Problem 40 - Thomas Calculus

Question

Let \(S\) be the portion of the cylinder \(y=\ln x\) in the first octant whose projection parallel to the \(y\) -axis onto the \(x z\) -plane is the rectangle \(R_{x :} : 1 \leq x \leq e, 0 \leq z \leq 1 .\) Let \(n\) be the unit vector normal to \(S\) that points away from the \(x z\) -plane. Find the flux of \(\mathbf{F}=2 y \mathbf{j}+z \mathbf{k}\) through \(S\) in the direction of \(\mathbf{n} .\)

Step-by-Step Solution

Answer

The flux through the surface is -2.

Step 1: Understand the Surface and Limits

We are given the surface defined by the equation \(y = \ln x\) within the specified region in the first octant. This corresponds to a portion of the cylinder whose projection based on the conditions given, is a rectangle in the \(xz\)-plane. Specifically, this rectangle is defined for \(1 \leq x \leq e\) and \(0 \leq z \leq 1\). Thus, the surface \(S\) is bounded by this range and represents the curved surface of the cylinder.

Step 2: Establish the Parameterization of the Surface

To parameterize the surface \(S\), we use the variables \(x\) and \(z\) as parameters since the problem is naturally suited for them:\[\mathbf{r}(x, z) = \langle x, \ln x, z \rangle\]where \(x\) ranges from 1 to \(e\) and \(z\) ranges from 0 to 1.This parameterization properly represents the surface in terms of position vectors.

Step 3: Calculate Partial Derivatives

To find the normal vector, compute the partial derivatives of \(\mathbf{r}\) with respect to both \(x\) and \(z\):\[\frac{\partial \mathbf{r}}{\partial x} = \langle 1, \frac{1}{x}, 0 \rangle\]\[\frac{\partial \mathbf{r}}{\partial z} = \langle 0, 0, 1 \rangle\]

Step 4: Find the Normal Vector by Cross Product

The normal vector \(\mathbf{n}\) to the surface is found using the cross product:\[\mathbf{n} = \frac{\partial \mathbf{r}}{\partial x} \times \frac{\partial \mathbf{r}}{\partial z} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \1 & \frac{1}{x} & 0 \0 & 0 & 1\end{vmatrix} = \langle \frac{1}{x}, -1, 0 \rangle\]Make sure this vector points away from the \(xz\)-plane, which it does.

Step 5: Calculate the Flux Integral

The flux \(\Phi\) of \(\mathbf{F}\) through the surface \(S\) can be computed using the surface integral:\[\Phi = \iint_{S} \mathbf{F} \cdot \mathbf{n} \ dS\]Substitute \(\mathbf{F} = 2y\mathbf{j} + z\mathbf{k}\) and \(\mathbf{n} = \langle \frac{1}{x}, -1, 0 \rangle\):\[\Phi = \int_{0}^{1} \int_{1}^{e} (2 \ln x)(-1) \ dx \ dz\]\[\Phi = -2 \int_{0}^{1} \int_{1}^{e} \ln x \ dx \ dz\]

Step 6: Compute the Double Integral

First, solve the inner integral: \[\int_{1}^{e} \ln x \ dx\]Using integration by parts where \(u = \ln x\) and \(dv = dx\),\[u = x \quad; \quad du = \frac{1}{x} dx\]\[\int \ln x \ dx = x \ln x - \int x \cdot \frac{1}{x} \ dx = x \ln x - x\]Evaluating from 1 to \(e\):\[[e(\ln e - 1) - (1(\ln 1 - 1)) = e - e + 1] = 1\]\[\int_{0}^{1} (-2 \cdot 1) \ dz = -2z \bigg|_{0}^{1} = -2\]

Step 7: Interpret Results - Compute Final Flux

With the calculations completed, the total flux \(\Phi\) of \(\mathbf{F}\) through \(S\) becomes:\[\Phi = -2 \times 1 = -2\]The negative sign indicates the direction is as expected.

Key Concepts

Parameterization of Surfaces

When dealing with surfaces like the segment of a cylinder described in our problem, parameterization is a key concept. It helps in translating a geometrical shape into mathematical expressions that we can easily manipulate. Parameterization involves using a few variables to describe points on the surface.

In this exercise, the surface is specified by the equation \(y = \ln x\), confined within a particular region. We choose \(x\) and \(z\) as parameters due to the constraints \(1 \leq x \leq e\) and \(0 \leq z \leq 1\). This rectilinear domain in the \(xz\)-plane simplifies our computations.

The parameterized form becomes \(\mathbf{r}(x, z) = \langle x, \ln x, z \rangle\). This vector function seamlessly traces every point on the surface by considering each permissible pair \((x, z)\). It is essentially our way of converting a curved, complicated surface into a more analytically friendly form, bridging the gap between spatial geometry and calculus.

Partial Derivatives

Partial derivatives represent the rate of change of a multivariable function when one of the variables changes, while all others remain constant. When studying surface parameterization, partial derivatives help us understand how the surface bends and twists.

In the given task, we were tasked with finding partial derivatives of the position vector \(\mathbf{r}(x, z) = \langle x, \ln x, z \rangle \) with respect to \(x\) and \(z\).

• For \(x\), the partial derivative is \(\frac{\partial \mathbf{r}}{\partial x} = \langle 1, \frac{1}{x}, 0 \rangle\). This describes how small changes in \(x\) affect the surface structure when \(z\) is kept constant.


• For \(z\), it's \(\frac{\partial \mathbf{r}}{\partial z} = \langle 0, 0, 1 \rangle\), illustrating how changes in \(z\) influence the surface when \(x\) is fixed.

These derivatives are fundamental in finding the normal vector to the surface, which aids in flux calculations.

Cross Product

The cross product is a mathematical operation that finds a vector perpendicular to two given vectors in three-dimensional space. It is crucial for determining normal vectors needed for evaluating surface integrals.

In our exercise, we applied the cross product to the partial derivatives \(\frac{\partial \mathbf{r}}{\partial x}\) and \(\frac{\partial \mathbf{r}}{\partial z}\):

\[\mathbf{n} = \frac{\partial \mathbf{r}}{\partial x} \times \frac{\partial \mathbf{r}}{\partial z} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & \frac{1}{x} & 0 \ 0 & 0 & 1\end{vmatrix} = \langle \frac{1}{x}, -1, 0 \rangle\]

This normal vector \(\mathbf{n}\) points away from the \(xz\)-plane, confirming the correctness of our orientation. The cross product not only provides the direction of this normal prodding from the surface but also interprets the physical interaction described by the surface's geometry, essential for calculating flux integration.

Surface Integrals

Surface integrals extend the concept of integration to functions defined on surfaces. They represent the accumulation of quantities across a surface, such as flux passing through it.

The flux integral provided in the exercise was needed to find the flow of vector field \(\mathbf{F} = 2y\mathbf{j} + z\mathbf{k}\) through the surface \(S\). The flux \(\Phi\) calculation is given by:

\[ \Phi = \iint_{S} \mathbf{F} \cdot \mathbf{n} \ dS \]

Here, \(\mathbf{n}\) is the normal vector derived earlier. After substituting the expressions for \(\mathbf{F}\) and \(\mathbf{n}\), and integrating over the defined parameters, we evaluated the surface integral:

\[-2 \int_{0}^{1} \int_{1}^{e} \ln x \ dx \ dz = -2\]

The integral reflected how the vector field \(\mathbf{F}\) interacts with the surface \(S\). In essence, surface integrals like this enable us to compute how vector fields affect surfaces they intersect or envelop, showcasing a vital application of calculus in physics and engineering.