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 $$ \begin{aligned} \mathbf{r}(u, \boldsymbol{v})=&((R+r \cos u) \cos v) \mathbf{i} \\ &+((R+r \cos u) \sin v) \mathbf{j}+(r \sin u) \mathbf{k} \end{aligned} $$ 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 parametrization is verified, and the surface area is shown to be \(A = 4 \pi^{2} R r\).

Step 1: Analyze the Circle and Its Rotation

The circle, with radius \(r\) and center at \((R, 0, 0)\), is initially in the \(x z\)-plane. By rotating this circle about the \(z\)-axis, we generate the torus of revolution. We need to express this circle parametrically prior to rotation.

Step 2: Parametrize the Circle

The circle in the \(x z\)-plane can be parametrized as: \((R + r \cos u, 0, r \sin u)\). Here, \(u\) represents the angle swept out by a point on the circle as it rotates.

Step 3: Rotate the Circle Around the z-axis

To form the torus, rotate this circle around the \(z\)-axis. If \(v\) is the angle of rotation about the \(z\)-axis, then the point on the torus is given by: \(x=(R+r\cos u)\cos v\), \(y=(R+r\cos u)\sin v\), and \(z=r\sin u\).

Step 4: Write the Parametric Equations

Combine the coordinates from the rotation to form the vector parametrization: \(\mathbf{r}(u, v) = ((R+r\cos u) \cos v) \mathbf{i} + ((R+r\cos u) \sin v) \mathbf{j} + (r\sin u) \mathbf{k}\). This represents the torus with parameters \(u\) and \(v\).

Step 5: Calculate the Surface Area

We find the surface area by calculating the integral of the magnitude of the cross product of the partial derivatives of \(\mathbf{r}(u, v)\) with respect to \(u\) and \(v\): \(A = \int_0^{2\pi}\int_0^{2\pi} \left| \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} \right| \ du \ dv\).

Step 6: Compute Partial Derivatives

First, find \(\frac{\partial \mathbf{r}}{\partial u}\) and \(\frac{\partial \mathbf{r}}{\partial v}\):- \(\frac{\partial \mathbf{r}}{\partial u} = -r\sin u \cos v\ \mathbf{i} - r\sin u \sin v\ \mathbf{j} + r\cos u\ \mathbf{k}\)- \(\frac{\partial \mathbf{r}}{\partial v} = -(R + r\cos u)\sin v\ \mathbf{i} + (R + r\cos u)\cos v\ \mathbf{j}\)

Step 7: Evaluate the Cross Product

Calculate the cross product:\[ \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -r\sin u \cos v & -r\sin u \sin v & r\cos u \ -(R + r\cos u)\sin v & (R + r\cos u)\cos v & 0 \end{vmatrix} \]Resulting in magnitude: \(r(R + r\cos u)\).

Step 8: Integrate Over the Parameters

Integrate the magnitude: \[ A = \int_0^{2\pi}\int_0^{2\pi} r(R + r\cos u) \ du \ dv = 4\pi^{2}Rr \] completing the solution.

Key Concepts

Surface Area Calculation

Calculating the surface area of a torus involves using integrals to sum up tiny surface patches across the entire structure. The torus can be visualized as a doughnut-shaped object. Its surface area can be mathematically found using a specific process. The surface area calculation relies heavily on parametrization and integration of these parameters. From the given equation, \[ A = \int_0^{2\pi} \int_0^{2\pi}| \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}| \ du \ dv \], we can determine the total area covering the torus.

The double integral here represents summing up the surface elements over the parameters \( u \) and \( v \), representing angles around the circle and around the torus, respectively. This formula gives us the desired surface area of the torus which comes out to be \(4\pi^2 Rr\). This result confirms that the surface area depends on both the major radius \( R \) and the minor radius \( r \) of the torus.

Parametric Equations

A parametric equation describes a torus in terms of two parameters, \(u\) and \(v\), both of which range from 0 to \(2\pi\). This equation lets us map every point on the torus by varying these two parameters. The parametric representation is \[ \mathbf{r}(u, v) = ((R + r \cos u) \cos v) \mathbf{i} + ((R + r \cos u) \sin v) \mathbf{j} + (r \sin u) \mathbf{k} \], where:

• \( R \) is the distance from the center of the tube to the center of the torus.
• \( r \) is the radius of the tube.
• \( u \) and \( v \) are the angles that can vary from \(0\) to \(2\pi\).

The parameters \( u \) and \( v \) effectively "control" movement within the torus, allowing us to traverse both around the circular cross-section of the tube (specified by \( u \)) and the circular path of the torus (specified by \( v \)). Through these equations, we can plot or visualize every point on the torus.

Cross Product

The cross product is vital when calculating the surface area of our torus. It allows us to determine a vector perpendicular to two vectors in 3D space, which in turn helps in determining areas. Here, we specifically use the cross product of the partial derivatives to find the infinitesimal area element of the torus surface.

We calculate \( \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} \), which provides the perpendicular vector to the surface element spanned by \( \frac{\partial \mathbf{r}}{\partial u} \) and \( \frac{\partial \mathbf{r}}{\partial v} \). The magnitude of this vector gives us the area of an infinitesimally small patch on the torus: \[ r(R + r \cos u) \]. This calculation highlights how the cross product is essential in finding areas associated with parametrically defined surfaces like the torus.

Partial Derivatives

Partial derivatives help us understand how a function changes with respect to changes in just one of its variables, while keeping others constant. In the context of our torus, they are crucial in calculating the cross-product used for surface area computation.

For the given parametric equations, \( \frac{\partial \mathbf{r}}{\partial u} \) and \( \frac{\partial \mathbf{r}}{\partial v} \) represent instantaneous rates of change of the torus points as \( u \) and \( v \) vary, respectively. These derivatives are:

• \( \frac{\partial \mathbf{r}}{\partial u} = -r\sin u \cos v\ \mathbf{i} - r\sin u \sin v\ \mathbf{j} + r\cos u\ \mathbf{k} \)
• \( \frac{\partial \mathbf{r}}{\partial v} = -(R + r\cos u)\sin v\ \mathbf{i} + (R + r\cos u)\cos v\ \mathbf{j} \)

These derivatives help build the vectors needed for the cross product calculation, thereby leading to the surface area of the torus. Understanding partial derivatives is fundamental in addressing how geometric surfaces deform and stretch.