Question

Given \(\mathbf{V}=x^{2} \mathbf{i}+y^{2} \mathbf{j}+z^{2} \mathbf{k},\) integrate \(\mathbf{V} \cdot \mathbf{n} d \sigma\) over the whole surface of the cube of side 1 with four of its vertices at \((0,0,0),(0,0,1),(0,1,0),(1,0,0) .\) Evaluate the same integral by means of the divergence theorem.

Short answer

The integral over the cube’s surface is 3. Using the divergence theorem confirms this result.

Step-by-step solution

Understand the Problem and Given Vector Field

Given the vector field \( \mathbf{V}=x^{2} \mathbf{i}+y^{2} \mathbf{j}+z^{2} \mathbf{k},\) and you need to integrate \( \mathbf{V} \cdot \mathbf{n} d \sigma \)overthe surface of a cube with side length 1 and vertices at \(0,0,0\), \(0,0,1\), \(0,1,0\), and \(1,0,0\). The exercise also requires evaluating the integral using the divergence theorem.

Define the Cube's Surface

The cube has six faces with each face oriented perpendicular to one of the coordinate axes. Recall the normal vectors for each face:For faces perpendicular to the x-axis:1. \( x = 0 : \mathbf{n} = -\mathbf{i} \)2. \( x = 1 : \mathbf{n} = \mathbf{i} \)For faces perpendicular to the y-axis:3. \( y = 0 : \mathbf{n} = -\mathbf{j} \)4. \( y = 1 : \mathbf{n} = \mathbf{j} \)For faces perpendicular to the z-axis:5. \( z = 0 : \mathbf{n} = -\mathbf{k} \)6. \( z = 1 : \mathbf{n} = \mathbf{k} \)

Break Down the Surface Integrals

Evaluate \( \int_{S} \mathbf{V} \cdot \mathbf{n} \, d \sigma \) for each face of the cube separately. For a face where the normal vector points in the \( \mathbf{i}, \mathbf{j}, or \mathbf{k} \)direction, the expression simplifies to the respective component of \( \mathbf{V} \)at that surface.

Calculate Flux Through Each Surface

1. For \( x = 0 \, \mathbf{n} = -\mathbf{i} \): \( \mathbf{V} \cdot \mathbf{n} = -x^{2} \), integral from 0 to 1 for \( y \)and integral from 0 to 1 for \( z \): \( \int_{0}^{1} \int_{0}^{1} 0 \, dy \, dz = 0\)2. For \( x = 1 \, \mathbf{n} = \mathbf{i} \): \( \mathbf{V} \cdot \mathbf{n} = x^{2} = 1 \), integral from 0 to 1 for \( y \)and integral from 0 to 1 for \( z \): \( \int_{0}^{1} \int_{0}^{1} 1 \, dy \, dz = 1\)3. For \( y = 0 \, \mathbf{n} = -\mathbf{j} \): \( \mathbf{V} \cdot \mathbf{n} = -y^{2} \), integral from 0 to 1 for \( x \)and integral from 0 to 1 for \( z \): \( \int_{0}^{1} \int_{0}^{1} 0 \, dx \, dz = 0\)4. For \( y = 1 \, \mathbf{n} = \mathbf{j} \): \( \mathbf{V} \cdot \mathbf{n} = y^{2} = 1 \), integral from 0 to 1 for \( x \)and integral from 0 to 1 for \( z \): \( \int_{0}^{1} \int_{0}^{1} 1 \, dx \, dz = 1\)5. For \( z = 0 \, \mathbf{n} = -\mathbf{k} \): \( \mathbf{V} \cdot \mathbf{n} = -z^{2} \), integral from 0 to 1 for \( x \)and integral from 0 to 1 for \( y \): \( \int_{0}^{1} \int_{0}^{1} 0 \, dx \, dy = 0\)6. For \( z = 1 \, \mathbf{n} = \mathbf{k} \): \( \mathbf{V} \cdot \mathbf{n} = z^{2} = 1 \), integral from 0 to 1 for \( x \)and integral from 0 to 1 for \( y \): \( \int_{0}^{1} \int_{0}^{1} 1 \, dx \, dy = 1\)

Sum All Surface Integrals

Adding all the non-zero surface integrals:\(0 + 1 + 0 + 1 + 0 + 1 = 3 \)

Verify Using the Divergence Theorem

The divergence theorem states:\( \int_{S} \mathbf{V} \cdot \mathbf{n} \, d \sigma = \int_{V} \mathbf{abla} \cdot \mathbf{V} \, dV\)First, calculate the divergence of \mathbf{V}:\( \mathbf{abla} \cdot \mathbf{V} = \frac{\partial}{\partial x}(x^{2}) + \frac{\partial}{\partial y}(y^{2}) + \frac{\partial}{\partial z}(z^{2}) = 2x + 2y + 2z \)Now, integrate over the volume of the cube with side length 1:\( \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} (2x+2y+2z) \, dx \, dy \, dz = 2 \int_{0}^{1} x \, dx + 2 \int_{0}^{1} y \, dy + 2 \int_{0}^{1} z \, dz = 2 \left[ \frac{x^{2}}{2} \right]_{0}^{1} + 2 \left[ \frac{y^{2}}{2} \right]_{0}^{1} + 2 \left[ \frac{z^{2}}{2} \right]_{0}^{1} = 1 + 1 + 1 = 3 \)

Conclusion

The result from directly integrating over the surface and using the divergence theorem match. Therefore, the final result is 3.

Key concepts

vector field

A vector field is a mathematical function that associates a vector to every point in a space. In this exercise, the vector field provided is \(\textbf{V} = x^{2} \textbf{i} + y^{2} \textbf{j} + z^{2} \textbf{k} \). This means at any point \((x, y, z) \) in the space, the vector \(\textbf{V}\) has components \(x^{2} \), \(y^{2} \), and \(z^{2} \) along the \(\textbf{i} \), \(\textbf{j} \text{, and } k \) directions, respectively.
Vector fields are used in physics to model various quantities like velocity fields in fluid dynamics, electromagnetic fields, and more.
Understanding the behavior of the vector field is essential because it helps in calculating other quantities, such as the flux integral over a surface.

flux integral

A flux integral calculates how much of a vector field passes through a surface. In our case, we need to integrate \( \textbf{V} \cdot \textbf{n} \ d \sigma \), where \( \textbf{V} \) is the vector field, and \( \textbf{n} \) is the unit normal vector to the surface, over the entire surface of a cube.
This involves computing the dot product of the vector field and the normal vector for each face of the cube and then integrating this over the area of each face.

• The result of this integral tells us the net 'flow' of the vector field through the surface.
• Positive values mean the vector field is exiting the surface, and negative values mean it's entering.

surface integration

Surface integration involves summing up values over a two-dimensional surface. For example, to find the flux integral over a cube, we sum contributions from each face of the cube.
In our exercise, the cube has six faces:

• Two parallel to the yz-plane
• Two parallel to the xz-plane
• Two parallel to the xy-plane

We break down the integral over the entire surface into integrals over these individual faces. Each face lies in a coordinate plane with its own normal vectors.
By calculating the surface integrals individually for each face and then summing them, we obtain the total flux through the cube.

normal vectors

Normal vectors are vectors that are perpendicular to a surface at a given point. For each face of the cube, these vectors indicate the direction out of the surface. Identifying normal vectors for each face is crucial for setting up flux integrals properly.
For the cube in this exercise, the normal vectors are:

• Perpendicular to x-axis: \( \textbf{n} = -\textbf{i}, \ \textbf{i} \)
• Perpendicular to y-axis: \( \textbf{n} = -\textbf{j}, \ \textbf{j} \)
• Perpendicular to z-axis: \( \textbf{n} = -\textbf{k}, \ \textbf{k} \)

These vectors help simplify the dot product \( \textbf{V} \cdot \textbf{n} \) because, on each face, only one component of \( \textbf{V} \) is non-zero. Calculating the flux integral then becomes a simpler task of integrating a single function over each face.

cube surface

The cube surface in this exercise has a side length of 1 with vertices at (0,0,0), (0,0,1), (0,1,0), (1,0,0) among others. This specific configuration means that our cube is conveniently aligned with the coordinate axes.
To integrate over this cube surface, you need to assess each of the six faces independently:

• Face at \( x = 0, \textbf{n} = -\textbf{i} \)
• Face at \( x = 1, \textbf{n} = \textbf{i} \)
• Face at \( y = 0, \textbf{n} = -\textbf{j} \)
• Face at \( y = 1, \textbf{n} = \textbf{j} \)
• Face at \( z = 0, \textbf{n} = -\textbf{k} \)
• Face at \( z = 1, \textbf{n} = \textbf{k} \)

Each face integrates a component of \( \textbf{V} \), yielding contributions to the total flux. Summing these contributions gives the total flux over the cube's surface.
Understanding this is pivotal for ensuring accurate flux calculations using the divergence theorem.