Problem 11 - Thomas Calculus

Question

In Exercises \(7-12,\) use the surface integral in Stokes' Theorem to calculate the circulation of the field \(\mathbf{F}\) around the curve \(C\) in the indicated direction. \begin{equation} \begin{array}{l}{\mathbf{F}=\left(y^{2}+z^{2}\right) \mathbf{i}+\left(x^{2}+y^{2}\right) \mathbf{j}+\left(x^{2}+y^{2}\right) \mathbf{k}} \\ {C : \text { The square bounded by the lines } x=\pm 1 \text { and } y=\pm 1 \text { in the }} \\ {x y \text { -plane, counterclockwise when viewed from above }}\end{array} \end{equation}

Step-by-Step Solution

Answer

The circulation of \( \mathbf{F} \) around the curve \( C \) is 0.

Step 1: Understand Stokes' Theorem

Stokes' Theorem relates a surface integral over a surface \( S \) to a line integral around the boundary \( C \) of \( S \). It states: \[ \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (abla \times \mathbf{F}) \cdot \mathbf{n} \, dS, \] where \( \mathbf{n} \) is the unit normal vector to the surface.

Step 2: Parameterize the Surface \( S \)

The curve \( C \) is the boundary of a surface \( S \), which in this problem is the square plane defined by \( -1 \leq x \leq 1 \) and \( -1 \leq y \leq 1 \) in the \( xy \)-plane. Parameterization is not necessary here because \( z = 0 \) constant plane is defined by these bounds.

Step 3: Compute \( \nabla \times \mathbf{F} \)

The curl of the vector field \( \mathbf{F} \) is \( abla \times \mathbf{F} = \left( \frac{\partial}{\partial y}(x^2 + y^2) - \frac{\partial}{\partial z}(x^2 + y^2) \right) \mathbf{i} + \left( \frac{\partial}{\partial z}(y^2 + z^2) - \frac{\partial}{\partial x}(x^2 + y^2) \right) \mathbf{j} + \left( \frac{\partial}{\partial x}(x^2 + y^2) - \frac{\partial}{\partial y}(y^2 + z^2) \right) \mathbf{k}. \) Evaluate this to find \( abla \times \mathbf{F} = -2y \mathbf{k}. \)

Step 4: Set Up the Surface Integral

Since the surface \( S \) is in the \( xy \)-plane, the unit normal vector \( \mathbf{n} \) is \( \mathbf{k} = (0, 0, 1) \). The surface integral simplifies to \[ \iint_S (-2y \mathbf{k}) \cdot \mathbf{k} \, dS = \iint_S -2y \, dS. \]

Step 5: Evaluate the Double Integral

Compute the double integral over the rectangle \( -1 \leq x \leq 1 \), \( -1 \leq y \leq 1 \): \[ \iint_S -2y \, dS = \int_{-1}^{1} \int_{-1}^{1} -2y \, dx \, dy. \] The integral with respect to \( x \) is \( \int_{-1}^{1} dx = 2 \), leading to \[ \int_{-1}^{1} -2y \, (2) \, dy = -4 \int_{-1}^{1} y \, dy. \] Integrate \( y \) from \( -1 \) to \( 1 \): \[ = -4 \left[ \frac{y^2}{2} \right]_{-1}^{1} = -4 \left( \frac{1}{2} - \frac{1}{2} \right) = 0. \]

Step 6: Conclusion from Stokes' Theorem

Since the surface integral \( \iint_S (abla \times \mathbf{F}) \cdot \mathbf{n} \, dS = 0 \), by Stokes' Theorem, the line integral \( \oint_C \mathbf{F} \cdot d\mathbf{r} \) also equals zero. This means that the circulation of \( \mathbf{F} \) around \( C \) is 0.

Key Concepts

Surface Integral

A surface integral extends the concept of a double integral to compute the flux of a vector field across a surface. In Stokes' Theorem, the surface integral over a smooth surface relates to a line integral around its boundary. Here, instead of measuring the circulation along the boundary, we look at how the field behaves across the surface itself.

• The surface integral requires a surface, denoted as \( S \), and a vector field \( \mathbf{F} \).
• The result is affected by the unit normal vector \( \mathbf{n} \), which points perpendicular to the surface.
  

In our exercise, the surface integral is simplified by considering \( S \) to be plane \( z = 0 \), which means that calculating \((abla \times \mathbf{F}) \cdot \mathbf{n}\) gives direct information about circulation of \( F \) around \( C \). This leads to a straightforward evaluation showing there's zero net circulation across this flat surface.

Line Integral

When we perform a line integral, we're effectively adding up a function along a curve. With vector fields like \( \mathbf{F} \), a line integral might measure flow along a path or the work done by a force field along a trajectory.

• In Stokes' Theorem, the line integral appears on the left side, showing the circulation of \( \mathbf{F} \) around a closed curve \( C \).
• It's often calculated as \( \oint_C \mathbf{F} \cdot d\mathbf{r} \), where \( d\mathbf{r} \) represents tiny steps along the curve.

In the given exercise, \( C \) specifies a square in the \( xy \)-plane. Instead of directly computing this line integral, Stokes' Theorem allows substituting with a typically simpler surface integral, revealing that the circulation—or line integral—is zero, demonstrating no net rotation around the curve.

Vector Field

A vector field assigns a vector to every point in space. Vector fields are visualized as arrows scattered through the region of interest, representing directions and magnitudes of various effects such as magnetic or velocity fields. In studying Stokes' Theorem, understanding \( \mathbf{F} \) is crucial.

• For our vector field \( \mathbf{F} = (y^2 + z^2) \mathbf{i} + (x^2 + y^2) \mathbf{j} + (x^2 + y^2) \mathbf{k} \), each component represents how the field changes in \( x, y, \) and \( z \) directions.
• The behavior of the field is witnessed through these components, aligning with how the surface and curve are defined.
  

By working through the exercise, we assess how the vector field's effects add up around point-to-point circulations, especially crucial when applying Stokes' Theorem to transition mathematically from the given surface to boundary.

Curl of a Vector Field

The curl of a vector field measures its rotation or swirling at a point. In imposing physically, it illustrates how a particle at that point would spin if caught in the flow.

• Mathematically, it's represented as \( abla \times \mathbf{F} \), calculated through partial derivatives focusing on changes within the vector field components.
• In the context of Stokes' Theorem, the curl is critical because it directly contributes to the surface integral used in the proved equivalence to line integrals.

For \( \mathbf{F} \) in this exercise, finding its curl \( abla \times \mathbf{F} = -2y \mathbf{k} \) shows us that rotation happens purely within the plane, lacking \( i \) or \( j \) component influence, simplifying the dynamics drastically. This implies no net circling effect on \( C \) when considering its imposed geometry.