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Chapter 16: Problem 31

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 \(A=4 \pi^{2} R r\)

Short Answer

Expert verified
Parametrization is verified, and the surface area is \(A=4\pi^2 R r\).

Step by step solution

01

Understand the Geometry of the Torus

A torus is created when a circle of radius \( r \) centered at \((R, 0, 0)\) is revolved around the \( z \)-axis. The variable \( u \) represents the angle on the circle, which describes points on the path of the circle itself, while \( v \) represents the angle around the \( z \)-axis, capturing the revolution of the circle around the torus.
02

Parametrize the Circle

Consider a point \((x, z)\) on the circle in the \( xz \)-plane, with the circle's center at \((R, 0)\). The parametric equations of the circle are given by \( x = R + r \cos u \) and \( z = r \sin u \), for \(0 \leq u \leq 2\pi\).
03

Revolve the Circle About the z-axis

When the circle is revolved about the \( z \)-axis, each point \((x, z)\) is rotated around the \( z \)-axis by the angle \( v \). This results in the point becoming \(( x \cos v, x \sin v, z)\).
04

Write the Parametrization of the Torus

From Step 3, substitute the expression for \( x \) and \( z \) from the circle into the rotation transformation: \[ \mathbf{r}(u, v) = ((R + r \cos u) \cos v) \mathbf{i} + ((R + r \cos u) \sin v) \mathbf{j} + (r \sin u) \mathbf{k} \]This is the parametric form of the torus, with \( 0 \leq u, v \leq 2\pi \).
05

Calculate the Surface Area Differential

The surface area \( dA \) of the torus is determined by the cross product of the partial derivatives \( \frac{\partial \mathbf{r}}{\partial u} \) and \( \frac{\partial \mathbf{r}}{\partial v} \), and is given by: \[ dA = \left\| \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} \right\| du \, dv \]
06

Compute Partial Derivatives

Calculate \( \frac{\partial \mathbf{r}}{\partial u} \) and \( \frac{\partial \mathbf{r}}{\partial v} \).- \( \frac{\partial \mathbf{r}}{\partial u} = (-r \sin u \cos v) \mathbf{i} + (-r \sin u \sin v) \mathbf{j} + (r \cos u) \mathbf{k} \) - \( \frac{\partial \mathbf{r}}{\partial v} = (-(R + r \cos u) \sin v) \mathbf{i} + ((R + r \cos u) \cos v) \mathbf{j} \)
07

Calculate the Cross Product

Compute \( \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} \).The magnitude of this cross product is \( r(R + r \cos u) \).
08

Integrate to Find Surface Area

The surface area \( A \) of the torus is obtained by integrating the magnitude of the cross product over the full range of \( u \) and \( v \): \[A = \int_{0}^{2\pi} \int_{0}^{2\pi} r(R + r \cos u) \, du \, dv = 4 \pi^2 R r\] This confirms that the surface area is \( A = 4 \pi^2 R r \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Surface Area of Torus
Understanding the surface area of a torus involves applying both geometry and calculus. A torus is a three-dimensional shape that resembles a doughnut, formed by revolving a circle around an axis鈥攊n this case, the z-axis. The circle has a smaller radius, denoted as \( r \), and its center is located at a distance \( R \) from the axis.
  • Visualize the circle in the \( xz \)-plane with center \((R, 0, 0)\), revolving around the z-axis.
  • The angles \( u \) and \( v \) describe positions on the torus: \( u \) follows the path of the circle, while \( v \) traces its revolution around the z-axis.
  • The surface area calculation employs the cross product of partial derivatives in parametric form, integrating over the defined range of angles.
  • The resulting formula for the surface area is \(4 \pi^2 R r\), derived from multiplying the inclination of the circle by the distance traveled during one full revolution.
Differential Geometry
Differential geometry frames the study of curves and surfaces using calculus concepts. It explains how geometric shapes are depicted, analyzed, and interpreted through mathematical principles. The torus exemplifies differential geometry as:
  • Its shape needs parametrization鈥攁 method for describing every point on its surface using two parameters \(u\) and \(v\).
  • The parameters \(u\) and \(v\) map each point in three-dimensions using trigonometric functions.
  • By considering differential changes, such as partial derivatives, the surface area and other important geometrical properties can be investigated.
  • These derivatives capture how the orientation and position of the surface change with a small movement in \(u\) or \(v\), ultimately aiding surface area calculations.
Parametric Equations
Parametric equations define curves and surfaces using parameters rather than the traditional coordinate equations. For the torus, the parametric equations are vital:
  • The circle in the \( xz \)-plane is parametrized, with its center at \((R, 0)\) and radius \( r \): \(x = R + r \cos u\), \(z = r \sin u\).
  • Upon revolving around the z-axis, each parametric point becomes: \(( (R + r \cos u) \cos v, (R + r \cos u) \sin v, r \sin u )\).
  • These equations essentially plot a continuous path in space, creating the entire torus shape when \(u\) and \(v\) cover their full intervals from \(0\) to \(2 \pi\).
  • Parametric equations offer a smooth and comprehensive way to describe complex 3D shapes.
Multivariable Calculus
Multivariable calculus allows the extension of calculus to functions of several variables, crucial for understanding how to calculate quantities like surface area over a shape like a torus.
  • By considering both the parameters \(u\) and \(v\), one uses partial derivatives to assess change along each path independently.
  • The resulting partial derivatives of the torus's parametric equations help calculate tangent vectors, leading to the cross product that gives the area differential \(dA\).
  • The integral of \(dA\) over the torus provides the total surface area, linking geometry with calculus by illustrating how infinitesimal shapes combine to form the whole.
  • This approach highlights the power of calculus in transforming how we compute and understand spatial properties of objects.

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Most popular questions from this chapter

Find the flux of the field \(\mathbf{F}(x, y, z)=z^{2} \mathbf{i}+x \mathbf{j}-3 z \mathbf{k}\) outward through the surface cut from the parabolic cylinder \(z=4-y^{2}\) by the planes \(x=0, x=1,\) and \(z=0\).

Use a CAS to perform the following steps for finding the work done by force \(\mathbf{F}\) over the given path: a. Find \(d \mathbf{r}\) for the path \(\mathbf{r}(t)=g(t) \mathbf{i}+h(t) \mathbf{j}+k(t) \mathbf{k}.\) b. Evaluate the force \(\mathbf{F}\) along the path. c. Evaluate \(\int_{C} \mathbf{F} \cdot d \mathbf{r}.\) $$\begin{array}{ll} \mathbf{F}=x y^{6} \mathbf{i}+3 x\left(x y^{5}+2\right) \mathbf{j} ; & \mathbf{r}(t)=(2 \cos t) \mathbf{i}+(\sin t) \mathbf{j} \\ 0 \leq t \leq 2 \pi \end{array}$$

Let \(C\) be the ellipse in which the plane \(2 x+3 y-z=0\) meets the cylinder \(x^{2}+y^{2}=12 .\) Show, without evaluating either line integral directly, that the circulation of the field \(\mathbf{F}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\) around \(C\) in either direction is zero.

Suppose that \(f(t)\) is differentiable and positive for \(a \leq t \leq b .\) Let \(C\) be the path \(\mathbf{r}(t)=t \mathbf{i}+f(t) \mathbf{j}, a \leq t \leq b\) and \(\mathbf{F}=y \mathbf{i} .\) Is there any relation between the value of the work integral $$ \int_{C} \mathbf{F} \cdot d \mathbf{r} $$ and the area of the region bounded by the \(t\) -axis, the graph of \(f\) and the lines \(t=a\) and \(t=b ?\) Give reasons for your answer.

Find a parametrization of the hyperboloid of two sheets \(\left(z^{2} / c^{2}\right)-\left(x^{2} / a^{2}\right)-\left(y^{2} / b^{2}\right)=1\)
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