Problem 21 - Thomas Calculus

Question

Use a parametrization to express the area of the surface as a double integral. Then evaluate the integral. (There are many correct ways to set up the integrals, so your integrals may not be the same as those in the back of the book. They should have the same values, however.) Circular cylinder band The portion of the cylinder \(x^{2}+y^{2}=1\) between the planes \(z=1\) and \(z=4\)

Step-by-Step Solution

Answer

The area of the surface is \(6\pi\).

Step 1: Parametrize the Surface

The surface of the cylinder can be parameterized by using cylindrical coordinates. For a circular cylinder defined by the equation \(x^2 + y^2 = 1\), a common parametrization is \( \mathbf{r}(u, v) = (\cos u, \sin u, v) \) where \(u\) is the angular parameter ranging from 0 to \(2\pi\) and \(v\) ranges from the lower z-bound to the upper z-bound, which are 1 and 4 respectively.

Step 2: Compute the Partial Derivatives

Calculate the partial derivatives of the parametrization \( \mathbf{r}(u, v) = (\cos u, \sin u, v) \). The partial derivative with respect to \(u\) is \( \frac{\partial \mathbf{r}}{\partial u} = (-\sin u, \cos u, 0) \), and with respect to \(v\) is \( \frac{\partial \mathbf{r}}{\partial v} = (0, 0, 1) \).

Step 3: Compute the Cross Product

The cross product of the partial derivatives \( \frac{\partial \mathbf{r}}{\partial u} \) and \( \frac{\partial \mathbf{r}}{\partial v} \) will give us the normal vector to the surface. Calculating the cross product, we have: \[ \mathbf{N} = \left( -\sin u, \cos u, 0 \right) \times \left( 0, 0, 1 \right) = (\cos u, \sin u, 0) \].

Step 4: Determine the Magnitude of the Normal Vector

The magnitude of the normal vector \( \mathbf{N} = (\cos u, \sin u, 0) \) is given by \( \sqrt{\cos^2 u + \sin^2 u} = 1 \).

Step 5: Set Up the Double Integral for Surface Area

The surface area \(S\) of the cylinder is given by the double integral \( \int \int_D \| \mathbf{N} \| \, du \, dv \). Since the magnitude \(\| \mathbf{N} \| = 1\), the area integral becomes \( \int_{0}^{2\pi} \int_{1}^{4} \; dv \, du \).

Step 6: Evaluate the Double Integral

Evaluate the double integral: First integrate with respect to \(v\) over the interval [1, 4]: \[ \int_{1}^{4} \, dv = (v) \Big|_{1}^{4} = 4 - 1 = 3 \]. Then integrate the result with respect to \(u\) over [0, 2\pi]: \[ \int_{0}^{2\pi} 3 \, du = 3u \Big|_{0}^{2\pi} = 3(2\pi) = 6\pi \].

Key Concepts

Cylindrical Coordinates

Cylindrical coordinates are a three-dimensional coordinate system. They extend polar coordinates by adding a height component, making them useful for parametrizing surfaces like cylinders. Just as in polar coordinates, a point in the plane is determined by a radius and an angle. In cylindrical coordinates, a third component, the height, is added.
1. **Radius**: The distance from the origin to the projection of the point on the xy-plane, denoted as r.
2. **Angle**: The angle between the positive x-axis and the line from the origin to the point's projection, typically denoted as \( \theta \) (though in this case, we use \( u \)).
3. **Height**: The z-coordinate of the point, which remains unchanged from Cartesian coordinates.
For our circular cylinder, defined by \( x^2 + y^2 = 1 \), the transformation into cylindrical coordinates is simple. The x and y coordinates translate to \( (r \cos u) \) and \( (r \sin u) \) respectively, where \( r = 1 \) is constant, and u varies from 0 to \( 2\pi \). The height is directly represented by the z-coordinate ranging from the plane at z=1 to z=4. Thus, the parametrization becomes \( (\cos u, \sin u, v) \).

Double Integral

A double integral is a mathematical tool used to calculate the volume under a surface or the surface area over a region. It involves integrating a function over a two-dimensional area. To compute the surface area of the cylinder segment, the double integral method is specifically applied.
In our example, we integrate over the region defined by the angular parameter \( u \) and the height \( v \). The double integral is denoted as \( \int \int_D f(u, v) \, du \, dv \).
The function \( f(u, v) \) is the magnitude of the normal vector to the surface, and the limits of integration are:

• For \( u \): 0 to \( 2\pi \)
• For \( v \): 1 to 4

This results in the calculation of surface area when the normal vector magnitude is constant, as it is here. Simplifying the formula, this integral becomes \( \int_{0}^{2\pi} \int_{1}^{4} 1 \, dv \, du \) leading to the final surface area value.

Surface Area

Surface area refers to the total area covering the outside of a three-dimensional object. Calculating the surface area may vary depending on the object's shape and characteristics. For instance, a circular cylinder has a specific method beyond simple geometric shapes.
Here, we approach the surface area calculation by considering a section of a cylinder with two defining planes. The integration limits specify the portion of the cylinder determined by \( z = 1 \) and \( z = 4 \).
By setting up a parametric equation using cylindrical coordinates and computing the double integral, we specifically focus on the lateral surface area of the cylinder, as the top and bottom areas at z=1 and z=4 are not included in this setup. Thus, the double integral derived \, \( \int_{0}^{2\pi} \int_{1}^{4} \, dv \, du \), effectively captures this lateral area only.

Circular Cylinder

A circular cylinder is a surface generated by moving a straight line, perpendicular to the traces on a circular base, along another line parallel to one of the axes in space. Its surface can be described in a variety of coordinate systems, but cylindrical coordinates are particularly simple and effective.
Defined by equations like \( x^2 + y^2 = 1 \), the circular cylinder's cross-section is a circle. The height of the cylinder extends between specific z-values, which, in this exercise, are from 1 to 4.
Visualizing the segment obtained when sliced by planes; it provides an intuitive understanding of its boundaries. As you calculate the surface area, remember we're only considering this cylindrical shell, excluding the base and top covers, and focusing on its direct lateral surface between the designated planes.