Question

Use either Stokes' theorem or the divergence theorem to evaluate each of the following integrals in the easiest possible way. \(\iint(\operatorname{curl} \mathbf{V}) \cdot \mathbf{n} d \sigma\) over any surface whose bounding curve is in the \((x, y)\) plane, where $$\mathbf{V}=\left(x-x^{2} z\right) \mathbf{i}+\left(y z^{3}-y^{2}\right) \mathbf{j}+\left(x^{2} y-x z\right) \mathbf{k}$$

Short answer

The surface integral evaluates to zero.

Step-by-step solution

Identify the vector field \mathbf{V}

The given vector field is \[ \mathbf{V} = (x - x^2z) \mathbf{i} + (yz^3 - y^2) \mathbf{j} + (x^2 y - xz) \mathbf{k} \]

Recall Stokes' Theorem

Stokes' Theorem states that for a vector field \mathbf{V}, \[ \iint_S (\operatorname{curl} \mathbf{V}) \cdot \mathbf{n} \ d \sigma = \oint_C \mathbf{V} \cdot d\mathbf{r} \] where \mathbf{n} is the unit normal vector to the surface, \(S\), and \(C\) is the boundary curve of \(S\).

Identify the surface and its bounding curve

The problem specifies a surface whose bounding curve is in the \(x, y\) plane.

Apply Stokes' Theorem

Using Stokes' Theorem, we convert the surface integral \[ \iint_S (\abla \times \mathbf{V}) \cdot \mathbf{n} \ d \sigma \] into a line integral \[ \oint_C \mathbf{V} \cdot d\mathbf{r} \] over the bounding curve in the \(x, y\) plane.

Evaluate the line integral

Since the bounding curve \(C\) is in the \(x, y\) plane, \(z = 0\) along this curve. Substitute \(z = 0\) into \mathbf{V}:\[ \mathbf{V} = x \mathbf{i} - y^2 \mathbf{j} \] Now, perform the line integral \[ \oint_C (x \mathbf{i} - y^2 \mathbf{j}) \cdot d\mathbf{r} \].

Select suitable parameterization for the curve

Parameterize the bounding curve \(C\) in the \(x, y\) plane. For simplicity, assume \(C\) to be a circle of radius \(r\) centered at the origin. Then, use \( x = r\cos\theta \) and \( y = r\sin\theta \) with \( 0 \leq \theta \leq 2\pi \).

Compute the differential element \(d\mathbf{r}\)

Using the parameterization, \[ d\mathbf{r} = \frac{d\mathbf{r}}{d\theta} d\theta = (- r \sin\theta \mathbf{i} + r \cos\theta \mathbf{j}) d\theta \]

Substitute into the line integral

Substitute the parameterization and differential element into the line integral: \[ \oint_C \mathbf{V} \cdot d\mathbf{r} = \int_{0}^{2 \pi} (r \cos\theta \mathbf{i} - r^2 \sin^2 \theta \mathbf{j}) \cdot (- r \sin \theta \mathbf{i} + r \cos\theta \mathbf{j}) d\theta \]

Simplify the integrand

Evaluate the dot product: \[ (r \cos\theta \mathbf{i} - r^2 \sin^2 \theta \mathbf{j}) \cdot (- r \sin \theta \mathbf{i} + r \cos \theta \mathbf{j}) = -r^2 \cos\theta \sin \theta + r^3 \cos \theta \sin \theta = 0 \]

Conclude the result

Since the integrand simplifies to zero, the value of the line integral and thus the surface integral is:\[ \mathbf{0} \]

Key concepts

Vector Field

In mathematics and physics, a vector field represents a vector assigned to every point in a subset of space. Imagine the vector field as a collection of arrows pointing in various directions and with varying lengths, depending on their assigned vectors. For instance, in the given problem, the vector field is defined as \( \mathbf{V} = (x - x^2z) \mathbf{i} + (yz^3 - y^2) \mathbf{j} + (x^2 y - xz) \mathbf{k} \). This signifies that the vector at any point \( (x, y, z) \) within the space is computed from these components. Each arrow points in the direction defined by the vector \( (\mathbf{i}, \mathbf{j}, \mathbf{k}) \) and can visualize physical phenomena such as fluid flow or electromagnetic fields in space.

Surface Integral

A surface integral is a generalization of multiple integrals to integration over surfaces. It is the integral of a scalar field or a vector field over a surface. In this exercise, we are focusing on the surface integral of the curl of a vector field. Specifically, we need to evaluate \( \iint_S (\text{curl} \mathbf{V}) \cdot \mathbf{n} \ d \sigma \). Surface integrals are essential for understanding the flow or flux of a field through a given surface. In other words, it measures how much of the vector field is passing through the surface. The surface integral here can be translated into a line integral around the boundary of the surface using Stokes' Theorem.

Line Integral

A line integral, also known as a path integral, is the integral of a function along a curve. For vector fields, it is often used to measure the work done by a force field over a path. Using Stokes' Theorem in our problem converts a difficult surface integral into a line integral around its boundary curve. We end up with the expression \( \oint_C \mathbf{V} \cdot d\mathbf{r} \). This integral is over the curve in the \( x, y \) plane, which simplifies our problem greatly. The significance of line integrals lies in their ability to translate complex field information along one-dimensional paths. In physics, this could represent, for example, the total magnetic flux around a loop.

Parameterization

Parameterization is a technique of defining a curve, surface, or volume using parameters. It is especially useful in evaluating integrals. In this exercise, the boundary curve in the \( x, y \) plane is parameterized as a circle of radius \( r \), centered at the origin, using the equations \( x = r \cos \theta \) and \( y = r \sin \theta \), with \( 0 \leq \theta \leq 2 \pi \). Parameterizing the bounding curve helps in simplifying the integral by transforming it into a form where \( d \mathbf{r} \) can be readily computed as \( ( - r \sin \theta \mathbf{i} + r \cos \theta \mathbf{j} ) \ d\theta \). This step is crucial for transforming the integrand into terms of the parameter, making the evaluation straightforward.