Problem 19 - Thomas Calculus

Question

In Exercises \(19-24,\) use the surface integral in Stokes' Theorem to calculate the flux of the curl of the field \(F\) across the surface \(S\) in the direction of the outward unit normal \(\mathbf{n}\) . \begin{equation} \begin{array}{l}{\mathbf{F}=2 \mathbf{z}+3 x \mathbf{j}+5 y \mathbf{k}} \\ {S : \mathbf{r}(r, \theta)=(r \cos \theta) \mathbf{i}+(r \sin \theta) \mathbf{j}+\left(4-r^{2}\right) \mathbf{k}} \\ {0 \leq r \leq 2, \quad 0 \leq \theta \leq 2 \pi}\end{array} \end{equation}

Step-by-Step Solution

Answer

The flux is \( 12\pi \).

Step 1: Understand Stokes' Theorem

Stokes' Theorem relates the surface integral of the curl of a vector field over a surface \( S \) to the line integral of the vector field over the boundary \( C \) of \( S \). It is given by:\[\iint_{S} (abla \times \mathbf{F}) \cdot \mathbf{n} \, dS = \oint_{C} \mathbf{F} \cdot d\mathbf{r} \]where \( \mathbf{n} \) is the outward unit normal to the surface, and \( C \) is positively oriented.

Step 2: Parametrize the Surface

The given surface \( S \) is parametrized by \( \mathbf{r}(r, \theta) = (r \cos \theta) \mathbf{i} + (r \sin \theta) \mathbf{j} + (4 - r^2) \mathbf{k} \). This describes a paraboloid surface for \( 0 \leq r \leq 2 \) and \( 0 \leq \theta \leq 2 \pi \).

Step 3: Find the Boundary of \( S \)

The boundary \( C \) of the surface \( S \) occurs at the edges of the defined parameter range. At \( r = 2 \), the curve becomes a circle in the \( x-y \) plane, i.e.,\[ \mathbf{r}(2, \theta) = 2 \cos \theta \mathbf{i} + 2 \sin \theta \mathbf{j} + 0 \mathbf{k} \].

Step 4: Compute the Vector Field Curl

The curl of \( \mathbf{F} = 2\mathbf{z} + 3x \mathbf{j} + 5y \mathbf{k} \) is computed as:\[ abla \times \mathbf{F} = \left(\frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z}\right) \mathbf{i} - \left(\frac{\partial F_3}{\partial x} - \frac{\partial F_1}{\partial z}\right) \mathbf{j} + \left(\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}\right) \mathbf{k} \].Substituting the components \( F_1 = 0, F_2 = 3x, F_3 = 5y \), we get:\[ abla \times \mathbf{F} = (5 - 0) \mathbf{i} - (0 - 2) \mathbf{j} + (3 - 0) \mathbf{k} = 5 \mathbf{i} - 2 \mathbf{j} + 3 \mathbf{k} \].

Step 5: Parametrize the Boundary

The curve \( C \) lying in the \( x-y \) plane is:\[ \mathbf{r}(\theta) = 2 \cos \theta \mathbf{i} + 2 \sin \theta \mathbf{j} + 0 \mathbf{k} \].Thus, the differential \( d\mathbf{r} \) is:\[ d\mathbf{r} = (-2 \sin \theta) \mathbf{i} + (2 \cos \theta) \mathbf{j} \, d\theta \].

Step 6: Calculate the Line Integral Over \( C \)

Compute the line integral \( \oint_{C} \mathbf{F} \cdot d\mathbf{r} \):\[ \int_{0}^{2\pi} (0 \cdot (-2 \sin \theta) + 3(2\cos\theta) \cdot (2\cos\theta) + 5(2\sin\theta) \cdot (2\cos\theta)) \, d\theta = \]\[ \int_{0}^{2\pi} (12\cos^2\theta + 20\sin\theta\cos\theta) \, d\theta \].Using trigonometric identities, solve the integral to find the flux.

Step 7: Evaluate the Trigonometric Integral

Simplify and solve:For \( \int_{0}^{2\pi} 12\cos^2\theta \, d\theta \), use the identity \( \cos^2\theta = \frac{1 + \cos(2\theta)}{2} \):\[ \int_{0}^{2\pi} 6(1 + \cos 2\theta) \, d\theta = 12\pi\].For \( \int_{0}^{2\pi} 20\sin\theta\cos\theta \, d\theta \), use \( \sin 2\theta = 2\sin\theta\cos\theta \):\[ 10 \int_{0}^{2\pi} \sin 2\theta \, d\theta = 0 \].Therefore, the line integral evaluates to \(12\pi\).

Step 8: Conclusion Using Stokes' Theorem

By Stokes' Theorem, the surface integral of the curl over \( S \) equals the line integral over \( C \). We found \( \oint_{C} \mathbf{F} \cdot d\mathbf{r} = 12\pi \), so the flux of \( abla \times \mathbf{F} \) across \( S \) is also \( 12\pi \).

Key Concepts

Surface Integral

In multivariable calculus, a surface integral allows us to integrate over a surface in three-dimensional space. It is similar to a line integral, but instead of integrating along a curve, we integrate over a surface. In the context of Stokes' Theorem, we use the surface integral to calculate the flux of the curl of a vector field across a surface. This involves integrating the dot product of the curl of the vector field and the outward unit normal vector over the given surface.
The formula for a surface integral in Stokes' Theorem is given by:

• \( \iint_{S} (abla \times \mathbf{F}) \cdot \mathbf{n} \, dS \)

This expression accounts for both the magnitude and direction of the vector field as it interacts with the surface. It provides the net circulation along the surface boundary.

Curl of a Vector Field

The curl of a vector field is a measure of the circulation of the field around a given point. If you imagine a small paddle wheel placed in the vector field, the curl would represent how much and in which direction the wheel would rotate. The formula for computing the curl of a vector field \( \mathbf{F} = P\mathbf{i} + Q\mathbf{j} + R\mathbf{k} \) in three dimensions is:

• \( abla \times \mathbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right) \mathbf{i} - \left( \frac{\partial R}{\partial x} - \frac{\partial P}{\partial z} \right) \mathbf{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \mathbf{k} \)

A non-zero curl indicates the presence of rotational motion in the field, and this value plays an essential role in Stokes' Theorem.

Line Integral

A line integral, also known as a contour integral, computes the integral of a scalar or vector field along a curve. It generalizes the concept of a one-dimensional integral by extending it to curves in a plane or space. In Stokes' Theorem, the line integral of a vector field \( \mathbf{F} \) over the boundary curve \( C \) is calculated as:

• \( \oint_{C} \mathbf{F} \cdot d\mathbf{r} \)

Here, \( d\mathbf{r} \) is an infinitesimal vector along the curve that represents both the direction and the magnitude of the path element. The line integral measures the total effect of the vector field along the curve, and in the context of Stokes' Theorem, it equates to the surface integral of the curl over the given surface.

Parameterization of Surfaces

Parameterization is a technique used to express a surface in terms of two parameters, typically denoted as \( u \) and \( v \). This method helps in converting a complex surface into a more manageable form using a mapping function. Given a surface \( S \), it can be parameterized as \( \mathbf{r}(u, v) = (x(u, v), y(u, v), z(u, v)) \).
For the surface defined in the exercise, the parameterization with polar coordinates \( (r, \theta) \) is:

• \( \mathbf{r}(r, \theta) = (r \cos \theta) \mathbf{i} + (r \sin \theta) \mathbf{j} + (4 - r^2) \mathbf{k} \)

This representation transforms the surface equations into a domain defined by \( r \) and \( \theta \), which simplifies the integration process in Stokes' Theorem. Appropriate parameterization is crucial for applying the theorem to calculate flux efficiently.