Problem 24 - Thomas Calculus

Question

In Exercises \(19-28,\) use a parametrization to find the flux \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d \sigma\) across the surface in the specified direction. Cylinder \(\mathbf{F}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\) outward through the portion of the cylinder \(x^{2}+y^{2}=1\) cut by the planes \(z=0\) and \(z=a\)

Step-by-Step Solution

Answer

The flux across the surface is \(2\pi a\).

Step 1: Understand the Surface

The surface is a cylinder with radius 1, centered along the z-axis. It is confined between two planes, namely, at the bottom where \(z=0\) and at the top where \(z=a\). We need to consider only the lateral surface (side) of the cylinder because the problem specifies flux through 'the portion of the cylinder'. The bottom and top surfaces are not included.

Step 2: Parametrize the Surface

To parametrize the lateral surface of the cylinder, we can use cylindrical coordinates. Let \(x = \cos \theta\), \(y = \sin \theta\), and \(z = z\) to parametrize the surface, where \(0 \leq \theta < 2\pi\) and \(0 \leq z \leq a\). Thus, the parametric equations are: \(\mathbf{r}(\theta, z) = (\cos \theta, \sin \theta, z)\).

Step 3: Find the Normal Vector

To find the normal vector, calculate the cross product of the partial derivatives of \(\mathbf{r}(\theta, z)\) with respect to \(\theta\) and \(z\). Compute \(\frac{\partial \mathbf{r}}{\partial \theta} = (-\sin \theta, \cos \theta, 0)\) and \(\frac{\partial \mathbf{r}}{\partial z} = (0, 0, 1)\). The normal vector is then the cross product: \(\mathbf{n} = \frac{\partial \mathbf{r}}{\partial \theta} \times \frac{\partial \mathbf{r}}{\partial z} = (\cos \theta, \sin \theta, 0)\).

Step 4: Calculate the Dot Product \(\mathbf{F} \cdot \mathbf{n}\)

Substitute \(\mathbf{F} = (x, y, z)\) with the parametrization into \(\mathbf{F}\) to get \(\mathbf{F}(\theta, z) = (\cos \theta, \sin \theta, z)\). Then, the dot product \(\mathbf{F} \cdot \mathbf{n}\) is: \(\mathbf{F} \cdot \mathbf{n} = (\cos \theta, \sin \theta, z) \cdot (\cos \theta, \sin \theta, 0) = \cos^2(\theta) + \sin^2(\theta) = 1\) since \(\cos^2(\theta) + \sin^2(\theta) = 1\).

Step 5: Set Up the Double Integral

The flux integral over the surface is \(\iint_S \mathbf{F} \cdot \mathbf{n} \, d\sigma = \int_0^{2\pi} \int_0^a 1 \, dz \, d\theta\).

Step 6: Evaluate the Integral

Evaluate the inner integral with respect to \(z\): \(\int_0^a 1 \, dz = z \Big|_0^a = a\). Then evaluate the outer integral with respect to \(\theta\): \(\int_0^{2\pi} a \, d\theta = a\theta \Big|_0^{2\pi} = a(2\pi - 0) = 2\pi a\). The total flux across the surface is \(2\pi a\).

Key Concepts

Parametrization

When dealing with complex surfaces, like the side of a cylinder, we need to express these surfaces using simpler mathematical formulations. Parametrization allows us to do just that by describing surfaces through functions of one or more variables. In this exercise, we work with the lateral surface of a cylinder. By parametrizing it using cylindrical coordinates, we can break it down into equations that are easier to manipulate and integrate.

In our example, we use parameters

• \(\theta\) to represent the angle around the z-axis, which is the rotational aspect of the cylinder
• \(z\) to denote the height on the z-axis, which spans from 0 to \(a\)

The parametric surface equation becomes \(\mathbf{r}(\theta, z) = (\cos \theta, \sin \theta, z)\). By using these equations, we go from a challenging 3D spatial understanding to simpler mathematical expressions.

Cylindrical Coordinates

Cylindrical coordinates are particularly advantageous when dealing with problems involving symmetry around an axis. The system extends the concept of polar coordinates by adding a third component, based on how far a point is along the vertical axis, which we refer to as the z-axis.

This coordinate system is suitable for problems involving cylinders or spiral shapes. Here, each point in space is represented by three values:

• \(r\) - radius from the origin in the x-y plane.
• \(\theta\) - angle around the z-axis.
• \(z\) - height along the z-axis.

For a cylinder with radius 1 centered along the z-axis, such as ours, we set \(r = 1\), and use \(x = \cos \theta\) and \(y = \sin \theta\). These formulations help in defining the surface of the cylinder precisely, constrained along the given planes.

Surface Integral

Surface integrals extend the concept of integrals from lines and areas to surfaces, allowing us to compute a quantity distributed over a curved plane. In our context,

• We use the surface integral to calculate the flux of a vector field through a surface.
• The integral \(\iint_S \mathbf{F} \cdot \mathbf{n} \, d\sigma\) represents the surface integral, with \(\mathbf{F}\) being the vector field and \(\mathbf{n}\) the normal vector.

The process involves two parts:

First is finding the dot product of the function and the normal vector. In this case, it simplifies to 1, thanks to trigonometric identities.

The second part is combining this dot product result with the surface area of each infinitesimal element across the surface, completing the integration over the given bounds. This gives us the full measure of flux through our specified surface.

Normal Vector

In geometry and vector calculus, the normal vector is crucial as it indicates a perpendicular direction to a surface. To complete our surface integral calculation, we find the normal vector through the cross product of partial derivatives. For our parametrization \(\mathbf{r}(\theta, z)\), this involves:

• Calculating the partial derivative with respect to \(\theta\), which gives \((-\sin \theta, \cos \theta, 0)\).
• Computing the partial derivative with respect to \(z\), resulting in \((0, 0, 1)\).

The normal vector \(\mathbf{n}\) is therefore determined by the cross product of these two derivatives. For the cylinder, it simplifies to \((\cos \theta, \sin \theta, 0)\). This vector, fundamental for flux calculation, points outwards in the radial direction, as required for our surface-oriented problem.