Problem 10 - Thomas Calculus

Question

In Exercises \(9-20,\) use the Divergence Theorem to find the outward flux of \(\mathbf{F}\) across the boundary of the region \(D .\) $$\mathbf{F}=x^{2} \mathbf{i}+y^{2} \mathbf{j}+z^{2} \mathbf{k}$$ $$\begin{array}{l}{\text { a. Cube } D :\quad \quad \quad \quad \text { The cube cut from the first octant by }} \quad\quad\quad\quad\quad\quad\quad\quad {\text { the planes } x=1, y=1, \text { and } z=1}\end{array} $$ $$b. Cube D :\quad\quad\quad\quad \quad The \quad cube \quad bounded \quad \quad \quad by \quad the \quad\quad\quad\quad\quad\quad\quad\quad\quad planes=\pm 1, y=\pm 1, \text { and } z=\pm 1$$ $$\begin{array}{c}{\text { c. Cylindrical can } D : \text { The region cut from the solid cylinder }} \quad\quad\quad\quad\quad\quad\quad\quad {x^{2}+y^{2} \leq 4 \text { by the planes } z=0 \text { and }} \\\ {z=1}\end{array}$$

Step-by-Step Solution

Answer

Flux: a) 3, b) 0, c) \(32\pi\).

Step 1: Understand the Divergence Theorem

The Divergence Theorem states that for a vector field \( \mathbf{F} \) and a closed surface \( S \) that bounds a region \( D \), the outward flux of \( \mathbf{F} \) across \( S \) is equal to the triple integral over \( D \) of the divergence of \( \mathbf{F} \). Mathematically, it is expressed as \( \oint_S \mathbf{F} \cdot d\mathbf{A} = \iiint_D (abla \cdot \mathbf{F}) \, dV \).

Step 2: Compute the Divergence of \( \mathbf{F} \)

Given \( \mathbf{F} = x^2 \mathbf{i} + y^2 \mathbf{j} + z^2 \mathbf{k} \), the divergence \( abla \cdot \mathbf{F} \) is calculated as:\[abla \cdot \mathbf{F} = \frac{\partial}{\partial x}(x^2) + \frac{\partial}{\partial y}(y^2) + \frac{\partial}{\partial z}(z^2) = 2x + 2y + 2z.\]

Step 3: Compute Outward Flux for Each Region

Compute the outward flux using the calculated divergence for each given region: **Part a:**\(D = [0,1]\times[0,1]\times[0,1]: \)\[\iiint_D (2x + 2y + 2z) \, dx \, dy \, dz = \int_0^1 \int_0^1 \int_0^1 (2x + 2y + 2z) \, dx \, dy \, dz.\]Evaluate to get the flux:\[= 3 \text{ (output from calculation of individual integrals)}.\]**Part b:** \(D = [-1,1] \times [-1,1] \times [-1,1]: \)\[\iiint_D (2x + 2y + 2z) \, dx \, dy \, dz = 0\text{ since}\ (\int_{-1}^1 x \, dx = 0, \int_{-1}^1 y \, dy = 0, \int_{-1}^1 z \, dz = 0).\]**Part c:**\(D: x^2+y^2 \leq 4, \, 0 \leq z \leq 1:\)Use cylindrical coordinates where \( x = r\cos{\theta}, \, y = r\sin{\theta}, \, dx \, dy \, dz = r \, dr \, d\theta \, dz.\)\[\iiint_D (2x + 2y + 2z) \, dx \, dy \, dz = \int_0^{1} \int_0^{2\pi} \int_0^{2} (2r\cos{\theta} + 2r\sin{\theta} + 2z) r \, dr \, d\theta \, dz = 32\pi.\]

Step 4: Conclusion

The outward flux is determined for each region using the calculated values: - **a.** Outward flux is 3.- **b.** Outward flux is 0.- **c.** Outward flux is \(32\pi\).

Key Concepts

Flux Calculation

When we talk about calculating flux, we're looking at how much of a vector field is passing through a surface. It's like measuring the flow of a river across a bridge. In mathematics, the flux of a vector field \( \mathbf{F} \) across a surface \( S \) is given by integrating the vector field over the surface. The Divergence Theorem provides a convenient way to calculate this for closed surfaces surrounding a volume. Instead of summing up tiny flows directly through the surface, we can calculate the entire flow out of a volume by a triple integral over the volume's divergence.

• The Divergence Theorem relates the flux through a closed surface to a triple integral over the volume.
• In our exercise, instead of calculating the flux manually, we use the theorem to simplify our work by looking inside the volume.

Understanding flux calculation helps students apply vector calculus in fields like fluid dynamics and electromagnetic theory.

Vector Fields

A vector field is a function that assigns a vector to every point in a region. Imagine a forest where each tree stands for a vector in our field; the arrows show direction and speed of the wind passing through the area. The vector field in our exercise is given by \( \mathbf{F} = x^2 \mathbf{i} + y^2 \mathbf{j} + z^2 \mathbf{k} \). This field varies from point to point, meaning each spot in space has a different vector based on its \(x, y,\) and \(z\) location.
In vector calculus, calculating the divergence of a vector field helps to understand how vectors behave collectively.

• The divergence \( abla \cdot \mathbf{F} \) is like a net measure of flow out from a point.
• For our vector field, we find its divergence by analyzing the changes in vector components.

Grasping vector fields is crucial for students studying physical processes like heat, fluid flow, and electromagnetism.

Triple Integrals

Triple integrals allow you to calculate a volume integral in three-dimensional space. When using the Divergence Theorem, we switch from a surface to a volume perspective using triple integrals.Triple integrals extend the idea of integration to functions of three variables, integrating over a 3D region. You can think of it like stacking slices of a loaf of bread to get its whole volume.

• A triple integral considers tiny volume elements \(dV = dx \, dy \, dz\).
• In our exercise, the triple integral \( \iiint_D (2x + 2y + 2z) \, dx \, dy \, dz \) sums up the divergence over the volume \(D\).

Triple integrals allow integrating concepts like mass, charge, and heat through a given volume, crucial for advanced physics and engineering.

Cylindrical Coordinates

Cylindrical coordinates are an alternative to Cartesian coordinates for dealing with problems featuring circular symmetry. Think of them as coordinates that wrap around a cylinder, where each point is defined using a radius, angle, and height (\( r, \theta, z \)). This system is perfect when your region involves circular structures.

• The relationship is \( x = r\cos{\theta}, \ y = r\sin{\theta} \).
• In our cylindrical integration, the volume element \( dx \, dy \, dz \) becomes \( r \, dr \, d\theta \, dz \).

For example, in Part (c) of the exercise, we work with a cylindrical volume, simplifying calculations by transforming it into \( (0 \leq r \leq 2, \ 0 \leq \theta \leq 2\pi, \ 0 \leq z \leq 1) \). Using cylindrical coordinates simplifies symmetry problems and is essential in fields like physics and engineering where cylindrical structures are common.