Problem 5 - Calculus Early Transcendentals

Question

Let \(\sigma\) be the cylindrical surface that is represented by the vector-valued function \(\mathbf{r}(u, v)=\cos v \mathbf{i}+\sin v \mathbf{j}+\) \(u \mathbf{k}\) with \(0 \leq u \leq 1\) and \(0 \leq v \leq 2 \pi\) (a) Find the unit normal \(\mathbf{n}=\mathbf{n}(u, v)\) that defines the positive orientation of \(\sigma .\) (b) Is the positive orientation inward or outward? Justify your answer.

Step-by-Step Solution

Answer

The positive orientation is inward, as the unit normal vector points toward the cylinder's axis.

Step 1: Understand the Surface

The surface described by \(\mathbf{r}(u, v) = \cos v \mathbf{i} + \sin v \mathbf{j} + u \mathbf{k}\) is a cylinder along the \(z\)-axis with a radius of 1 and spanning from \(z = 0\) to \(z = 1\). The parameter \(v\) defines the angle around the axis, while \(u\) defines the height of the cylinder.

Step 2: Find Partial Derivatives

Calculate the partial derivatives \(\frac{\partial \mathbf{r}}{\partial u}\) and \(\frac{\partial \mathbf{r}}{\partial v}\).\[ \frac{\partial \mathbf{r}}{\partial u} = \mathbf{k} \]\[ \frac{\partial \mathbf{r}}{\partial v} = -\sin v \mathbf{i} + \cos v \mathbf{j} \]

Step 3: Compute the Cross Product

The cross product of \(\frac{\partial \mathbf{r}}{\partial u}\) and \(\frac{\partial \mathbf{r}}{\partial v}\) gives the normal vector to the surface.\[\mathbf{N} = \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} = \mathbf{k} \times (-\sin v \mathbf{i} + \cos v \mathbf{j}) = -\cos v \mathbf{i} - \sin v \mathbf{j}\]

Step 4: Calculate the Unit Normal Vector

The unit normal vector is obtained by normalizing \(\mathbf{N}\).\[\| \mathbf{N} \| = \sqrt{(-\cos v)^2 + (-\sin v)^2} = \sqrt{\cos^2 v + \sin^2 v} = 1\]\[\mathbf{n}(u, v) = \frac{\mathbf{N}}{\| \mathbf{N} \|} = -\cos v \mathbf{i} - \sin v \mathbf{j}\]

Step 5: Determine Orientation

The unit normal vector \(\mathbf{n}(u, v) = -\cos v \mathbf{i} - \sin v \mathbf{j}\) points radially inward, because it is directed opposite to the radius \(\cos v \mathbf{i} + \sin v \mathbf{j}\). Thus, the positive orientation is inward. Since it points towards the axis, the normal is inward on the cylindrical surface.

Key Concepts

Vector-Valued Function

A vector-valued function is an essential tool in describing surfaces and curves in three-dimensional space. It assigns a vector to each point in the domain, usually defined by two or more parameters. In this context, the vector-valued function, \( \mathbf{r}(u, v) = \cos v \mathbf{i} + \sin v \mathbf{j} + u \mathbf{k} \), represents a cylindrical surface. Here, \( u \) and \( v \) are parameters where \( u \) determines the height along the cylinder's axis (often the z-axis), and \( v \) controls the angle around this axis, making the parameterization extremely useful for cylindrical coordinates.
The vector \( \mathbf{r}(u, v) \) gives the position of points on the cylinder by combining:

• \( \cos v \mathbf{i} \): which represents the x-coordinate,
• \( \sin v \mathbf{j} \): which represents the y-coordinate,
• \( u \mathbf{k} \): which represents the z-coordinate.

By varying \( v \) from 0 to \( 2\pi \), a full circle is wrapped around the z-axis, and by adjusting \( u \) from 0 to 1, the circle extends into a complete cylinder, making it a comprehensive way to visualize and calculate geometric properties.

Unit Normal Vector

The unit normal vector is a critical concept in vector calculus that indicates a direction perpendicular to a surface. The process of finding this vector begins by calculating the cross product of the surface's partial derivatives. In our case, we compute the partial derivatives of the vector function \( \mathbf{r} \), which results in the vectors \( \frac{\partial \mathbf{r}}{\partial u} = \mathbf{k} \) and \( \frac{\partial \mathbf{r}}{\partial v} = -\sin v \mathbf{i} + \cos v \mathbf{j} \).
Next, we find the cross product of these two vectors, which gives the normal vector \( \mathbf{N} = -\cos v \mathbf{i} - \sin v \mathbf{j} \). To obtain the unit normal vector \( \mathbf{n} \), we normalize \( \mathbf{N} \) by dividing it by its magnitude. Since \( \| \mathbf{N} \| = 1 \), \( \mathbf{n} = \mathbf{N} \).
This unit normal vector is vital for defining the surface's orientation, which forms the basis for understanding the directionality of surfaces in vector fields.

Partial Derivatives

Partial derivatives are fundamental in computing how a function changes as one of its inputs changes while the others are held constant. They provide insight into the behavior of vector-valued functions, particularly in relation to surfaces.
For our cylindrical surface, we need the partial derivatives with respect to both \( u \) and \( v \) of the function \( \mathbf{r}(u, v) \). This results in the derivatives \( \frac{\partial \mathbf{r}}{\partial u} = \mathbf{k} \) and \( \frac{\partial \mathbf{r}}{\partial v} = -\sin v \mathbf{i} + \cos v \mathbf{j} \).

• \( \frac{\partial \mathbf{r}}{\partial u} \) indicates changes along the height of the cylinder.
• \( \frac{\partial \mathbf{r}}{\partial v} \) reflects changes around the cylinder's circular section.

These derivatives help determine the surface's tangent planes and, consequently, the unit normal vector, which further aids in establishing the surface’s orientation.

Positive Orientation

Orientation of a surface refers to a consistent choice of normal direction over the surface. For the cylindrical surface in the exercise, we determine its orientation using the unit normal vector. A surface is said to have positive orientation if the set of unit normals is consistently directed outward or inward.
In this problem, the calculated unit normal vector \( \mathbf{n}(u, v) = -\cos v \mathbf{i} - \sin v \mathbf{j} \), obtained from normalizing the cross product, points radially inward. This is because it opposes the radius vector \( \cos v \mathbf{i} + \sin v \mathbf{j} \), indicating an inward direction towards the central axis of the cylinder.
Understanding whether the orientation is inward or outward is crucial for various applications, such as integrating over surfaces and examining vector fields' flux through the surface. Knowing the positive orientation helps in correctly applying the right-hand rule and other methods for surface integrals.