your waist. When considering the torus' surface area, imagine covering the entire torus with a thin layer of plastic wrap. The total area of this wrap is the torus' surface area.
To find the surface area of a torus, we first need to establish its geometry. Think about a circle with a small radius, denoted as \( r \), this is like the thickness of the doughnut. The center of this circle is at a distance \( R \) from the axis of rotation, which represents how wide the doughnut is. The formula to determine the surface area of a torus is based on these radii:

• Small radius \( r \) – distance from the center of the circle to its edge
• Large radius \( R \) – distance from the center of revolution to the center of the circle

The surface area \( A \) of a torus is given by the formula \( A = 4\pi^{2} R r \). Here, the \( 4\pi^{2} \) comes from the geometrical shape's rotational symmetry, ensuring that the full surface, encompassing every possible angle of the rotation, is accounted for.

Parametric Equations

Parametric equations are essentially a way of describing a surface using parameters, usually angles when dealing with curves and surfaces like a torus. For our torus, we use two parameters, \( u \) and \( v \), representing rotations around different axes.
The equation for a torus involves these angles:

• \( u \) defines a point on the initial circle, as it rotates within its plane
• \( v \) defines how this circle itself rotates around the central axis forming the torus

These parameters help to navigate through every point on the torus. The parametric equations presented are:

• \( x(u, v) = (R + r \, \cos u) \cos v \)
• \( y(u, v) = (R + r \, \cos u) \sin v \)
• \( z(u, v) = r \, \sin u \)

These equations succinctly describe how each point on the torus depends on the angles \( u \) and \( v \), where \( 0 \leq u \leq 2\pi \) and \( 0 \leq v \leq 2\pi \). This parametric form is crucial for analyzing and computing properties like the surface area.

Surface Area Integration

Calculating the surface area using integration involves a few steps. First, we derive partial derivatives of the parametric equations to determine how much the position changes with a small change in the parameters.
After calculating these partial derivatives, we take their cross product. This gives us a vector pointing perpendicularly to the surface of the torus, and its magnitude provides the differential area element. For a torus, this step simplifies to

\[ ||\mathbf{r}_u \times \mathbf{r}_v|| = r(R + r \cos u) \]

• \( \mathbf{r}_u \) – Partial derivative with respect to \( u \)
• \( \mathbf{r}_v \) – Partial derivative with respect to \( v \)

This magnitude allows us to perform surface area integration over our parameters \( u \) and \( v \). The integration:

\[ A = \int_0^{2\pi} \int_0^{2\pi} r(R + r \cos u) \, duduvelpi\]

is done in two stages:

• Integrate with respect to \( u \) from 0 to \( 2\pi \) where \( \int_0^{2\pi} (R + r \cos u) \, du = 2\pi R \)
• Integrate with respect to \( v \) resulting in \( \int_0^{2\pi} r \cdot 2\pi R \, dv = 4\pi^2 Rr \)

The result of this double integral gives the complete surface area of the torus, capturing every aspect of its geometry.