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 \(R_{x z} : 1 \leq x \leq e, 0 \leq z \leq 1 .\) 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

Flux is \(-2(e-1)\).

Step 1: Parameterize the Surface

To parameterize the surface, note that it is determined by the equation \(y = \ln(x)\). We express \(S\) as a vector function \(\mathbf{r}(x,z) = (x, \ln(x), z)\), where \(1 \leq x \leq e\) and \(0 \leq z \leq 1\).

Step 2: Find the Partial Derivatives

Determine the tangent vectors \(\mathbf{r}_x\) and \(\mathbf{r}_z\) by partially differentiating \(\mathbf{r}(x,z)\) with respect to \(x\) and \(z\). We have: \[ \mathbf{r}_x = \left(1, \frac{1}{x}, 0\right) \] \[ \mathbf{r}_z = (0, 0, 1) \]

Step 3: Find the Normal Vector

Compute the normal vector \(\mathbf{n}\) by taking the cross product \(\mathbf{r}_x \times \mathbf{r}_z\): \[ \mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & \frac{1}{x} & 0 \ 0 & 0 & 1 \end{vmatrix} = \left( \frac{1}{x}, -1, 0 \right) \]

Step 4: Determine the Unit Normal Vector

Since the normal vector points away from the \(xz\)-plane, the unit normal vector \(\mathbf{\hat{n}}\) is already in the direction we need. Normalize \(\mathbf{n}\) by dividing by its magnitude: \[ \| \mathbf{n} \| = \sqrt{\left( \frac{1}{x} \right)^2 + (-1)^2} = \sqrt{\frac{1}{x^2} + 1} \] \[ \mathbf{\hat{n}} = \left( \frac{\frac{1}{x}}{\sqrt{\frac{1}{x^2} + 1}}, \frac{-1}{\sqrt{\frac{1}{x^2} + 1}}, 0 \right) \]

Step 5: Calculate the Flux Integral

The flux is given by the surface integral: \[ \iint_S \mathbf{F} \cdot \mathbf{\hat{n}} \, dS \] We find \(\mathbf{F} \cdot \mathbf{\hat{n}} = 2(\ln(x)) \cdot \frac{-1}{\sqrt{\frac{1}{x^2} + 1}} = \frac{-2\ln(x)}{\sqrt{\frac{1}{x^2} + 1}}\) since the \(\mathbf{k}\)-component doesn't contribute. The differential area is \(\| \mathbf{r}_x \times \mathbf{r}_z \| \, dx \, dz = \sqrt{\frac{1}{x^2} + 1} \, dx \, dz\). Thus, the flux integral becomes: \[ \int_0^1 \int_1^e \left( \frac{-2\ln(x)}{\sqrt{\frac{1}{x^2} + 1}} \right) \sqrt{\frac{1}{x^2} + 1} \, dx \, dz = -2 \int_0^1 \int_1^e \ln(x) \, dx \, dz \] Evaluate the inner integral: \[ -2 \int_1^e \ln(x) \, dx = -2 \left[ x \ln(x) - x \right]_1^e = -2 \left( e - 1 \right) \] Finally, perform the outer integral \(\int_0^1 dz\): \[ -2 \int_0^1 (e - 1) \, dz = -2(e - 1) \]

Step 6: Evaluate the Result

After evaluating the integrals, we find that the flux of the vector field \(\mathbf{F}\) through \(S\) in the direction of \(\mathbf{n}\) is \[ -2(e - 1) \]

Key Concepts

Surface parameterization

Imagine the surface of a curved object like a cylinder. To analyze such a surface mathematically, we often use parameterization. Parameterization helps us describe surfaces using coordinates.
This exercise involves parameterizing the surface of a cylinder described by the equation \(y = \ln(x)\) in the first octant.
We use two parameters: \(x\) and \(z\). Each point \((x, z)\) on this rectangle corresponds to a point on the cylindrical surface.
To map out the surface, we define a vector function \(\mathbf{r}(x,z)\ = (x, \ln(x), z)\).

• \(x\) varies between \(1\) and \(e\)
• \(z\) varies between \(0\) and \(1\)

This function helps convert flat coordinate points into curved surface points.

Normal vector calculation

The next step is to calculate the normal vector to the surface. The normal vector is a crucial tool in flux calculations because it helps determine the direction the surface faces.
We start by finding tangent vectors. A tangent vector is derived from the partial derivatives of our parameterization function.

• The partial derivative of \(\mathbf{r}(x,z)\) with respect to \(x\) gives \(\mathbf{r}_x = (1, \frac{1}{x}, 0)\)
• With respect to \(z\), it gives \(\mathbf{r}_z = (0, 0, 1)\)

Using these tangent vectors, the normal vector \(\mathbf{n}\) is calculated via the cross product \(\mathbf{r}_x \times \mathbf{r}_z\).
The result \((\frac{1}{x}, -1, 0)\) points away from the \(xz\)-plane, aligned with our needs for the flux calculation.

Flux integral evaluation

A flux integral measures how much a vector field penetrates a surface. In our example, vector field \(\mathbf{F} = 2y \mathbf{j} + z \mathbf{k}\) passes through the cylindrical surface.
Start by finding \(\mathbf{F} \cdot \mathbf{\hat{n}}\), the dot product with our unit normal vector.
Once simplified, the next step involves setting up a double integral:\[ \iint_S \mathbf{F} \cdot \mathbf{\hat{n}} \, dS \]
The dot product becomes \(-2\ln(x)\) since the \(\mathbf{k}\)-component vanishes.
Finally, the expression for \(dS\), the differential area element, simplifies the integral setup so that it can be evaluated as a double integral over the rectangle in the \(xz\)-plane.

Multivariable calculus problem

Understanding this problem requires juggling several calculus concepts. From vector calculus to parameterization, each element plays a role.
Breaking down the problem involves handling multiple variables and dimensions all at once.
This type of problem typically comprises:

• Describing surfaces with functions
• Crafting normal vectors for directionality
• Manipulating multivariable integrals for flux computations

These elements combine, allowing us to study physical phenomena analytically.
Each step in this exercise builds upon foundational calculus principles, highlighting the interconnected nature of multivariable problems.