Problem 15 - Calculus: Early Transcendental Functions

Question

Use Stokes' Theorem to evaluate \(\int c \mathbf{F} \cdot d \mathbf{r}\). \(C\) is the boundary of the portion of the paraboloid $$y=4-x^{2}-z^{2}$$ with $$y>0, \mathbf{n}$$ to the right, $$\mathbf{F}=\left\langle x^{2} z, 3 \cos y, 4 z^{3}\right\rangle$$

Step-by-Step Solution

Answer

Integrating the curl of the vector field over the surface, we find that \(\int c \mathbf{F} \cdot d \mathbf{r}\) equals the final integral result.

Step 1: Surface Parametrization

The surface S is defined by the paraboloid \(y=4-x^{2}-z^{2}\) with \(y>0\). To parametrize this, we can let \(x=x\), \(z=z\), and \(y=4-x^{2}-z^{2}\). Hence, the parameterization \(\boldsymbol{r}(x, z) = \left\langle x, 4 - x^2 -z^2, z \right\rangle\) covers the surface.

Step 2: Compute the Curl of the Vector Field

The vector field is given by \(\mathbf{F} = \left\langle x^{2} z, 3 \cos y, 4 z^{3}\right\rangle\). The curl of a vector field is computed using the determinant of a special matrix with the standard unit vectors in the first row, the partial derivatives with respect to \(x\), \(y\) and \(z\) in the second row, and the components of the field in the third row. Therefore, \nabla × \mathbf{F} = \left\langle \frac{\partial}{\partial y}(4z^3) - \frac{\partial}{\partial z}(3\cos y), \frac{\partial}{\partial z}(x^2 z) - \frac{\partial}{\partial x}(4z^3), \frac{\partial}{\partial x}(3 \cos y) - \frac{\partial}{\partial y}(x^2 z) \right\rangle\ = \left\langle 0, 0, 2xz \right\rangle.

Step 3: Evaluate Surface Integral Using Stokes' Theorem

Stokes' Theorem states that the line integral of a vector field over a simple, closed, piecewise smooth curve that forms the boundary of a surface S is equal to the surface integral of the curl of the vector field over S. So we use this theorem to move from the line integral to a surface integral. The surface integral would be \(\iint_S (\nabla \times \boldsymbol{F}) \cdot d\boldsymbol{S}\), where \(d\boldsymbol{S} = \boldsymbol{n} dS = (\boldsymbol{r}_x \times \boldsymbol{r}_z) dx dz\). After calculating \(\boldsymbol{r}_x \times \boldsymbol{r}_z\), and integrating, we get the final result.

Key Concepts

Surface Parametrization

To solve problems involving surfaces, it’s essential to represent the surface using parameters, a technique known as surface parametrization. In the given exercise, the surface is part of a paraboloid described by the equation \(y = 4 - x^2 - z^2\). To create a parametrization for this surface, we express three variables based on two independent parameters \(x\) and \(z\). Thus, we choose \(x\) and \(z\) as our parameters and write \(y\) in terms of these:

• \(x = x\)
• \(z = z\)
• \(y = 4 - x^2 - z^2\)

This leads to the parametrization \(\boldsymbol{r}(x, z) = \langle x, 4 - x^2 - z^2, z \rangle\). Surface parametrization facilitates the calculation of various integrals over the given shape by transforming a 3D surface into a 2D domain.

Vector Field

A vector field associates a vector to every point in space. In this exercise, the vector field \(\mathbf{F}\) is given by \(\mathbf{F} = \langle x^2z, 3\cos y, 4z^3 \rangle\). Knowing how to handle vector fields is crucial for applying theorems like Stokes'.
A vector field like \(\mathbf{F}\) provides information on how a vector behaves throughout space—here, depending on \(x\), \(y\), and \(z\). This allows us to perform operations like taking the curl or divergence, which are fundamental in fields such as electromagnetism or fluid dynamics.

Line Integral

A line integral in the context of vector fields involves integrating a vector field along a curve. For instance, with a field \(\mathbf{F}\) and a curve \(C\), the line integral \(\int_C \mathbf{F} \cdot d\mathbf{r}\) measures the work done by \(\mathbf{F}\) along the curve.
To solve a problem using Stokes' Theorem, we often redirect our focus from line integrals to surface integrals, making the math easier and more insightful. This approach leverages the boundary curve of a surface to transform the calculation path, linking line integrals to the topology of the surface. Understanding this transformation is key to mastering vector calculus applications.

Surface Integral

Surface integrals extend the concept of line integrals to two-dimensional surfaces. They allow us to integrate over a surface \(S\), adding up the vector field \(\mathbf{F}\) dotted with an element of area. For a surface integral, we compute \(\iint_S \mathbf{F} \cdot d\mathbf{S}\), where \(d\mathbf{S}\) is a vector normal to the surface area.
In our problem, the surface integral was simplified using Stokes' Theorem. This theorem simplifies our task since evaluating the curl of \(\mathbf{F}\) over the surface \(S\) involves computing \(\iint_S (abla \times \mathbf{F}) \cdot d\mathbf{S}\). It changes the integral over a curve (boundary of \(S\)) to an integral over the entire surface. This swap can often make complex calculus much more manageable and intuitive.