Question

\(\iint \mathbf{V} \cdot \mathbf{n} d \sigma\) over the entire surface of a cube in the first octant with edges of length 2 along the coordinate axes, where $$\mathbf{V}=\left(x^{2}-y^{2}\right) \mathbf{i}+3 y \mathbf{j}-2 x z \mathbf{k}$$

Short answer

The integral evaluates to 24.

Step-by-step solution

- Understand the Problem

The exercise requires calculating the flux of vector field \(\textbf{V} = (x^2 - y^2) \textbf{i} + 3y \textbf{j} - 2xz \textbf{k}\) over the surface of a cube in the first octant with edge lengths of 2 along the coordinate axes. Transform this surface integral using the Divergence Theorem.

- State the Divergence Theorem

The Divergence Theorem states that for a vector field \(\textbf{V}\) and a surface \(\textbf{S}\): \(\[\begin{equation}\iint_{\textbf{S}} \textbf{V} \cdot \textbf{n} \, d \sigma = \iiint_{\textbf{V}} abla \cdot \textbf{V} \, dV \end{equation}\]\) where \(abla \cdot \textbf{V}\) is the divergence of \(\textbf{V}\).

- Compute the Divergence of \(\textbf{V}\)

Calculate the divergence of \(\textbf{V}\). For \(\textbf{V} = (x^2 - y^2) \textbf{i} + 3y \textbf{j} - 2xz \textbf{k}\), the divergence is: \(\[\begin{equation}abla \cdot \textbf{V} = \frac{\partial}{\partial x}(x^2 - y^2) + \frac{\partial}{\partial y}(3y) + \frac{\partial}{\partial z}(-2xz)\end{equation}\]\) Computing these partial derivatives, we get: \(\begin{array}{l}\frac{\partial}{\partial x}(x^2 - y^2) = 2x\frac{\partial}{\partial y}(3y) = 3\frac{\partial}{\partial z}(-2xz) = -2x\end{array}\) Therefore, \(abla \cdot \textbf{V} = 2x + 3 - 2x = 3\).

- Set Up the Volume Integral

Using the Divergence Theorem, transform the surface integral into a volume integral: \(\iiint_{\textbf{V}} 3 \, dV\) over the volume of the cube in the first octant with edge lengths of 2. The volume integral is: \(\iiint_0^2 \,\iiint_0^2 \,\iiint_0^2 3 \, dx \, dy \, dz\).

- Perform the Volume Integration

We can split the integration into three separate integrals: \(\int_0^2 \int_0^2 \int_0^2 3 \, dx \, dy \, dz = 3 \, \int_0^2 \int_0^2 \int_0^2 dx \, dy \, dz\). Evaluating: \(3 \, \int_0^2 \int_0^2 \int_0^2 \, dxdydz = 3 \, [(x)_{0}^{2}] [(y)_{0}^{2}] [(z)_{0}^{2}] = 3 \, (2) (2) (2) = 3 \, \times 8 = 24\).

Key concepts

flux calculation

Flux measures the flow of a vector field through a surface. To find the flux, we often use the surface integral, which involves integrating a vector field over a given surface. However, this can be simplified using the Divergence Theorem. For our exercise, we need to calculate the flux of the given vector field \( \textbf{V} = (x^2 - y^2) \textbf{i} + 3y \textbf{j} - 2xz \textbf{k} \) over a cube in the first octant with edge lengths of 2. Instead of directly calculating this surface integral, we apply the Divergence Theorem to convert it to a volume integral, making the computation more straightforward.

vector field

A vector field is a function that assigns a vector to each point in space. In this problem, we are working with the vector field given by \( \textbf{V} = (x^2 - y^2) \textbf{i} + 3y \textbf{j} - 2xz \textbf{k} \). This means that at each point \( (x, y, z) \) in space, there is a vector that dictates a specific direction and magnitude. Understanding the components of the vector field is crucial for later steps in applying the Divergence Theorem.

surface integral

A surface integral helps compute flux when a vector field interacts with a surface. In mathematical terms, it is represented by \( \iint \textbf{V} \cdot \textbf{n} \ d \sigma \). Here, \( \textbf{V} \) is the vector field, \( \textbf{n} \) is the unit normal to the surface, and \( d \sigma \) is the surface element. For our exercise, calculating this directly over a cube is complex. To simplify, we use the Divergence Theorem, which converts this surface integral into a more manageable volume integral.

volume integral

A volume integral allows measuring quantities over a 3D region. According to the Divergence Theorem, the flux through a surface can be converted to a volume integral using the divergence of a vector field. For our vector field \( \textbf{V} \), the divergence is calculated as \( abla \cdot \textbf{V} \). Simplifying, we get \( abla \cdot \textbf{V} = 3 \). We then integrate this over the volume of the cube, \( \iiint_0^2 \iiint_0^2 \iiint_0^2 3 \ dV = 3 \times 8 = 24 \). This transformation not only simplifies the problem but also highlights the power of the Divergence Theorem in flux calculations.